Answer:
a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids.
b. (32.1, 42.3)
c. p-value < .00001
d. The null hypothesis is rejected at the α=0.05 significance level
e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.
p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.
Step-by-step explanation:
a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.
b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:
(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })
Substituting in the values of X bar and s, the interval becomes:
(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })
or (32.1, 42.3)
c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:
t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.
Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.
d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly changes the plasma aluminum levels of children.
e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.
Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.
Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.
To test whether the mean plasma aluminum level of infants on antacids differs from those not on antacids, a null hypothesis (that the means are equal) is established alongside an alternative. A confidence interval and p-value are calculated to assess this hypothesis, and based on these results, a decision is made to reject or not reject the null hypothesis.
Explanation:Hypotheses Formulation and Test Statistics
To analyze the plasma aluminum levels in infants receiving antacids compared to those not receiving them, one would perform a hypothesis test. The steps include formulating a null hypothesis (H0) and an alternative hypothesis (Ha), calculating the test statistic, finding the p-value, and making a decision regarding H0 based on the p-value and the confidence interval.
Null Hypothesis (H0): The mean plasma aluminum level of infants receiving antacids is equal to the mean level of those not receiving antacids (H0: μ = 4.13 µg/L).
Alternative Hypothesis (Ha): The mean plasma aluminum level of infants receiving antacids is not equal to the mean level of those not receiving antacids (Ha: μ ≠ 4.13 µg/L).
To construct a 95% confidence interval, we use the sample mean (× = 37.20 µg/L), sample standard deviation (s = 7.13 µg/L), the sample size (n = 10), and the t-distribution since the population variance is unknown. The confidence interval provides a range of values within which the true mean is likely to lie.
For the p-value, we compare it against the alpha level α=0.05. If the p-value is less than α, we reject H0; otherwise, we do not reject H0. The p-value indicates the likelihood of obtaining a sample mean at least as extreme as the one observed if H0 were true.
If the confidence interval does not include the population mean of children not receiving antacids and the p-value is less than α, we reject H0 in favor of Ha. If a one-sided test is appropriate (for example, if we only want to test if the mean aluminum level is higher in the treated group), Ha would be reformulated accordingly (Ha: μ > 4.13 µg/L), potentially resulting in a different decision from the two-sided test.
Following Exercise 3.5.9, let p1, . . . , pk be a pairwise relatively prime set of naturals, each greater than 1. Let X be the set {0, 1, . . . , p1 −1}× . . . ×{0, 1, . . . , pk −1}. Define a function f from {0, 1, . . . , p1p2 . . . pk − 1} to X by the rule f(x) = x%p1, . . . , x%pk. Prove that f is a subject
Answer: see the pictures attached
Step-by-step explanation:
how many pounds of chamomile tea that costs 18.20 per pound must be mixed with 12lb of orange tea that costs 12.25 per pound to make a mixture that costs 14.63 per pound
8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
Step-by-step explanation:
We are given the following in the question:
Chamomile tea:
Unit cost = 18.20 per pound
Amount = x pounds
Total cost =
[tex]18.20\times x = 18.20x[/tex]
Orange tea:
Unit cost = 12.25 per pound
Amount = 12 pounds
Total cost =
[tex]12.25\times 12 = 147[/tex]
Total mixture:
Unit cost = 14.63 per pound
Amount = (12+x) pounds
Total cost =
[tex]14.63\times (12+x) = 175.56 + 14.63x[/tex]
We can write the equation that cost of mixture is equal to cost of chamomile tea and orange tea.
[tex]18.20x + 147 = 175.56 + 14.63x\\\Rightarrow 18.20x- 14.63x = 175.56-147\\\Rightarrow 3.57x = 28.56\\\Rightarrow x = 8[/tex]
Thus, 8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
A state insurance commission estimates that 13% of all motorists in its state areuninsured. Suppose this proportion is valid. Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured. You may assume that the normaldistribution applies.
Answer:
79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
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.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 50, p = 0.13[/tex]
So
[tex]\mu = E(X) = np = 50*0.13 = 6.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.13*0.87} = 2.38[/tex]
Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
This is 1 subtracted by the pvalue of Z when X = 4. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 6.5}{2.38}[/tex]
[tex]Z = -0.84[/tex]
[tex]Z = -0.84[/tex] has a pvalue of 0.2005
1 - 0.2005 = 0.7995
79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
Using the normal approximation to the binomial distribution, the probability that at least 5 out of 50 motorists will be uninsured when 13% of motorists are uninsured is approximately 74.22%.
First, we calculate the mean (μ) and the standard deviation (σ) of the number of uninsured motorists in a sample of 50. The mean number of uninsured motorists is μ = np, where n is the sample size and p is the probability of a motorist being uninsured. So, μ = 50 * 0.13 = 6.5.
The standard d eviation is σ = √(np(1-p)). Therefore, σ = √(50 * 0.13 * (1 - 0.13)) ≈ 2.31.
To find the probability of at least 5 uninsured motorists, we want P(X ≥ 5). We need to standardize this to the standard normal distribution to find the z-score. The z-score is z = (X - μ) / σ.
For X = 5, the z-score is z = (5 - 6.5) / 2.31 ≈ -0.65. We use the z-table or a calculator to find the probability corresponding to z = -0.65.
Since we want at least 5 uninsured, we find P(Z ≥ -0.65), which equals 1 - P(Z < -0.65). The value from the z-table for -0.65 is approximately 0.2578. Therefore, the probability we're looking for is 1 - 0.2578 = 0.7422, or 74.22%.
American car travels 32 miles on 1 gallon of gas European car travels 12.7 km on 1 L of gas which car gets better gas mileage explain your reasoning
Answer: the American car gets better gas mileage.
Step-by-step explanation:
American car travels 32 miles on 1 gallon of gas. This means that its gas mileage is 32 miles per gallon.
European car travels 12.7 km on 1 L of gas. We would convert 12.7 km to miles.
1 kilometer = 0.621 mile
12.7 kilometer = 12.7 × 0.621 = 7.8867 miles
We would also convert 1 L to gallons.
1 liter = 0.264 us gallons
Therefore, the gas mileage of the European car in miles per gallon would be
7.8867/0.264 = 29.87 miles per gallon.
Therefore, since 32 miles per gallon is greater than 29.87 miles per gallon, it means that the American car gets better gas mileage.
A survey among US adults of their favorite toppings on a cheese pizza reported that 43% favored pepperoni, 14% favored mushrooms, and 6% favored both pepperoni and mushrooms. What is the probability that a random adult favored pepperoni or mushrooms on their cheese pizza? Provide your answer as a whole number in the box below, i.e., .32 is 32% so you would enter 32. Round as needed.
Answer:
51% of US adults favored pepperoni or mushrooms on their cheese pizza.
Step-by-step explanation:
We are given the following in the question:
Percentage of US adults that favored pepperoni = 43%
[tex]P(P) = 0.43[/tex]
Percentage of US adults that favored mushroom = 14%
[tex]P(M) = 0.14[/tex]
Percentage of US adults that favored both pepperoni and mushroom = 6%
[tex]P(M\cap P) = 0.06[/tex]
We have to evaluate the probability that a random adult favored pepperoni or mushrooms on their cheese pizza.
Thus, we have to evaluate:
[tex]P(M\cup P) = P(M) + P(P) - P(M\cap P)\\P(M\cup P) = 0.43 + 0.14 - 0.06\\P(M\cup P) = 0.51 = 51\%[/tex]
Thus, 51% of US adults favored pepperoni or mushrooms on their cheese pizza.
For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.
Santa Fe black-on-white is a type of pottery commonly found at archaeological excavations at a certain monument. At one excavation site a sample of 572 potsherds was found, of which 363 were identified as Santa Fe black-on-white.
(a) Let p represent the proportion of Santa Fe black-on-white potsherds at the excavation site. Find a point estimate for p. (Round your answer to four decimal places.)
(b) Find a 95% confidence interval for p. (Round your answers to three decimal places.)
lower limit
upper limit
Answer:
a) p = 0.6346
b) 95% confidence interval
Lower limit: 0.5951
Upper limit: 0.6741
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 572
Number of Santa Fe black-on-whitepots , x = 363
a) proportion of Santa Fe black-on-white potsherds
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{363}{572} = 0.6346[/tex]
b) 95% confidence interval
[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting the values, we get:
[tex]0.6346\pm 1.96(\sqrt{\frac{0.6346(1-0.6346)}{572}}) = 0.6346\pm 0.0395\\\\=(0.5951,0.6741)[/tex]
Lower limit: 0.5951
Upper limit: 0.6741
Write the equation of the line that passes through (3, 4) and (2, −1) in slope-intercept form. (2 points) a y = 3x − 7 b y = 3x − 5 c y = 5x − 11 d y = 5x − 9
Answer: y = 5x − 11
Step-by-step explanation:
The equation of a straight line can be represented in the slope-intercept form, y = mx + c
Where c = intercept
Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent
change in the value of y = y2 - y1
Change in value of x = x2 -x1
y2 = final value of y
y 1 = initial value of y
x2 = final value of x
x1 = initial value of x
The line passes through (3,4) and (2, -1),
y2 = - 1
y1 = 4
x2 = 2
x1 = 3
Slope,m = (- 1 - 4)/(2 - 3) = - 5/- 1 = 5
To determine the y intercept, we would substitute x = 3, y = 4 and m= 5 into
y = mx + c. It becomes
4 = 5 × 3 + c
4 = 15 + c
c = 4 - 15 = - 11
The equation becomes
y = 5x - 11
Give (in percents) the two marginal distributions, for marital status and for income. Do each of your two sets of percents add to exactly 100%? If not, why not?
Answer:
Answer is attached
Step-by-step explanation:
6.11
Marginal Distribution of Marital Status
Single 4.1%
Married 93.9%
Divorced 1.5%
Widowed .5%
Marginal Distribution of Job Grade
Job Grade 1 11.6%
Job Grade 2 51.5%
Job Grade 3 30.2%
Job Grade 4 6.7%
Each of the marginal distributions sum up to exactly 100%. Depending on how one rounds the percentages, it is possible that the results will not be exactly 100%.
The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) Find the mean and standard deviation of the sample count X who are married. b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married? c.) What is the probability that we find less than 4 of the seniors are married? d.) What is the probability that we find at least 1 of the seniors are married?
Answer:
a) Mean 6, standard deviation 2.42
b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.
c) 14.85% probability that we find less than 4 of the seniors are married.
d) 99.77% probability that we find at least 1 of the seniors are married
Step-by-step explanation:
For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]n = 300, p = 0.02[/tex]
a.) Find the mean and standard deviation of the sample count X who are married.
Mean
[tex]E(X) = np = 300*0.02 = 6[/tex]
Standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42[/tex]
b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?
This is P(X = 8).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040[/tex]
10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.
c.) What is the probability that we find less than 4 of the seniors are married?
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023[/tex]
[tex]P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143[/tex]
[tex]P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436[/tex]
[tex]P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485[/tex]
14.85% probability that we find less than 4 of the seniors are married.
d.) What is the probability that we find at least 1 of the seniors are married?
Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
From c), we have that [tex]P(X = 0) = 0.0023[/tex]. So
[tex]0.0023 + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 0.9977[/tex]
99.77% probability that we find at least 1 of the seniors are married
In this problem, the mean and standard deviation of a binomial distribution, with probability of success 0.02 and sample size 300, are found. Subsequently, the probabilities that 8, less than 4, and at least 1 of the seniors are married are computed using the binomial formula.
Explanation:This problem deals with the Binomial distributions in statistics. Since we know the proportion of high school seniors who are married is 0.02, and the sample size is 300, we can use these values to calculate the mean and the standard deviation.
a.) The mean (mean = np) of the sample count X who are married is 0.02*300=6, and the standard deviation would be sqrt(n*p*(1-p)) = sqrt(300*0.02*0.98) = √5.88≈2.43.
b.) The probability that, in our sample of 300, we find that 8 of the seniors are married is given by the binomial formula P(X=k) = C(n,k)(p^k)(1-p)^(n-k). Plugging n=300, k=8, p=0.02 into the formula, we get the desired probability.
c.) The probability that we find less than 4 of the seniors are married is sum of P(X=k) from k=0 to 3. This could be computed using the aforementioned binomial formula. Remember, you're summing the probabilities for each k.
d.) The probability that we find at least 1 of the seniors are married can be found by subtracting the probability that none of the seniors are married from 1 (i.e., P(X >=1) = 1 - P(X=0)).
Learn more about Binomial Distribution here:https://brainly.com/question/39749902
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An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique," (2005, Vol. 13, pp. 273-279) cites a success rate more than 90% for meniscal tears with a rim width of less than 3 mm, but only a 67% success rate for tears of 3-6 mm. If you are unlucky enough to suffer a meniscal tear of less than 3 mm on your left knee, and one of width 3-6 mm on your right knee, what are the mean and variance of the number of successful surgeries?
Answer:
Mean = 1.57
Variance=0.31
Step-by-step explanation:
To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:
1) Both surgeries are successful (X=2).
[tex]P(e_1)=0.90*0.67=0.603[/tex]
2) Left knee unsuccessful and right knee successful (X=1).
[tex]P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067[/tex]
3) Right knee unsuccessful and left knee successful (X=1).
[tex]P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297[/tex]
4) Both surgeries are unsuccessful (X=0).
[tex]P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033[/tex]
Then, the mean can be calculated as the expected value:
[tex]M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57[/tex]
The variance can be calculated as:
[tex]V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111[/tex]
The mean and variance of the number of successful surgeries for both knees combined are:
Mean: [tex]\({1.57}\)[/tex]
Variance: [tex]\({0.3111}\)[/tex]
The mean and variance of the number of successful surgeries for the given meniscal tears can be calculated using the information provided about the success rates.
For a meniscal tear with a rim width of less than 3 mm, the success rate is more than 90%. For simplicity, let's assume the success rate is exactly 90% (since we don't have the exact number above 90%). For a tear of 3-6 mm, the success rate is 67%.
Let's denote the success of a surgery as a random variable [tex]\( X \)[/tex], which takes the value 1 if the surgery is successful and 0 if it is not. The probability of success [tex]\( P(X = 1) \)[/tex] is the success rate, and the probability of failure [tex]\( P(X = 0) \)[/tex] is [tex]\( 1 - P(X = 1) \)[/tex].
For the left knee (tear less than 3 mm):
- [tex]\( P(X = 1) = 0.90 \)[/tex] (success rate)
- [tex]\( P(X = 0) = 1 - 0.90 = 0.10 \)[/tex] (failure rate)
For the right knee (tear 3-6 mm):
- [tex]\( P(X = 1) = 0.67 \)[/tex] (success rate)
- [tex]\( P(X = 0) = 1 - 0.67 = 0.33 \)[/tex] (failure rate)
The mean (expected value) of the number of successful surgeries for each knee is calculated as follows:
For the left knee:
[tex]\[ E(X) = \sum_{i=0}^{1} x_i \cdot P(X = x_i) = 1 \cdot 0.90 + 0 \cdot 0.10 = 0.90 \][/tex]
For the right knee:
[tex]\[ E(X) = \sum_{i=0}^{1} x_i \cdot P(X = x_i) = 1 \cdot 0.67 + 0 \cdot 0.33 = 0.67 \][/tex]
The variance of the number of successful surgeries for each knee is calculated using the formula for the variance of a binary random variable:
For the left knee:
[tex]\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]
[tex]\[ E(X^2) = \sum_{i=0}^{1} x_i^2 \cdot P(X = x_i) = 1^2 \cdot 0.90 + 0^2 \cdot 0.10 = 0.90 \][/tex]
[tex]\[ \text{Var}(X) = 0.90 - (0.90)^2 = 0.90 - 0.81 = 0.09 \][/tex]
For the right knee:
[tex]\[ E(X^2) = \sum_{i=0}^{1} x_i^2 \cdot P(X = x_i) = 1^2 \cdot 0.67 + 0^2 \cdot 0.33 = 0.67 \][/tex]
[tex]\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]
[tex]\[ \text{Var}(X) = 0.67 - (0.67)^2 = 0.67 - 0.4489 = 0.2211 \][/tex]
Now, assuming the surgeries on the two knees are independent events, the mean and variance for both knees combined can be calculated as follows:
Mean for both knees:
[tex]\[ E(X_{\text{left}} + X_{\text{right}}) = E(X_{\text{left}}) + E(X_{\text{right}}) = 0.90 + 0.67 = 1.57 \][/tex]
Variance for both knees:
Since the surgeries are independent, the variance of the sum is the sum of the variances:
[tex]\[ \text{Var}(X_{\text{left}} + X_{\text{right}}) = \text{Var}(X_{\text{left}}) + \text{Var}(X_{\text{right}}) = 0.09 + 0.2211 = 0.3111 \][/tex]
Therefore, the mean and variance of the number of successful surgeries for both knees combined are:
Mean: [tex]\({1.57}\)[/tex]
Variance: [tex]\({0.3111}\)[/tex]
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night. Assume that room rates are normally distributed with a standard deviation of $55. What is the minimum cost that captures the 20% most expensive hotel rooms in New York City?
Answer:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 250.31.
Step-by-step explanation:
Previous concepts
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".
Solution to the problem
Let X the random variable that represent the room hotel rate of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(204,55)[/tex]
Where [tex]\mu=204[/tex] and [tex]\sigma=55[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.20[/tex] (a)
[tex]P(X<a)=0.80[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.842[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.842[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 250.31.
Suppose 10000 people are given a medical test for a disease. About1% of all people have this condition. The test results have a 15% false positive rate and a 10% false negative rate. What percent of the people who tested positive actually have the disease?
Answer:
The percent of the people who tested positive actually have the disease is 38.64%.
Step-by-step explanation:
Denote the events as follows:
X = a person has the disease
P = the test result is positive
N = the test result is negative
Given:
[tex]P(X)=0.01\\P(P|X^{c})=0.15\\P(N|X)=0.10[/tex]
Compute the value of P (P|X) as follows:
[tex]P(P|X)=1-P(P|X^{c})=1-0.15=0.85[/tex]
Compute the probability of a positive test result as follows:
[tex]P(P)=P(P|X)P(X)+P(P|X^{c})P(X^{c})\\=(0.85\times0.10)+(0.15\times0.90)\\=0.22[/tex]
Compute the probability of a person having the disease given that he/she was tested positive as follows:
[tex]P(X|P)=\frac{P(P|X)P(X)}{P(P)}=\frac{0.85\times0.10}{0.22} =0.3864[/tex]
The percentage of people having the disease given that he/she was tested positive is, 0.3864 × 100 = 38.64%.
In choosing what music to play at a charity fund raising event, Shaun needs to have an equal number of string quartets from Mendelssohn, Beethoven, and Haydn. If he is setting up a schedule of the 66 string quartets to be played, and he has 66 Mendelssohn, 1616 Beethoven, and 6868 Haydn string quartets from which to choose, how many different schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
Answer:
64.69e221
Step-by-step explanation:
When choosing, the combination formula for selection is used. That is when selecting "r" number of items from a possible "n" items, then the number of ways is denoted as:
nCr = n! / (n-r)! × r!
Since 66 string quartet have to be chosen and the 3 genres must be equally represented in the string quartet, then we must have 22 number of each genre in it.
Number of ways to select 22 mendelssohn from possible 66 = 66C22 = 1.82 × 10^17
Number of ways to select 22 Beethoven from possible 1616 = 1616C22 = 2.97 × 10^49
Number of ways to select 22 Haydn from possible 6868 = 6868C22 = 2.2 × 10^63
Total number of ways to arrange these 66 schedules = 66! = 5.44 × 10^92
Number of possible schedule = [1.82 * 10^17] * [2.97*10^49] * [2.2*10^63] * [5.44*10^92]
=64.69 ×10^221. ≈64.69e221
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. Suppose the skater has a mass of 64 kg . One eighth of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. The rest of her mass is approximately in the form of a 20-cm-radius cylinder.
Answer:
Step-by-step explanation:
Given data:
Mass of the one arm of the skater, m = (1/16) x 64 = 4 kg
Rest mass of the skater in the form of cylinder, M = (7 / 8) x 64 kg = 56 kg
Radius of the cylinder, R = 20 cm = 0.20 m
The parallel axis theorem:
According to a 2013 study by the Pew Research Center, 15% of adults in the United States do not use the Internet (Pew Research Center website, December, 15, 2014). Suppose that 10 adults in the United States are selected randomly.
a. Is the selection of the 10 adults a binomial experiment? Explain.
b. What is the probability that none of the adults use the Internet (to 4 decimals)?
c. What is the probability that 3 of the adults use the Internet (to 4 decimals)? If you calculate the binomial probabilities manually, make sure to carry at least 4 decimal digits in your calculations.
d. What is the probability that at least 1 of the adults uses the Internet (to 4 decimals)?
Answer:
a) For this case we can use the binomial model since we assume independent events and the same probability for each trial is the same p =0.15
b) [tex]P(X=0)=(10C0)(0.15)^0 (1-0.15)^{10-0}=0.1969[/tex]
c) [tex]P(X=3)=(10C3)(0.15)^3 (1-0.15)^{10-3}=0.1298[/tex]
d) [tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)[/tex]
And using the result from part a we got:
[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)= 1-0.1969 =0.8031[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n p)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Solution to the problem
Part a
For this case we can use the binomial model since we assume independent events and the same probability for each trial is the same p =0.15
Part b
For this case we want this probability:
[tex] P(X=0)[/tex]
And replacing we got:
[tex]P(X=0)=(10C0)(0.15)^0 (1-0.15)^{10-0}=0.1969[/tex]
Part c
For this case we want this probability:
[tex] P(X=3)[/tex]
And replacing we got:
[tex]P(X=3)=(10C3)(0.15)^3 (1-0.15)^{10-3}=0.1298[/tex]
Part d
For this cae we want thi probability:
[tex] P(X \geq 1)[/tex]
And we can use the complment rule and we got:
[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)[/tex]
And using the result from part a we got:
[tex] P(X \geq 1)= 1-P(X <1) = 1-P(X=0)= 1-0.1969 =0.8031[/tex]
A company determines that its marginal cost, in dollars, for producing x units of a product is given by Upper C prime (x )equals4500 x Superscript negative 1.9, where xgreater than or equals1.11. Suppose that it were possible for the company to make infinitely many units of this product. What would the total cost be?
Answer:
Total Cost = Fixed Cost as x --> ∞
Step-by-step explanation:
C'(x) = 4500 x⁻¹•⁹ where x ≥ 1
Marginal Cost = C'(x) = (dC/dx)
C(x) = ∫ (marginal cost) dx
C(x) = ∫ (4500 x⁻¹•⁹)
C(x) = (-5000 x⁻⁰•⁹) + k
where k = constant of integration or in economics term, K = Fixed Cost.
C(x) = [-5000/(x⁰•⁹)] + Fixed Cost
The company wants to make infinitely many units, that is, x --> ∞
C(x --> ∞) = [-5000/(∞⁰•⁹)] + Fixed Cost
(∞⁰•⁹) = ∞
C(x --> ∞) = [-5000/(∞)] + Fixed Cost
But mathematically, any number divide by infinity = 0;
(-5000/∞) = 0
C(x --> ∞) = 0 + Fixed Cost = Fixed Cost.
Total Cost of producing infinite number of units for this cost function is totally the Fixed Cost.
Please help me find the answer.
Answer:
b/a = c/b
if a = b, then b = c
Answer: the second one (b/a = c/b) and the last one (if a = b then b = c) are the only ones that are true
Step-by-step explanation:
The number of views of a page on a Web site follows aPoisson distribution with a mean of 1.5 per minute.a. What is the probability of no views in a minute?b. What is the probability of two or fewer views in10 minutes?c. Does the answer to the previous part depend on whetherthe 10-minute period is an uninterrupted interval? Explain.d. Determine the length of a time interval such that the probabilityof no views in an interval of this length is 0.001.
Answer:
a. P (X=0) = 0.223
b. P(X≤2) = 3.93 * 10^-5
C. No
D. t=4.6 miutes
Step-by-step explanation:
a. P (X=0) = [tex]e^{-1.5}[/tex] = 0.223
b. P(X≤2) =[tex]e^{1.5 * 10} (1 + 1.5*10 + \frac{1.5^{2} * 10^{2} }{2} )[/tex]
= 3.93 * 10^-5
C. No, the answer to the previous part does not depend on whether the 10-minute period is an uninterrupted interval, it only depend on the uniformity of the density of views per minute ad independency of the disjoint time intervals.
D. P (X=0) =0.001
[tex]e^{-1.5t} = 0.001\\\\-1.5t=ln(0.001)\\\\t =\frac{-6.9}{-1.5}[/tex]
t=4.6 miutes
The Poisson distribution is used to calculate the probabilities of certain counts of discrete events occurring in a given time interval. By using the average rate of occurrence, we can determine probabilities relevant to web traffic, call intervals, or urgent care visits.
Explanation:The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. Various scenarios using Poisson distributions involve understanding the probability of a certain number of events occurring over a specified period.
Probability of no views in one minute: This can be computed using the formula for the Poisson distribution where the expected mean number of events (λ) is 1.5. The probability P(X=0) is given by (e^-1.5 * 1.5^0) / 0!.Probability of two or fewer views in 10 minutes: Given that events occur at a rate of 1.5 per minute, over 10 minutes the average rate would be 15. Calculate the cumulative probability for 0, 1, and 2 events and sum them up.Dependence of the interval: The answer to part b does not depend on the interval being uninterrupted, as the Poisson model assumes independence of intervals.Length of interval for a 0.001 probability of no views: Find the interval length 't' such that P(X=0) = 0.001 using the Poisson formula and solving for 't'.You can now sell 80 cups of lemonade per week at 40¢ per cup, but demand is dropping at a rate of 4 cups per week each week. Assuming that raising the price does not affect demand, how fast do you have to raise your price if you want to keep your weekly revenue constant? HINT [Revenue = Price × Quantity.]
Final answer:
To keep the weekly revenue constant while demand drops, we can set up an equation using the revenue formula. By equating the original revenue with the new revenue, we can find the rate at which the price needs to be raised. Taking the derivative, we can determine the rate of change of the price.
Explanation:
To keep the weekly revenue constant, we need to find the rate at which the price has to be raised to offset the drop in demand. Currently, the price is 40¢ per cup and demand is dropping at a rate of 4 cups per week. Since revenue is equal to price times quantity, we can set up the equation:
Revenue = Price × Quantity.
Initially, we have 80 cups of lemonade sold at 40¢ per cup, resulting in a revenue of $32 (80 x 40¢). As demand drops by 4 cups per week each week, the new quantity sold can be represented by 80 - 4t, where t represents the number of weeks. Let P be the new price per cup that needs to be raised. The new revenue equation can be written as:
Revenue = P(80 - 4t).
To find the value of P, we equate the original revenue ($32) with the new revenue:
$32 = P(80 - 4t).
Simplifying the equation, we get:
32 = 80P - 4Pt.
Moving the terms around, we have:
4Pt = 80P - 32.
Dividing both sides by 4P, we get:
t = (80P - 32)/(4P).
So, the rate at which the price needs to be raised to keep the weekly revenue constant is given by the derivative of t with respect to P. Taking the derivative, we get:
t' = (4(80P - 32) - 4P(80))/(4P)^2.
Simplifying further, we have:
t' = (320P - 128 - 320P)/(4P)^2.
Simplifying again, we get:
t' = -128/(4P)^2.
Thus, the rate of change of t with respect to P is given by -128/(4P)^2. This represents the rate at which the price needs to be raised in order to keep the weekly revenue constant.
Draw a rectangle that shows 8 equal parts . Shade more than 3/8 of the rectangle but less than 5/8 .what fraction did you model? Use multiplication and division to write two equivalent fractions for your model.
Answer:
4/8 more than 3/ but less than 5/8
Answer: I modeled 4/8 because it is greater than 3 less than 5 2 equivalent fractions are 8/16 12/24
Step-by-step explanation:
An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.
a) Write out the hypotheses.
b) Calculate the test statistic.
c) Find the p-value.
d) Do we reject the null hypothesis? Explain.
e) What can you conclude based on this evidence
Answer:
An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.
a) Write out the hypotheses.
Fewer than 75% of claims settled within two months of filing.
b) Calculate the test statistic.
Test statistic = percent of claims settled in two months = 23/35 = 65.7%
c) Find the p-value.
we need to use the z-score with a table
=
standard deviation = s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}, =
√(1/34) [(0.657)(12) + (0.343)(23)] = √0.463911765 = 0.681
let 1 = "solved" and 0 = "unresolved"
thus our mean is 0.657
z = (0.75 - 0.657)/(0.681) - 0.137
p = 0.55172
d) Do we reject the null hypothesis? Explain.
Yes, because our p value is not below 0.05 and is not substantial to prove our null hypothesis
e) What can you conclude based on this evidence?
That further testing is needed.
Brian is filling a conic container with water. He has the container half full. The radius of the container is 5 inches and the height is 20 inches. What is the current volume of the water?
The current volume of the water is 261.66 square inches.
Solution:
The container is in cone shape.
Radius of the container = 5 inch
Height of the container = 20 inch
Volume of the container = [tex]\frac{1}{3} \pi r^2 h[/tex]
[tex]$=\frac{1}{3}\times 3.14 \times 5^2 \times 20[/tex]
Volume of the container = 523.33 square inch
Current volume of the water = Half of the volume of container
[tex]$=\frac{1}{2}\times523.33[/tex]
= 261.66 square inch
The current volume of the water is 261.66 square inches.
The altitude of a triangle is increasing at a rate of 2.5 2.5 centimeters/minute while the area of the triangle is increasing at a rate of 2 2 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.5 7.5 centimeters and the area is 96 96 square centimeters
Answer:
The base of the triangle is decreasing at a rate 8 centimeter per minute.
Step-by-step explanation:
We are given the following in the question:
[tex]\dfrac{dh}{dt} = 2.5\text{ cm per minute}\\\\\dfrac{dA}{dt} = 2\text{ square cm per minute}[/tex]
Instant height = 7.5 cm
Instant area = 96 square centimeters
Area of triangle =
[tex]A=\dfrac{1}{2}\times b\times h[/tex]
where b is the base and h is the height of the triangle.
[tex]96 = \dfrac{1}{2}\times b \times 7.5\\\\b = \dfrac{96\times 2}{7.5} = 25.6[/tex]
Rate of change of area of triangle =
[tex]\dfrac{dA}{dt} = \dfrac{d}{dt}(\dfrac{bh}{2})\\\\\dfrac{dA}{dt} =\dfrac{1}{2}(b\dfrac{dh}{dt} + h\dfrac{db}{dt})[/tex]
Putting values, we get,
[tex]2 = \dfrac{1}{2}(7.5\dfrac{db}{dt}+25.6(2.5))\\\\4 = 7.5\dfrac{db}{dt} + 64\\\\-60 = 7.5\dfrac{db}{dt} \\\\\Rightarrow \dfrac{db}{dt} = \dfrac{-60}{7.5} = -8[/tex]
Thus, the base of the triangle is decreasing at a rate 8 centimeter per minute.
ALGEBRA 2 HELP!!!
Find all the linear factors
f(x)=[tex]3x^{4}-7x^{3}-10x^{2} +28x-8[/tex]
Answer:
x + 2
3x − 1
x − 2
Step-by-step explanation:
The order of the polynomial is 4, so there are 4 roots.
Use rational root theorem to find possible rational roots.
±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3
By trial and error, -2, ⅓, and 2 are all roots. The fourth root can't be imaginary (complex roots come in pairs), so one of the real roots must be repeated.
Answer: the factors are (x - 2)(x + 2)(3x - 1)
Step-by-step explanation:
The given polynomial function is expressed as
f(x) = 3x⁴ - 7x³ - 10x² + 28x - 8
The first step is to test for any value of x that satisfies the polynomial when
3x⁴ - 7x³ - 10x² + 28x - 8 = 0
Assuming x = 2, then
3(2)⁴ - 7(2)³ - 10(2)² + 28 × 2 - 8 = 0
48 - 56 - 40 + 56 - 8 = 0
0 = 0
It means that x - 2 is a factor.
To determine the other factors, we would apply the long division method. The steps are shown in the attached photo. Looking at the photo, we would factorize the quadratic equation which is expressed as
3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x + 2) - 1(x + 2)
(3x - 1)(x + 2) = 0
One mole of nickel (6 1023 atoms) has a mass of 59 grams, and its density is 8.9 grams per cubic centimeter, so the center-to-center distance between atoms is 2.23 10-10 m. You have a long thin bar of nickel, 2.1 m long, with a square cross section, 0.08 cm on a side. You hang the rod vertically and attach a 69 kg mass to the bottom, and you observe that the bar becomes 1.11 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in nickel.
Final answer:
Nickel does not crystallize in a simple cubic structure because the hypothetical density of a simple cubic arrangement is 2.23 g/cm³, significantly less than the actual density of nickel, which is 8.90 g/cm³.
Explanation:
To determine if nickel crystallizes in a simple cubic structure, we can compare the calculated density of a simple cubic arrangement of nickel atoms to the actual density of nickel. Given the edge length of the unit cell for nickel is 0.3524 nm (3.524 x 10⁻⁸ cm) and knowing the properties of a simple cubic lattice, the volume of a single cubic unit cell would be V = a³ = (3.524 x 10⁻⁸ cm)³ = 4.376 x 10⁻²³ cm³. With the mass of a single nickel atom (58.693 g/mol divided by Avogadro's number, 6.022 x 10²³ atoms/mol), if the nickel crystallized in a simple cubic structure, the predicted density would be:
density = mass/volume = (9.746 x 10⁻²³ g) / (4.376 x 10⁻²³ cm³) = 2.23 g/cm³. However, because the actual density of nickel is 8.90 g/cm³, we can conclude that nickel does not form a simple cubic structure.
A salesperson obtained a systematic sample of size 2525 from a list of 500500 clients. To do so, he randomly selected a number from 1 to 2020, obtaining the number 1313. He included in the sample the 1313th client on the list and every 2020th client thereafter. List the numbers that correspond to the 2525 clients selected.
Question
A salesperson obtained a systematic sample of size 25 from a list of 500 clients. To do so, he randomly selected a number from 1 to 20, obtaining the number 13. He included in the sample the 13th client on the list and every 20th client thereafter. List the numbers that correspond to the 25 clients selected.
Answer:
13, 33, 53, ...... 393
Step-by-step explanation:
The Question is making reference to arithmetic progression
Where the first term is 13
And the difference between each interval is 20 (i.e number of clients)
Using the following formula, well obtained the client in any position
Tn = a + (n - 1) d
Where Tn = nth term
a = T1 = 1st term = 13
n = number of terms = 1 to 25
d = common difference = 20
Calculating T2
T2 = 13 + (2 - 1) * 20
T2 = 13 + 20
T2 = 33
Calculating T3
T3 = 13 + (3 - 1) * 20
T3 = 53
.......
Calculating the 20th term
T3 = 13 + (20 - 1) * 20
T20 = 393
1. A manufacturer of a printer determines that the mean number of days before a cartridge runs out of ink is 75 days, with a standard deviation of 6 days. Assuming a normal distribution, what is the probability that the number of days will be less than 67.5 days?
Answer:
[tex]P(X<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.25)=0.106[/tex]
Step-by-step explanation:
Previous concepts
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".
Solution to the problem
Let X the random variable that represent the number of days before cartridge runs out of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(75,6)[/tex]
Where [tex]\mu=75[/tex] and [tex]\sigma=6[/tex]
We are interested on this probability
[tex]P(X<67.5)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.25)=0.106[/tex]
A random sample of 1800 NAU students in Flagstaff found 1134 NAU students who love their MAT114 class. Find a 95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class. Express your results to the nearest hundredth of a percent. .
Answer:
95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (60.77% , 65.23%)
Step-by-step explanation:
Among 1800 NAU students, 1134 students love their class. We have to find the 95% confidence interval of students who love their class.
We will use the concept of confidence interval of population proportion for this problem.
The proportion of students who love the class = p = [tex]\frac{1134}{1800}=0.63[/tex]
Proportion of students who do not love the class = q = 1 - p = 1 - 0.63 = 0.37
Total number of students in the sample = n = 1800
Confidence Level = 95%
The z values associated with this confidence level(as seen from z table) = 1.96
The formula to calculate the confidence interval for population proportion is:
[tex](p-z\times\sqrt{\frac{p \times q}{n}},p+z\times\sqrt{\frac{p \times q}{n}})[/tex]
Using the values in this expression gives:
[tex](0.63-1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}}, 0.63+1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}})\\\\ =(0.6077,0.6523)[/tex]
Thus, 95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (0.6077 ,0.6523) or (60.77% , 65.23%
Use Green's Theorem to find the work done by the force F(x, y) = x(x + y) i + xy2 j in moving a particle from the origin along the x-axis to (6, 0), then along the line segment to (0, 6), and then back to the origin along the y-axis.
Final answer:
The question involves using Green's Theorem to calculate work done by a specific force field along a triangular path. Green's Theorem links the line integral around a closed curve to a double integral over the region it encloses. However, direct calculation of line integrals may be more applicable for this problem.
Explanation:
The question asks to use Green's Theorem to calculate the work done by a force F(x, y) = x(x + y) i + xy^2 j moving a particle along a specified path. Green's Theorem relates a line integral around a simple closed curve C, to a double integral over the plane region D bounded by C. It's expressed in the form ∑_C (M dx + N dy) = ∫∫_D (∂N/∂x - ∂M/∂y) dA where M and N are components of a vector field.
To use Green's Theorem, we identify M = x(x + y) and N = xy^2. Thus, ∂N/∂x = y^2 and ∂M/∂y = x. The path described (origin to (6,0), to (0,6), and back to origin) encloses a triangle, the area of which is easily integrated over. However, it is crucial to note that direct integration methods or alternative strategies may sometimes be more straightforward for such paths, especially when they do not form a typical closed curve as in standard Green's Theorem applications.
In the context of this problem, the key would be computing the line integrals directly due to the specific nature of the force field and the path involved. The use of Green's Theorem hints at setting up an integration over the area, but with the given vector field and triangular path, the execution would involve careful calculation of respective line integrals or assessing the area directly since the theorem simplifies to a calculation over the region enclosed by the path.
Use inverse trigonometric functions to solve the following equations. If there is more than one solution, enter all solutions as a comma-separated list (like "1, 3"). If an equation has no solutions, enter "DNE".solve tan ( θ ) = 1 tan(θ)=1 for θ θ (where 0 ≤ θ < 2 π 0≤θ< 2π).
The solutions to the equation tan(θ) = 1 within the specified range 0 ≤ θ < 2π: θ = 0.7854, 3.9270
Apply the inverse tangent function:
We begin by applying the inverse tangent function (arctan) to both sides of the equation: arctan(tan(θ)) = arctan(1)
Since arctan is the inverse of tangent, they cancel each other out on the left side, leaving us with: θ = arctan(1)
Determine the reference angle:
arctan(1) = π/4, which is the reference angle in the first quadrant where tangent is 1.
Find solutions in other quadrants:
The tangent function has a period of π, meaning it repeats its values every π radians.
Since tangent is also positive in the third quadrant, we add π to the reference angle to find the solution in that quadrant: θ = π/4 + π = 5π/4
Consider the specified range:
We're given the range 0 ≤ θ < 2π. Both π/4 and 5π/4 fall within this range, so they are the valid solutions.
Therefore, the solutions to the equation tan(θ) = 1 within the specified range are θ = 0.7854 (π/4) and θ = 3.9270 (5π/4).
Final answer:
To solve the equation tan(θ) = 1 for θ, we need to use the inverse tangent function. The solution to the equation is θ = π/4.
Explanation:
To solve the equation tan(θ) = 1 for θ, we need to use the inverse trigonometric function. In this case, we will use the inverse tangent function, also known as arctan or atan.
Applying the inverse tangent function to both sides of the equation, we get θ = atan(1).
Using a calculator, we find that atan(1) = π/4. Therefore, the solution to the equation is θ = π/4.