The probability that the number in a sample of 130 who do so differs from the expected value is 0.6198.
How to calculate probability?From the information, the critical value is given as 84. Therefore, the mean will be:
= np = 130 × 2/3 = 86.67
The standard deviation is also 5.3748. The corresponding z score will be:
= (84 - 86.67)/5.3748
= -0.50.
Therefore, the left tailed area will be:
P(z < -0.50) = 0.3099
Since, it's two tailed, we'll multiply by 2. This will be:
= 2 × 0.3099
= 0.6198
In conclusion, the probability is 0.6198.
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The probability that the number differs from the expected value by at least as much is approximately 0.620 (rounded to three decimal places).
To determine the probability that the number in a sample of 130 kissing couples who lean to the right differs from the expected value by at least as much as observed, we will conduct a hypothesis test using the binomial distribution. We can then approximate this with a normal distribution since the sample size is large.
Given:
- Sample size [tex](\( n \)) = 130[/tex]
- Observed number of right-leaning couples [tex](\( X \)) = 84[/tex]
- Expected proportion of right-leaning couples [tex](\( p \)) = \(\frac{2}{3}\)[/tex]
- Significance level [tex](\( \alpha \))[/tex] = 0.05
Step 1: Calculate the expected number and standard deviation
The expected number of right-leaning couples is:
[tex]\[ \mu = n \cdot p = 130 \cdot \frac{2}{3} \approx 86.67 \][/tex]
The standard deviation is:
[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{130 \cdot \frac{2}{3} \cdot \frac{1}{3}} \approx \sqrt{130 \cdot \frac{2}{9}} \approx \sqrt{28.89} \approx 5.375 \][/tex]
Step 2: Calculate the z-score for the observed value
The z-score measures how many standard deviations the observed value is away from the expected value:
[tex]\[ z = \frac{X - \mu}{\sigma} = \frac{84 - 86.67}{5.375} \approx \frac{-2.67}{5.375} \approx -0.497 \][/tex]
Since we are interested in the probability that the number differs from the expected value by at least as much as observed, we need to consider both tails of the distribution.
Step 3: Calculate the probability using the standard normal distribution
The probability for a z-score of [tex]\( \pm 0.497 \)[/tex] can be found using the standard normal distribution table or a calculator. For z = 0.497 :
[tex]\[ P(Z \leq 0.497) \approx 0.690 \][/tex]
Since we are considering both tails:
[tex]\[ P(|Z| \geq 0.497) = 2 \cdot (1 - 0.690) = 2 \cdot 0.310 = 0.620 \][/tex]
Step 4: Conclusion
The probability that the number of right-leaning couples in a sample of 130 differs from the expected value by at least as much as observed is approximately 0.620.
So, the rounded answer to three decimal places is:
[tex]\[ \boxed{0.620} \][/tex]
A researcher wishes to estimate the proportion of adults who have high-speed internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 95% confidence ifa) she uses a previous estimate of 0.36?b) she does not use any prior estimates?
Answer: a) 984 b) 1068
Step-by-step explanation:
When the prior estimate of the population proportion(p) is available .
Then the formula to find the sample size :-
[tex]n=p(1-p)(\dfrac{z^*}{E})^2[/tex]
, where E = margin of error
and z* = Critical z-value .
a) p= 0.36
E= 0.03
Critical value for 95% confidence level = z*= 1.96
Required sample size=[tex]n= 0.36(1-0.36)(\dfrac{1.960}{0.03})^2[/tex]
[tex]n= 0.36(0.64)(65.3333333333)^2[/tex]
[tex]n=(0.2304)(4268.44444444)=983.4496\approx984[/tex]
Hence, the required sample size is 984.
b) When the prior estimate of the population proportion is unavailable .
Then we use formula to find the sample size :-
[tex]n= 0.25(\dfrac{z^*}{E})^2[/tex]
, where E = margin of error
and z* = Critical z-value
Put E= 0.03 and z*= 1.960
Required sample size =[tex]n= 0.25(\dfrac{1.960}{0.03})^2[/tex]
[tex]n= 0.25(65.3333333333)^2[/tex]
[tex]n= 0.25(4268.44444444)=1067.11111111\approx1068[/tex]
Hence, the required sample size is 1068.
A coupon for $5 off any lunch price states that a 15% tip will be added to the price before the $5 is subtracted. So, C(x) = x - 5 represents the price after the coupon reduction. T(x) = 1.15 x represents the price after the tip is applied. Write the simplified composite functions C(T(x)) and T(C(x)). Which composite function represents the conditions of the coupon?
Answer:
[tex]C(T(x))=1.15x-5[/tex] and [tex]T(C(x))=1.15x-5.75[/tex]
C(T(x)) represents the conditions of the coupon.
Step-by-step explanation:
The price after the coupon reduction is represented by the function
[tex]C(x)=x-5[/tex]
The price after the tip is applied is represented by the function
[tex]T(x)=1.15x[/tex]
We need to find the composite functions C(T(x)) and T(C(x)).
[tex]C(T(x))=C(1.15x)[/tex] [tex][\because T(x)=1.15x][/tex]
[tex]C(T(x))=1.15x-5[/tex] [tex][\because C(x)=x-5][/tex]
This function represents that 15% tip will be added first after that $5 is subtracted.
Similarly,
[tex]T(C(x))=T(x-5)[/tex] [tex][\because C(x)=x-5][/tex]
[tex]T(C(x))=1.15(x-5)[/tex] [tex][\because T(x)=1.15x][/tex]
[tex]T(C(x))=1.15x-5.75[/tex]
This function represents that $5 is subtracted first after that 15% tip will be added.
It is given that a coupon for $5 off any lunch price states that a 15% tip will be added to the price before the $5 is subtracted.
It means 15% tip will be added first after that $5 is subtracted. So, C(T(x)) represents the conditions of the coupon.
The functions C(T(x)) and T(C(x)) represent the application of a tip and a coupon to a lunch price, respectively, in different order. The function that correctly represents the specific conditions given by the coupon in the problem statement is T(C(x)).
Explanation:The composite function C(T(x)) is found by substituting T(x) into the function C(x). So, C(T(x)) = T(x) - 5 = 1.15x - 5.
The composite function T(C(x)) is calculated by substituting C(x) into the function T(x). So, T(C(x)) = 1.15 * (x - 5) = 1.15x - 5.75.
In the context of the coupon conditions, the right composite function is T(C(x)). This composite function first applies the $5 coupon reduction (C(x)) and then the 15% tip (T(x)), which exactly follows the procedure described by the coupon.
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Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body bone mineral content (TBBMC). A highly skilled operator is required to take the measurements. Recently, a new DXA machine was purchased by a research lab and two operators were trained to take the measurements. TBBMC for eight subjects was measured by both operators. The units are grams (g). A comparison of the means for the two operators provides a check on the training they received and allows us to determine if one of the operators is producing measurements that are consistently higher than the other. Here are the data:
Subject
Operator 1 2 3 4 5 6 7 8
1 1.326 1.337 1.079 1.229 0.936 1.009 1.179 1.289
2 1.323 1.322 1.073 1.233 0.934 1.019 1.184 1.304
Take the difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2. (Operator 1 minus Operator 2. Round your answers to four decimal places.)
X-bar=
S=
Answer:
[tex]\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095[/tex]
-The sample is too small to make judgments about skewness or symmetry.
H0:[tex]\mu_{1}=\mu_{2}[/tex]
H1:[tex]\mu_{1} \neq \mu_{2}[/tex]
[tex]t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013[/tex]
[tex]p_v =2*P(t_{(14)}<-0.0133)=0.990[/tex]
So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.
Step-by-step explanation:
First we need to find the difference defined as:
(Operator 1 minus Operator 2)
d1=1.326-1.323=0.003 d2=1.337-1.322=0.015
d3=1.079-1.073=0.006 d4=1.229-1.233=-0.004
d5=0.936-0.934=0.002 d6=1.009-1.019=-0.01
d7=1.179-1.184=-0.005 d8=1.289-1.304=-0.015
Now we can calculate the mean of differences given by:
[tex]\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001[/tex]
And for the sample deviation we can use the following formula:
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095[/tex]
Describe the distribution of these differences using words. (which one is correct)
We can plot the distribution of the differences with the folowing code in R
differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)
hist(differences)
And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.
-The sample is too small to make judgments about skewness or symmetry.
Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)
[tex]\bar X_{1}=1.173[/tex] represent the mean for the operator 1
[tex]\bar X_{2}=1.174[/tex] represent the mean for the operator 2
[tex]s_{1}=0.1506[/tex] represent the sample standard deviation for the operator 1
[tex]s_{2}=0.1495[/tex] represent the sample standard deviation for the operator 2
[tex]n_{1}=8[/tex] sample size for the operator 1
[tex]n_{2}=8[/tex] sample size for the operator 2
t would represent the statistic (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:
H0:[tex]\mu_{1}=\mu_{2}[/tex]
H1:[tex]\mu_{1} \neq \mu_{2}[/tex]
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
We can replace in formula (1) like this:
[tex]t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013[/tex]
Statistical decision
For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{1}+n_{2}-2=8+8-2=14[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{(14)}<-0.0133)=0.990[/tex]
So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.
Which is a reasonable first step that can be used to solve the equation 2(x+6)=3(x-4)+5? Combine like terms, 3 and 5, on the right side of the equation. Distribute 2 to (x + 6) and 3 to (x – 4). Distribute 2 to x on the left side of the equation and 3 to x on the right side of the equation. Apply the subtraction property of equality to move the 6 from the left side of the equation to the right side.
Answer:
Distribute 2 to (x + 6) and 3 to (x – 4)
Step-by-step explanation:
A good first step for an equation such as this is to eliminate parentheses. That is accomplished by using the distributive property on the left and right. That property requires each term inside parentheses be multiplied by the factor outside parentheses. The process can be described as ...
Distribute 2 to (x + 6) and 3 to (x – 4)
_____
Comment on further steps
The result of the first step above would be ...
2x + 12 = 3x - 12 + 5
19 = x . . . . . . . . . . . next step: add 7-2x to both sides and collect terms
_____
Comment on answer wording
The last answer choice uses the wording "move the 6 from the left side of the equation to the right side". This sort of description is often invoked when a constant needs to be removed from one side of the equation or the other.
There is no property of equality that supports this action.
If you want to "move" a constant, you can invoke the addition property of equality that lets you add the same number to both sides of the equation. The number you would choose to add is the additive inverse of the one you want to "move."
In the above solution we have the sum -12+5 on the right after the first step. We want to eliminate that sum, so in the next step we show adding the additive inverse of the value of that sum. The sum of -12+5 is -7, and we add 7 to make that sum into the additive identity element, 0. (Likewise, we add -2x to turn the term 2x into the additive identity element, 0.) After these additions, we have ...
2x -2x +12 +7 = 3x -2x -12 +5 +7
0 + 19 = x + 0
19 = x
The relevant property of equality is the addition property of equality, which says I can add the same amount to both sides of the equation. I can't "move" anything without violating the equal sign, but I can use the addition and multiplication properties of equality to do the same thing to both sides of the equation.
Answer:
The correct answer is Distribute 2 to (x + 6) and 3 to (x – 4).
Step-by-step explanation:
Jill and Kim each improved their yards by planting hostas and geraniums. They bought their supplies from the same store. Jill spent $101 on 11 hostas and 2 geraniums. Kim spent $80 on 8 hostas and 2 geraniums. What is the cost of one hosta and one geramium?
Answer: the cost of one hosta is $7
the cost of one geramium is $12
Step-by-step explanation:
Let x represent the cost of one hosta.
Let y represent the cost of one geramium.
They bought their supplies from the same store. Jill spent $101 on 11 hostas and 2 geraniums. This means that
11x + 2y = 101 - - - - - - - -1
Kim spent $80 on 8 hostas and 2 geraniums. This means that
8x + 2y = 80 - - - - - - - - -2
We will eliminate y by subtracting equation 2 from equation 1, it becomes
3x = 21
x = 21/3 = 7
Substituting x = 7 into equation 2, it becomes
8 × 7 + 2y = 80
56 + 2y = 80
2y = 80 - 56 = 24
y = 24/2 = 12
A certain species of alligators is to be introduced into a swamp, and wildlife experts estimate the population will grow to P(t)=(177)4^t/2, where t represents the number of years from the time of introduction. What is the doubling-time for this population of alligators?
Answer:
1 year
Step-by-step explanation:
The population as a function of time is:
[tex]P(t) =177*4^{\frac{t}{2}}[/tex]
First, find the initial population at t=0:
[tex]P(0) =177*4^{\frac{0}{2}}\\P(0) = 177[/tex]
Then, double it:
[tex]2P(0) = 2*177 = 354[/tex]
Finally, find the value of 't' for which the population is 354:
[tex]354 =177*4^{\frac{t}{2}}\\4^{\frac{t}{2}}=2\\log(4^{\frac{t}{2}})=log(2)\\\frac{t}{2} *log(4) = log(2)\\t=2*\frac{log(2)}{log(4)}\\t=1[/tex]
The alligator population will double afer 1 year.
To find the doubling time for the population of alligators, we set the population growth function equal to twice the original population and solve for t.
Explanation:To find the doubling time for the population of alligators, we need to determine the time it takes for the population to double. In this case, the population growth function is given by P(t) = 177 * 4^(t/2), where t represents the number of years from the time of introduction. To find the doubling time, we set P(t) equal to twice the original population: 2 * P(0) = P(t).
Substituting the given population growth function, we have: 2 * 177 = 177 * 4^(t/2). To solve for t, we divide both sides of the equation by 177: 2 =
Next, we take the logarithm (base 4) of both sides of the equation to solve for t: log4(2) = t/2. Multiplying both sides by 2 gives us the doubling time: t = 2 * log4(2). Use a calculator to approximate this value.
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The regression equation NetIncome = 2,159 + .0312 Revenue was estimated from a sample of 100 leading world companies (variables are in millions of dollars).
(a-1) Calculate the residual for the x, y pair ($45,533, $2,697). (A negative value should be indicated by a minus sign. Round your answer to 4 decimal places.) Residual
(a-2) Did the regression equation underestimate or overestimate the net income? The regression equation overestimated the net income. The regression equation underestimated the net income.
(b-1) Calculate the residual for the x, y pair ($60,417, $3,497). (A negative value should be indicated by a minus sign. Round your answer to 4 decimal places.) Residual
(b-2) Did the regression equation underestimate or overestimate the net income? The regression equation overestimated the net income. The regression equation underestimated the net income.
Answer:
a) -882.6296
b) The regression equation overestimated the net income.
c) -547.0104
d) The regression equation overestimated the net income.
Step-by-step explanation:
We are given the following information in the question:
The regression equation for net income is given by:
[tex]\text{Net Income} = y = 2159 + .0312(\text{Revenue})[/tex]
a) residual for the (x, y) pair ($45,533, $2,697)
Calculated net income =
[tex]2159 + .0312(45533) = 3579.6296[/tex]
Residual = Observed net income - Calculated net income
[tex]\text{Residual} = 2697-3579.6296 = -882.6296[/tex]
b) The regression equation overestimated the net income.
c) residual for the (x, y) pair ($60,417, $3,497)
Calculated net income =
[tex]2159 + .0312(60417) = 4044.0104[/tex]
Residual = Observed net income - Calculated net income
[tex]\text{Residual} = 3497-4044.0104 = -547.0104[/tex]
d) The regression equation overestimated the net income.
The answers are : (a-1) The residual for the [tex](x,y)[/tex] pair [tex](\$45,533, $2,697)[/tex] is [tex]\[\text{Residual} = -881.6136\][/tex]. (a-2) The regression equation overestimated the net income. (b-1) The residual for the x, y pair ([tex]\$60,417, $3,497[/tex]) is [tex]\[\text{Residual} = -546.9704\][/tex]. (b-2) The regression equation overestimated the net income.
1. Calculate the predicted value [tex](\(\hat{y}\))[/tex] using the regression equation:
[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times x \][/tex]
2. Compute the residual by subtracting the predicted value from the actual value:
[tex]\[ \text{Residual} = y - \hat{y} \][/tex]
Part (a-1)
For the [tex]\( x, y \) pair \((45,533, 2,697)\)[/tex]
1. Calculate the predicted net income
[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times 45{,}533 \][/tex]
[tex]\[ \hat{y} = 2{,}159 + 1{,}419.6136 \][/tex]
[tex]\[ \hat{y} = 3{,}578.6136 \][/tex]
2. Compute the residual
[tex]\[ \text{Residual} = y - \hat{y} \][/tex]
[tex]\[ \text{Residual} = 2{,}697 - 3{,}578.6136 \][/tex]
[tex]\[ \text{Residual} = -881.6136 \][/tex]
Part (a-2)
The residual is negative, which means the actual net income is less than the predicted net income. Therefore, the regression equation overestimated the net income.
Part (b-1)
For the [tex]\( x, y \) pair \((60,417, 3,497)\)[/tex]
1. Calculate the predicted net income
[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times 60{,}417 \][/tex]
[tex]\[ \hat{y} = 2{,}159 + 1{,}884.9704 \][/tex]
[tex]\[ \hat{y} = 4{,}043.9704 \][/tex]
2. Compute the residual
[tex]\[ \text{Residual} = y - \hat{y} \][/tex]
[tex]\[ \text{Residual} = 3{,}497 - 4{,}043.9704 \][/tex]
[tex]\[ \text{Residual} = -546.9704 \][/tex]
Part (b-2)
The residual is negative, which means the actual net income is less than the predicted net income. Therefore, the regression equation overestimated the net income.
Suppose a sample of size 400 yields pˆ = .5. You'd like to construct a confidence interval with a margin of error only half as great as the one produced by this sample. What's the minimum sample size necessary to accomplish this?a. 400b. 800c. 1,600d. 1,200e. 2,400
Answer:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.0245}{1.96})^2}=1600[/tex]
c. 1600
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
In order to solve this problem we need to assume a confidence level. Let's assume that is 95%
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
First we need to find the margin of error from the original sample given by:
[tex] ME=1.96\sqrt{\frac{0.5 (1-0.5)}{400}}=0.049[/tex]
And on this case we have that [tex]ME =\pm 0.049/2=0.0245[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.0245}{1.96})^2}=1600[/tex]
c. 1600
The minimum sample size necessary to achieve a margin of error half as great as the original is Option c. 1,600.
To determine the minimum sample size necessary to achieve a margin of error that is half as great as the one produced by the initial sample, we need to understand the relationship between sample size and margin of error.
The margin of error for a confidence interval for a proportion is inversely proportional to the square root of the sample size, n. Specifically, if the margin of error for a sample size of 400 is E, then to achieve half that margin of error (E/2), we need a sample size n' such that:
E/2 = (E / √(n'))
Simplifying gives us √(n') = 2 × √(n). Squaring both sides, we get:
n' = 4 × n
Given n = 400, we find:
n' = 4 × 400 = 1600
Therefore, the minimum sample size necessary is Option c. 1,600.
Nicole and Kim each improved their yards by planting daylilies and ivy. They bought their supplies from the same store. Nicole spent $99 on 9 daylilies and 7 pots of ivy. Kim spent $144 on 9 daylilies and 12 pots of ivy. Find the cost of one daylily and the cost of one pot of ivy.
PLEASE HELP!!!
Answer: the cost of one daylilies is $4
the cost of one ivy is $9
Step-by-step explanation:
Let x represent the cost of one daylilies.
Let y represent the cost of one ivy.
Nicole and Kim bought their supplies from the same store. Nicole spent $99 on 9 daylilies and 7 pots of ivy. This means that
9x + 7y = 99 - - - - - - - -1
Kim spent $144 on 9 daylilies and 12 pots of ivy. This means that
9x + 12y = 144 - - - - - - - - -2
We will eliminate x by subtracting equation 2 from equation 1, it becomes
- 5y = - 45
y = - 45/-5 = 9
Substituting y = 9 into equation 2, it becomes
9x + 12 × 9 = 144
9x + 108 = 144
9x = 144 - 108 = 36
x = 36/9 = 4
Write the integral that gives the length of the curve y = f (x) = ∫0 to 4.5x sin t dt on the interval [0,π].
Answer:
Arc length [tex]=\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx[/tex]
Arc length [tex]=9.75053[/tex]
Step-by-step explanation:
The arc length of the curve is given by [tex]\int_a^b \sqrt{1+[f'(x)]^2}\ dx[/tex]
Here, [tex]f(x)=\int_0^{4.5x}sin(t) \ dt[/tex] interval [tex][0, \pi][/tex]
Now, [tex]f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \int_0^{4.5x}sin(t) \ dt[/tex]
[tex]f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( [-cos(t)]_0^{4.5x} \right )[/tex]
[tex]f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos(4.5x)+1 \right )[/tex]
[tex]f'(x)=4.5sin(4.5x)[/tex]
Now, the arc length is [tex]\int_0^{\pi} \sqrt{1+[f'(x)]^2}\ dx[/tex]
[tex]\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx[/tex]
After solving, Arc length [tex]=9.75053[/tex]
Which of the following statements concerning the linear correlation coefficient are true? I: If the linear correlation coefficient for two variables is zero, then there is no relationship between the variables. II: If the slope of the regression line is negative, then the linear correlation coefficient is negative. III: The value of the linear correlation coefficient always lies between -1 and 1. IV: A linear correlation coefficient of 0.62 suggests a stronger linear relationship than a linear correlation coefficient of -0.82. A) II and III B) I and II C) I and IV D) III and IV Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05. r = 0.523, n = 25 A) Critical values: r = plusminus 0.396, significant linear correlation B) Critical values: r = plusminus 0.487, no significant linear correlation C) Critical values: r = plusminus 0.396, no significant linear correlation D) Critical values: r = plusminus 0.487, significant linear correlation Write the word or phrase that best completes each statement or answers the question. Construct a scatterplot for the given data. Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary. Managers rate employees according to job performance and attitude. The results for several randomly selected employees are given below.
Answer:
1) A) II and III
2) A) Critical values: r = plusminus 0.396, significant linear correlation
3) Yi= 0.41 + 0.37Xi
Step-by-step explanation:
Hello!
The objective of the linear correlation analysis is to test if there is an association between two study variables (X₁ and X₂).
Pearson's Coefficient of correlation
For Variables with a bivariate normal distribution (X₁, X₂)~N₂(μ₁; μ₂; σ₁²; σ₂²; ρ)
To do so, the study parameter is the population coefficient of correlation (ρ) - Rho- (If you were to make a graphic of the correlation line, Rho represents the slope)
Sample coefficient of correlation: r
It takes values between -1 and 1
This coefficient gives an idea of the degree of correlation between the variables.
If ρ = 0 then there is no linear correlation between X₁ and X₂ Graphically, the slope is cero
If ρ < 0 then there is a negative association between X₁ and X₂ (i.e. when one variable increases the other one decreases) In a graphic, the slope of the line is negative.
If ρ > 0 then there is a positive association between X₁ and X₂ (i.e. Both variables increase and decrease together)
The closer to 1 or -1 the coefficient is, the stronger the association between variables. Using the absolute value of the correlation coefficients you can compare them, the greater the value, the stronger is the association between variables. For example, if you were to have two coefficients r₁= -0.24 and r₂= 0.67 then the absolute values are Ir₁I= 0.24 and Ir₂I= 0.67 you can see that the coefficient of the second sample is bigger than the first sample, that means that there is a stronger correlation in the second sample than the first one.
The non-parametric coefficient of correlation has the same characteristics.
1) Statements:
I: If the linear correlation coefficient for the two variables is zero, then there is no relationship between the variables. FALSE, when r=0 then there is no linear association between the two variables, this doesn't mean that there isn't any other type of association between them.
II: If the slope of the regression line is negative, then the linear correlation coefficient is negative. TRUE
The regression and correlation analyses are closely linked because for a regression equation to be reasonable, the sample points must be linked to the equation and the correlation coefficient between both variables must be large when the degree of association is high and small when The degree of association is low in addition to being independent of the units.
The regression analysis tests whether or not there is an association between both variables and the correlation analysis indicates the degree of that association.
If the slope of the regression is negative, then the correlation coefficient is negative.
III: The value of the linear correlation coefficient always lies between -1 and 1. TRUE, it is one of the characteristics of the correlation coefficient.
0.62 suggests a stronger linear relationship than a linear correlation coefficient of -0.82. FALSE, to check wich correlation coefficient shows a stronger correlation look at their absolute values, the one that is closer to 1 is the stronger, Ir₁I= 0.62 < Ir₂I= 0.82
Correct answer:
A) II and III
2) Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05. r = 0.523, n = 25
For this, you have to use a Table of cumulative probabilities for the linear correlation coefficient. (I've used Pearson)
For a two-tailed test (H₀: ρ=0)
[tex]r_{n-2; \alpha/2}= r_{23; 0.025}=[/tex] ± 0.396
Against r = 0.523, the decision is to reject the null hypothesis. There is a linear correlation between the two study variables.
Correct answer:
A) Critical values: r = plus-minus 0.396, significant linear correlation
3) Construct a scatterplot for the given data. Check 1st attachment for Data and Scatterplot.
Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.
Equation of regression:
Yi= a + bXi
a= [tex](\frac{sum Yi}{n})[/tex]+b[tex](\frac{sum Xi}{n})[/tex]
b= [tex]\frac{sum XiYi*\frac{(sum Xi)(sum Yi)}{n} }{/(sumXi^2)-\frac{(sumXi)^2}{n} }[/tex]
Using the given Data:
∑Xi= -11
∑Xi²= 201
∑Yi= 0
∑Yi²= 176
Mean X= -1.10
Mean Y= 0
a= 0.41
b= 0.37
Yi= 0.41 + 0.37Xi
4) Managers rate empoyees acording to job performance and attitude. The results fro several randomly selected empoyees are given below.
Performance: 59; 63; 65; 69; 58; 77; 76; 69; 70; 64
Attitude: 72; 67; 78; 82; 73; 87; 92; 83; 87; 78
No question found?
I hope it helps!
Final answer:
The linear correlation coefficient ranges from -1 to 1, indicating strength and direction of a linear relationship. Statements II and III are true concerning the correlation coefficient. A calculated coefficient of 0.523 with a sample of 25 is significant given critical values of ±0.396 at a 0.05 significance level.
Explanation:
1. The linear correlation coefficient, known as the Pearson correlation coefficient and symbolized by r, provides a measure of the strength and direction of the linear relationship between two variables. Statement II is true: If the slope of the regression line is negative, the linear correlation coefficient is also negative. Statement III is accurate as well because the value of the linear correlation coefficient lies between -1 and 1, inclusive.
For Statement I, although a correlation coefficient of zero indicates no linear relationship, there could be a non-linear relationship present. Regarding Statement IV, the strength of the relationship is determined by the absolute value of the correlation coefficient, so -0.82 indicates a stronger relationship than 0.62.
Therefore, the correct answer is A) II and III.
2. Using a Table of Critical Values for the Pearson correlation coefficient or technology like a calculator's LinRegTTest function, we compare the computed value of r to the critical values at a given significance level to determine whether the correlation is significant. For r = 0.523 and n = 25, the degrees of freedom are 23 (n - 2), and by referencing a table or using software, we find the critical values at a significance level of 0.05.
The correct answer is A) Critical values: r = ±0.396, significant linear correlation, because the reported r value of 0.523 exceeds the critical value.
To play the lottery in a certain state, a person has to correctly select 5 out of 45 numbers, paying $1 for each five-number selection. If the five numbers picked are the same as the ones drawn by the lottery, an enormous sum of money is bestowed. What is the probability that a person will one combination of five numbers will win? What is the probability of winning if 100 different lottery tickets are purchased?
Answer:
Step-by-step explanation:
Given
There are 45 numbers out of which 5 numbers are required to win an enormous sum of money
No of ways in which 5 numbers can be selected out of 45 numbers.
[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]
here n=45, r=5
[tex]^{45}C_5=\frac{45!}{40!5!}[/tex]
[tex]^{45}C_5=1221759[/tex]
out of which there is only combination foe contest winning
[tex]P=\frac{1}{1221759}[/tex]
When 100 different tickets are bought then
Probability of winning[tex]=\frac{100}{1221759}[/tex]
The probability of winning with one combination of five numbers is [tex]\( \frac{1}{1,221,759} \)[/tex] .
The probability of winning if 100 different lottery tickets are purchased is [tex]\( 1 - \left(1 - \frac{1}{1,221,759}\right)^{100} \)[/tex], approximately [tex]\( 0.0000182 \)[/tex].
Step 1
The probability of winning the lottery with one combination of five numbers can be calculated using the formula for combinations:
[tex]\[ P(\text{win with one ticket}) = \frac{1}{\binom{45}{5}} \][/tex] Where [tex]\( \binom{45}{5} \)[/tex]represents the number of ways to choose 5 numbers from 45 without regard to the order. Calculating [tex]\( \binom{45}{5} \)[/tex] :
[tex]\[ \binom{45}{5} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1,221,759 \][/tex]
So, the probability [tex]\( P(\text{win with one ticket}) \)[/tex] is:
[tex]\[ P(\text{win with one ticket}) = \frac{1}{1,221,759} \approx 8.19 \times 10^{-7} \][/tex]
If 100 different lottery tickets are purchased, the probability of winning at least once is calculated using the complement rule:
[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{\binom{45}{5}} \right)^{100} \][/tex]
Step 2
Substituting [tex]\( \binom{45}{5} \)[/tex] :
[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{1,221,759} \right)^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (1 - 8.19 \times 10^{-7})^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (0.999999181)^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - 0.9999818 \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 0.0000182 \][/tex]
The probability of winning the lottery with one ticket is approximately [tex]\( 8.19 \times 10^{-7} \)[/tex], while the probability of winning if 100 different tickets are purchased increases to approximately 0.0000182. Purchasing more tickets improves the chances of winning, but it remains a very low probability event due to the large number of possible combinations.
The mean income per person in the United States is $50,000, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of $60,000 with a standard deviation of $10,000. At the 0.05 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?
a. State the null hypothesis and the alternate hypothesis.b. State the decision rule for 0.05 significance level.Reject H0 if t > ____c. Compute the value of the test statistic.d. Is there enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.05 significance level?
Answer:
We conclude that the residents of Wilmington, Delaware, have more income than the national average.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $50,000
Sample mean, [tex]\bar{x}[/tex] = $60,000
Sample size, n = 10
Alpha, α = 0.05
Sample standard deviation, s = $10,000
a) First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 50000\text{ dollars}\\H_A: \mu > 50000\text{ dollars}[/tex]
We use one-tailed(right) t test to perform this hypothesis.
c) Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{60000 - 50000}{\frac{10000}{\sqrt{10}} } = 3.162[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]
b) Rejection Rule:
If the calculated t-statistic is greater than the the critical value, we rect the null hypothesis.
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.
d) There is enough evidence to conclude that the residents of Wilmington, Delaware, have more income than the national average.
Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straight lines joining the points (1,0,1), (0,1,0) and (0,0,1). In particular, compute the unit normal vector and the curl of FF as well as the value of the integral
Stokes' theorem equates the line integral of [tex]\vec F[/tex] along the curve to the surface integral of the curl of [tex]\vec F[/tex] over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).
Parameterize this triangle (call it [tex]T[/tex]) by
[tex]\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)[/tex]
[tex]\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]T[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)[/tex]
Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of [tex]T[/tex] is traversed in the counterclockwise direction when viewed from above.
Compute the curl of [tex]\vec F[/tex]:
[tex]\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)[/tex]
Then by Stokes' theorem,
[tex]\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S[/tex]
where
[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS[/tex]
[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv[/tex]
The integral thus reduces to
[tex]\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}[/tex]
The question requires the use of Stokes' theorem to compute the integral of a given vector field around a defined curve. This involves calculating the curl of the vector field and a surface integral. However, specific computations were not provided.
Explanation:The question is asking for the use of the Stokes' theorem to compute the integral of the vector field F=(2x,2y,2x+2z)F=(2x,2y,2x+2z) around the curve consisting of the straight lines joining the points (1,0,1), (0,1,0) and (0,0,1). In Stokes theorem, we're interested in the curl of the vector field and the surface integral of that curl over some surface S, with orientation determined by a unit normal vector. The problem also requires the computation of the curl of F which is a vector field whose components are the partial derivatives of the components of F. However, with this given information, we're unable to proceed with computations as detailed integral computations weren't provided. Therefore, a complete answer can't be provided at this time.
Learn more about Stokes' Theorem here:https://brainly.com/question/35538480
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Find a solution to the initial value problem, y′′+18x=0,y(0)=5,y′(0)=1.
We want to find a solution to the initial value problem:
[tex]y'' + 18x = 0 \qquad,\qquad y(0) = 5 \qquad,\qquad y'(0)=1.[/tex]
We can start by integrating the equation once:
[tex]\dfrac{\textrm{d}^2 y}{\textrm{d}x^2} + 18 x = 0 \iff \dfrac{\textrm{d}^2 y}{\textrm{d}x^2} = -18 x \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -18\displaystyle\int x\textrm{ d}x \iff \dfrac{\textrm{d}y}{\textrm{d}x}=-18\dfrac{x^2}{2} + C \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + C.[/tex]
Using the initial condition [tex]y'(0) = 1[/tex], we can determine the integration constant [tex]C[/tex]:
[tex]\dfrac{\textrm{d}y}{\textrm{d}x}\Big\vert_{x= 0} = 1 \iff -9 \times 0^2 + C = 1 \iff C = 1.[/tex]
Therefore, we have:
[tex]\dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + 1[/tex]
We can now integrate again:
[tex]y(x) = \displaystyle\int\dfrac{\textrm{d}y}{\textrm{d}x}\textrm{ d}x = \int\left(-9x^2+1\right)\textrm{d}x = -9\int x^2\textrm{ d}x + \int\textrm{d}x =\\\\= -9\dfrac{x^3}{3} + x + K = -3x^3 + x + K.[/tex]
The integration constant [tex]K[/tex] is determined by using [tex]y(0) = 5[/tex]:
[tex]y(0) = 5 \iff -3 \times 0^3 + 0 + K = 5 \iff K = 5.[/tex]
Finally, the solution is:
[tex]\boxed{y(x) = -3x^3 + x + 5}.[/tex]
By separation of variables, the solution is given by:
[tex]y(x) = -3x^3 + x + 5[/tex]
The differential equation is:
[tex]y^{\prime\prime}(x) + 18x = 0[/tex]
[tex]y^{\prime\prime}(x) = -18x[/tex]
Applying separation of variables:
[tex]\int y^{\prime\prime}(x) = -\int 18x dx[/tex]
[tex]y^{\prime}(x) = -9x^2 + K[/tex]
Since [tex]y^{\prime}(0) = 1, K = 1[/tex]
Thus:
[tex]y^{\prime}(x) = -9x^2 + 1[/tex]
To find y, another separation of variables is appled:
[tex]\int y^{\prime}(x) = \int(-9x^2 + 1)dx[/tex]
[tex]y(x) = -3x^3 + x + K[/tex]
Since y(0) = 5, K = 5, thus, the solution is:
[tex]y(x) = -3x^3 + x + 5[/tex]
A similar problem is given at https://brainly.com/question/13244107
Based on historical data, your manager believes that 40% of the company's orders come from first-time customers. A random sample of 91 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.26 and 0.43? Answer = (Enter your answer as a number accurate to 4 decimal places.)
Answer:
[tex]P(0.26 \leq p \leq 0.43)=0.7204-0.0032=0.7172[/tex]
Step-by-step explanation:
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
The population proportion have the following distribution
[tex]p \sim N(p=0.4,\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.4(1-0.4)}{91}}=0.0514)[/tex]
And we can solve the problem using the z score on this case given by:
[tex]z=\frac{p_o -p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
We are interested on this probability:
[tex]P(0.26 \leq p \leq 0.43)[/tex]
And we can use the z score formula, and we got this:
[tex]P(\frac{0.26 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}} \leq Z \leq \frac{0.43 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}})[/tex]
[tex]P(-2.726 \leq Z \leq 0.584)[/tex]
And we can find this probability like this:
[tex]P(-2.726 \leq Z \leq 0.584)=P(Z<0.584)-P(Z<-2.726)=0.7204-0.0032=0.7172[/tex]
An electronics company produces transistors, resistors, and computer chips. Each transistor requires 1 unit of copper, 2 units of zinc, and 3 units of glass. Each resistor requires 2 units of copper and 3 units of zinc. Each computer chip requires 1 unit of zinc, and 5 units of glass. How many of each product can be made knowing that the company has 104 units of copper, 169 units of zinc, and 74 units of glass?
Answer:
The capacity is 18 transistors, 43 resistors and 4 computer chips.
Step-by-step explanation:
This problem can be solved by a system of equations.
I am going to say that:
x is the number of transistors that are made.
y is the number of resistors that are made.
z is the number of computer chips that are made.
Building the system:
A transistor requires 1 unit of copper. A resistor requires 2 units of copper. There are 104 units of copper. So [tex]x + 2y = 104[/tex]
A transistor requires 2 units of zinc. A resistor requires 3 units of zinc. A chip requires 1 unit of zing. There are 169 units of zinc. So [tex]2x + 3y + z = 169[/tex].
A transistor requires 3 units of glass. A chip requires 5 units of glass. There are 74 units of glass. So [tex]3x + 5z = 74[/tex]
We have the following system:
[tex]x + 2y = 104[/tex]
[tex]2x + 3y + z = 169[/tex]
[tex]3x + 5z = 74[/tex]
The solution is:
[tex]x = 18, y = 43, z = 4[/tex]
The capacity is 18 transistors, 43 resistors and 4 computer chips.
The functions f and g are differentiable for all real numbers x. The table below gives values for the functions and their first derivatives at selected values of x.
x f(x) f'(x) g(x) g'(x)
1 4 -3 5 2
2 -3 -1 4 6
3 π 8 -1 4
4 -5 unknown 0 3
a. If the function h is given by h(x) = f (x) / g(x)' find h'(1).
b. If the function r is given by r(x) = f (x)g(x), find the equation of the tangent line to r(x) at x = 2.
Answer:
13/9
y =-22x+32
Step-by-step explanation:
Given that the functions f and g are differentiable for all real numbers x. The table below gives values for the functions and their first derivatives at selected values of x.
x f(x) f'(x) g(x) g'(x)
1 4 -3 5 2
2 -3 -1 4 6
3 π 8 -1 4
4 -5 unknown 0 3
a) [tex]h(x) = \frac{f(x)}{g(x)} \\h'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}[/tex]
(using quotient rule)
Substitute 1 for x
[tex]h'(1) = \frac{g(1)f'(1)-f(1)g'(1)}{(g(1))^2}\\=\frac{9-(-4)}{9} \\=\frac{13}{9}[/tex]
b) [tex]r(x) = f(x) g(x)\\r'(x) = f(x) g'(x)+g(x)f'(x)[/tex]
when [tex]x =2, r(x) = r(2) = f(2) g(2) = -12[/tex]
point of contact is (2,-12)
Slope of tangent =[tex]r'(2) = f(2) g'(2)+g(2)f'(2)\\=-18+(-4) \\=-22[/tex]
Using point slope form, tangent is
[tex]y+12 = -22(x-2)\\y = -22x +32[/tex]
a. h'(1) = -23/25.
b. The equation of the tangent line to r(x) = f(x)g(x) at x = 2 is y = -22x + 32.
a. To find h'(1), you need to apply the quotient rule.
The quotient rule states that if you have a function h(x) = f(x) / g(x), then the derivative h'(x) is given by:
h'(x) = (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2
In your case, h(x) = f(x) / g(x), and you want to find h'(1), so:x = 1
f(1) = 4
f'(1) = -3
g(1) = 5
g'(1) = 2
Now, plug these values into the quotient rule formula:
h'(1) = (f'(1) * g(1) - f(1) * g'(1)) / (g(1))^2
h'(1) = (-3 * 5 - 4 * 2) / (5^2)
h'(1) = (-15 - 8) / 25
h'(1) = -23 / 25
So, h'(1) = -23/25.
b. To find the equation of the tangent line to r(x) = f(x)g(x) at x = 2, you need to follow these steps:
Find r(2):
r(2) = f(2) * g(2)
r(2) = (-3) * 4
r(2) = -12
Find r'(x) using the product rule. The product rule states that if you have a function r(x) = f(x) * g(x), then the derivative r'(x) is given by:
r'(x) = f(x) * g'(x) + f'(x) * g(x)
Plugging in the given values:
r'(2) = (-3) * 6 + (-1) * 4
r'(2) = -18 - 4
r'(2) = -22
Now, you have r(2) and r'(2). You can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is:
y - y1 = m(x - x1)
Where (x1, y1) is the point on the curve (in this case, (2, -12)), and m is the slope (in this case, m = r'(2)).
Plug in the values:
y - (-12) = -22(x - 2)
Simplify and solve for y:
y + 12 = -22(x - 2)
y + 12 = -22x + 44
y = -22x + 44 - 12
y = -22x + 32
So, the equation of the tangent line to r(x) at x = 2 is y = -22x + 32.
For similar question on tangent line.
https://brainly.com/question/7252502
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Suppose a repairman wants to determine the current percentage of customers who keep up with regular house maintenance. How many customers should the repairman survey in order to be 98% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion of customers who keep up with regular house maintenance?
Answer:
n=543
Step-by-step explanation:
1) Notation and definitions
[tex]n[/tex] random sample (variable of interest)
[tex]\hat p[/tex] estimated proportion of interest
[tex]p[/tex] true population proportion of interest
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
2) Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2 =0.01[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-2.33, t_{1-\alpha/2}=2.33[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.05[/tex] (5% points means 0.05 on fraction) and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
Since we don't have a prior estimate for [tex]\hat p[/tex] we can use 0.5 as the prior estimate, and replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.33})^2}=542.89[/tex]
And rounded up we have that n=543
g "Motor vehicle accidents are rare events, but it is always better to have fewer. Vehicle speed may be related to the number of accidents, and in general it is believed that slower speeds will lead to fewer accidents. On a stretch of highway 42, the average number of accidents per day was 0.23. A month ago, the speed limit was changed from 60 miles per hour to 50 miles per hour. In the one month period following the change in the speed limit, three accidents were observed. Explain in detail how you would test whether this is a significant decrease (3pts). Include a description of your null and alternative hypotheses."
Answer:
To test wether this is a significant decrease we have to perform a hypothesis test on the proportions.
The null hypothesis represents the past condition (the proportion of 0.23 accidents/day is equal or bigger) and the alternative hypothesis is what we claim that is happening now (the proportion have lowered).
We want to perform the test in order to know if there is enough evidence that it has changed. The result can be:
- The null hypothesis is rejected: there is enough evidence with this sample that the rate of accidents has lowered from 0.23.
- The null hypothesis failed to be rejected: there is not enough evidence to say that the rate of accidents has lowered, although the sample proportion is lower.
Step-by-step explanation:
Numerical solution:
We have to perform a hypothesis test on proportions. We want to know if there is enough evidence to claim that the number of accidents per day has lowered from 0.23.
The null and alternative hypothesis are:
[tex]H_0: \pi\geq0.23\\\\H_1:\pi<0.23[/tex]
The significance level assumed is 0.05.
The proportion of the sample is:
[tex]p=\frac{3}{30}=0.1[/tex]
The standard deviation is calculated from the population proportion
[tex]\sigma=\sqrt{\pi(1-\pi)/N} =\sqrt{0.23*(1-0.23)/30} =0.077[/tex]
The z-value now can be calculated as
[tex]z=\frac{p-\pi+0.5/N}{\sigma}=\frac{0.10-0.23+0.5/30}{0.077} =-1.475[/tex]
The P-value for z=-1.475 is P=0.07011. The P-value is greater than the significant level, so the effect is not significant and it failed to reject the null hypothesis.
Find an equation of the slant asymptote. Do not sketch the curve. y = 5x4 + x2 + x x3 − x2 + 5
Final answer:
The equation of the slant asymptote is y = 5x + 4. The equation of the slant asymptote for the given function is found by performing polynomial long division and taking the quotient without the remainder.
Explanation:
To find the equation of the slant asymptote for the function y = (5x4 + x2 + x) / (x3 - x2 + 5), we need to divide the numerator by the denominator using polynomial long division or synthetic division. The quotient, without the remainder, will give us the equation of the slant asymptote because as x approaches infinity, the remainder becomes insignificant compared to the terms in the quotient.
Step 1: Perform the division. (This will be specific to the given function.)
Step 2: The quotient (without the remainder) is the equation of the slant asymptote.
To find the equation of the slant asymptote, we need to divide the numerator (5x^4 + x^2 + x) by the denominator (x^3 - x^2 + 5) using long division.
The quotient is 5x + 4, so the equation of the slant asymptote is y = 5x + 4. It is important to note that the slant asymptote represents the behavior of the function as x approaches positive or negative infinity.
The time T that technician requires to perform preventive maintenance on an air conditioning unit has an unknown probability distribution. However, its mean is known to be 2 hours, with standard deviation 1 hour. Suppose the company maintains 70 of these units and that the conditions of the Central Limit Theorem apply. What is the probability that a maintenance operation will take more than 2 hours and 15 minutes?
Answer:
0.4
Step-by-step explanation:
To calculate the probability that a maintenance operation will take more than 2 hours and 15 minutes. We can first calculate the probability that ALL maintenance operation on 70 of the units will take less than 2 hours and 15 minutes, then subtract it from 1.
So the probability of a maintenance operation that would take less than 2 hours and 15 minutes, or 135 minutes is:
[tex]P(X \leq 135, \mu = 120, \sigma = 60) = 0.6 [/tex]
So the probability that a maintenance operation will take more than 2 hours and 15 minutes is:
[tex] 1 - 0.6 = 0.4[/tex]
In this exercise, we consider strings made from uppercase letters in the English alphabet and decimal digits.
How many strings of length 10 can be constructed in each of the following scenarios?
(a) The first and last characters of the string are letters.
(b) The first character is a vowel, the second character is a consonant, and the last character is a digit.
(c) Vowels (not necessarily distinct) appear in the third, sixth, and eighth positions and no other positions.
(d) Vowels (not necessarily distinct) appear in exactly two positions.
(e) Precisely four characters in the string are digits and no digit appears more than one time.
Answer:
a) [tex]26^2 36^8 [/tex]b) [tex]21\cdot10\cdot36^7[/tex] c) [tex]5^3 31^7 [/tex] d) [tex]10\cdot 9\cdot 8 \cdot 7 \cdot 26^6[/tex]Step-by-step explanation:
We will use the product rule from combinatorics.
a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are [tex] 26\cdot36\cdot 36\cdots 36\cdot 26=26^2 36^8 [/tex] strings.b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are [tex]21\cdot 10\cdot 36^7 [/tex] strings. c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are [tex]5^3 31^7 [/tex] strings.d) Remember that the binomial coefficient [tex] \binom{n}{k} [/tex] is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in [tex] \binom{10}{4} [/tex] ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are [tex]10\cdot 9\cdot 8 \cdot 7 \cdot 26^6[/tex] strings.The price of milk has been increasing over the last month. Audrey believes there is a positive correlation between the number of predicted storms and the price of milk. Number of Storms Predicted Milk Price 1 $2.70 3 $2.89 4 $3.50 6 $3.88 7 $3.91 Use the table to determine the average rate of change from 3 to 6 storms.
Answer:
0.33
Step-by-step explanation:
To solve this example we use this rule :
Δx/Δy
x is amount that changed. so Δx=0,99.
How we get 0.99.
Our storm is to 3 to 6 so we find difference for x : for 3rd and 6th member of table... 3.88-2.89=0.99
Now we have :
0.99/Δy
Because we need to find from 3 to 6 , Δy=3 ,
When we find both, we can find rate of change with :
0.99/3=0.33
Answer:
0.33 lol
Step-by-step explanation:
14 - (-21) + (-31) - (-25) - (-27)
Answer:
56 NOT -56
Step-by-step explanation:
That's just the answer
Answer:
56
Step-by-step explanation:
14 - (-21) + (-31) - (-25) - (-27)
= 14 +21 -31 + 25 +27
= 56
A manufacturer claims that the batteries it makes will last 18 hours, with a standard deviation of 1.5 hours. If the durations of the batteries are normally distributed, what proportion of batteries would be expected to last less than 16 hours?
A. 0.9082
B. 0.0918
C. 0.1134
D. 0.2537
E. 0.5918
Answer:
B. 0.0918
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 18, \sigma = 1.5[/tex]
What proportion of batteries would be expected to last less than 16 hours?
This is the pvalue of Z when X = 16. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16 - 18}{1.5}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918.
So the correct answer is:
B. 0.0918
Answer: the correct option is B
Step-by-step explanation:
the durations of the batteries are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = durations of the batteries in hours
u = mean time
s = standard deviation
From the information given,
u = 18 hours
s = 1.5 hours
We want to find the proportion or probability of batteries would be expected to last less than 16 hours. It is expressed as
P(x lesser than 16)
For x = 16,
z = (16 - 18)/1.5 = - 1.33
Looking at the normal distribution table, the probability corresponding to the z score is 0.09176
Approximately 0.0918
Nielsen Media Research wants to estimate the mean amount of time, in minutes, that full-time college students spend texting each weekday.Find the sample size necessary to estimate that mean with a 15 minute margin of error. Assume that a 96% confidence level is desired and that the standard deviation is estimated to be 112.2 minutes.
Answer:
n=237
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=112.2)[/tex]
We know that the margin of error for a confidence interval is given by:
[tex]Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.96=0.04[/tex] and [tex]\alpha/2=0.02[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=2.054[/tex]
If we solve for n from formula (1) we got:
[tex]\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}[/tex]
[tex]n=(\frac{z_{\alpha/2} \sigma}{Me})^2[/tex]
And we have everything to replace into the formula:
[tex]n=(\frac{2.054(112.2)}{15})^2 =236.05[/tex]
And if we round up the answer we see that the value of n to ensure the margin of error required [tex]\pm=15 min[/tex] is n=237.
Cyril, who is 68 years of age, received Social Security benefits of $12,000, wages of $5,000, interest and dividends of $4,000, unemployment compensation of $3,000 and municipal bond interest of $1,500. Calculate Cyril's adjusted gross income.
Answer:
Cyril's adjusted gross income is $12000.
Step-by-step explanation:
First, we have that the adjusted gross income is given by the sum of the income that is not taxable.
For our case, we have the following data:
Social security: 12000 (This is not taxable)
Wages: 5000 (This is taxable)
Interest and dividends: 4000 (This is taxable)
Unemployment compensation: 3000 (This is taxable)
municipal bond interest: $1,500 (This is not taxable)
Then, adding up the income that's deductible, we have that Cyril's adjusted gross income is:
AGI=5000+4000+3000=$12000
Guest ages at a ski mountain resort typically have a right-skewed distribution. Assume the standard deviation (σ) of age is 14.5 years. From a random sample of 40 guests the sample mean is 36.4 years. Calculate a 99 percent confidence interval for µ, the true mean age of guests.
Answer: (30.49 years, 42.31 years)
Step-by-step explanation:
The formula to find the confidence interval is given by :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}.[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
z* = Critical value.
[tex]\sigma[/tex] = Population standard deviation.
n= Sample size.
As per given , we have
[tex]\overline{x}=36.4[/tex]
[tex]\sigma=14.5[/tex]
n= 40
We know that the critical value for 99% confidence interval : z* = 2.576 (By z-table)
A 99 percent confidence interval for µ, the true mean age of guests will be :
[tex]36.4\pm (2.576)\dfrac{14.5}{\sqrt{40}}\\\\ 36.4\pm (2.576)2.29265130362\\\\=36.4\pm5.90586975813\\\\\approx36.4\pm5.91\\\\=(36.4-5.91,\ 36.4+5.91)\\\\=(30.49,\ 42.31) [/tex]
∴ a 99 percent confidence interval for µ, the true mean age of guests = (30.49 years, 42.31 years)
The 99% confidence interval for the true mean age of guests at the ski resort, with a sample mean of 36.4 years and standard deviation of 14.5 years from 40 guests, is approximately (30.5, 42.3) years.
Explanation:The question is about calculating a 99 percent confidence interval for the true mean age of guests at a ski mountain resort, where the ages have a right-skewed distribution, and the population standard deviation is 14.5 years. Given a sample mean of 36.4 years from a sample of 40 guests, we can use the following formula to calculate the confidence interval:
CI = µ ± (z * (σ / √n))
Where CI is the confidence interval, µ is the sample mean, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.
For a 99% confidence interval with a sample size of 40, we need to find the z-score that corresponds to the middle 99% of the normal distribution. The z-score for a 99% confidence level is typically 2.576. Then, the margin of error (ME) can be calculated as follows:
ME = 2.576 * (14.5 / √40) ≈ 2.576 * 2.29 ≈ 5.90
Thus, the confidence interval is:
CI = 36.4 ± 5.90
So, the 99% confidence interval for the true mean would be approximately (30.5, 42.3) years.
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A toy rocket is lunch vertically upward from ground level in into velocity of one 28 ft./s how long will it take for the rocket to return to the ground when is the rocket 32 feet above ground
Answer:
1.14 s
Step-by-step explanation:
Time, [tex]t=\frac {d}{s}[/tex]
Here, d is the distance and s is the speed/velocity
Since we're given the velocity, s as 28 ft/s and the distance between the position of the rocket and ground as 32 ft then
[tex]t=\frac {32}{28}=1.142857143\approx 1.14 s[/tex]
Therefore, it needs 1.14 seconds
Note: As you missed to mention the given equation for t seconds and height h, so I am taking a sample equation h(t) =-16t² + 28t + 40. So, I am explaining your question based on this equation, which would anyways clear your query.
Answer:
It will take 2 seconds for the rocket to return to the ground when is the rocket 32 feet above ground.
Note: Sample equation h(t) =-16t² + 28t + 40 was used to solve this problem, as you had not mentioned the equation.
Step-by-step explanation:
To determine:
How long will it take for the rocket to return to the ground when is the rocket 32 feet above ground?
Information Fetching and solution steps:
Initial Velocity = 28 ft/sThe equation for height h and second t is h(t) = -16t² + 28t +40So,
Let us consider the equation h(t) = -16t² + 28t + 40
32 = -16t² + 28t + 40
To find out how long will it take for the rocket to return to the ground when is the rocket 32 feet above ground, plug in h(t) = 32ft, rearrange into quadratic form, and solve:
32 = -16t² + 28t + 40
0 = -16t² + 28t + 8
Step 1: Factor right side of equation
0 = −4(4t + 1)(t − 2)
−4(4t + 1)(t − 2) = 0
Step 2: Set factors equal to 0
4t + 1 = 0 or t − 2 = 0
t = -1/4 or t = 2
As t can not be negative, so t = 2 seconds.
Hence, it will take 2 seconds for the rocket to return to the ground when is the rocket 32 feet above ground.
Keywords: time, height, velocity
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