An inequality to determine the possible length, x, of the rectangle is 3x - 6 ≥ 45/25.4.
The possible length, x is 2.5906 inches.
How to calculate the area of a rectangle?In Mathematics and Geometry, the area of a rectangle can be calculated by using the following mathematical equation:
A = xy
Where:
A represent the area of a rectangle.y represent the width of a rectangle.x represent the length of a rectangle.Conversion:
25.4 millimeters = 1 inches
45 millimeters = 45/25.4 inches.
Since the width of this rectangle is 6 inches shorter than 3 times it’s length and the width of the rectangle is at least 45 millimeters, an inequality to determine the possible length, x, of the rectangle can be written as follows;
3x - 6 ≥ 45/25.4
3x - 6 ≥ 1.7717
3x ≥ 1.7717 + 6
3x ≥ 7.7717
x ≥ 7.7717/3
x ≥ 2.5906 inches.
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Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 16 samples is 7.8 ppm with a standard deviation of 0.8. Assume the population is normally distributed. A level of significance of 0.01 will be used. Make the decision to reject or fail to reject the null hypothesis.
Answer:
We accept H₀ with the information we have, we can say level of ozone is under the major limit
Step-by-step explanation:
Normal Distribution
population mean = μ₀ = 7.5 ppm
Sample size n = 16 df = n - 1 df = 15
Sample mean = μ = 7.8 ppm
Sample standard deviation = s = 0.8
We want to find out if ozono level, is above normal level that is bigger than 7.5
1.- Hypothesis Test
null hypothesis H₀ μ₀ = 7.5
alternative hypothesis Hₐ μ₀ > 7.5
2.-Significance level α = 0.01 we will develop one tail-test (right)
then for df = 15 and α = 0,01 from t -student table we get
t(c) = 2.624
3.-Compute t(s)
t(s) = ( μ - μ₀ ) / s /√n ⇒ t(s) = ( 7.8 - 7.5 )*4/0.8
t(s) = 0.3*4/0.8
t(s) = 1.5
4.-Compare t(s) and t(c)
t(s) < t(c) 1.5 < 2.64
Then t(s) is inside the acceptance region. We accept H₀
(a) Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x)=x/(10 x**2+1) sum_(n=0)^infinity (-1)**n 10**n x**(2n+1) sum_(n=0)^infinity (-1)**n (x**(2n+1))/(10**n) sum_(n=0)^infinity (-1)**n 10**n x**(2n) 10 sum_(n=0)^infinity (-1)**n x**(2n+1) sum_(n=0)^infinity (-1)**n 10**(n+1) x**(n+1) Correct: Your answer is correct. (b) Determine the interval of convergence.
One way to do this is to recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
so that
[tex]\displaystyle\frac x{10x^2+1}=\frac x{1-(-10x^2)}=x\sum_{n=0}^\infty(-10x^2)^n=\sum_{n=0}^\infty(-10)^nx^{2n+1}[/tex]
(which seems to match the first option) so long as [tex]|-10x^2|=10x^2<1[/tex], or [tex]-\frac1{\sqrt{10}}<x<\frac1{\sqrt{10}}[/tex], which is the interval of convergence.
In the Country A legal system, a defendant is presumed innocent until proven guilty. Consider a null hypothesis, Upper H 0, that the defendant is innocent, and an alternative hypothesis, Upper H 1, that the defendant is guilty. A jury has two possible decisions: Convict the defendant (i.e., reject the null hypothesis) or do not convict the defendant (i.e., do not reject the null hypothesis). Explain the meaning of the risks of committing either a Type I or Type II error in this example.
(A) A Type I error would be incorrectly convicting the defendant when he is guilty. A Type II error would be incorrectly failing to convict the defendant when he is innocent.
(B) A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.
(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.
(D) A Type I error would be incorrectly failing to convict the defendant when he is innocent. A Type II error would be incorrectly convicting the defendant when he is guilty.
Answer:
(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.
Step-by-step explanation:
Type I error is rejecting the true null hypothesis and type II error is not rejecting the false null hypothesis. Hence in this scenario, it will be:
A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.
Option C is correct.
Euro Coin. Statistics students at the Akademia Podlaka conducted an experiment to test the hypothesis that the one-Euro coin is biased (i.e., not equally likely to land heads up or tails up). Belgian-minted one-Euro coins were spun on a smooth surface, and 140 out of 250 coins landed heads up. Does this result support the claim that one-Euro coins are biased
Answer:
[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]
[tex]p_v =2*P(z>1.897)=0.0578[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.
Step-by-step explanation:
1) Data given and notation
n=250 represent the random sample taken
X=140 represent the number of heads obtained
[tex]\hat p=\frac{140}{250}=0.56[/tex] estimated proportion of heads
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that that one-Euro coins are biased, so the correct system of hypothesis are:
Null hypothesis:[tex]p=0.5[/tex]
Alternative hypothesis:[tex]p \neq 0.5[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Check for the assumptions that he sample must satisfy in order to apply the test
a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.
b) The sample needs to be large enough
[tex]np_o =250*0.5=125>10[/tex]
[tex]n(1-p_o)=250*(1-0.5)=125>10[/tex]
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>1.897)=0.0578[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.
Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use α= .05 and test to see whether the consultant with more experience has the higher population mean service rating.
Consultant A: n = 16, x = 6.82, s = 0.64
Consultant B: n = 10, x = 6.25, s = 0.75
a. State the null and alternative hypotheses.
b. Compute the value of the test statistic.
c. What is the p-value?
d. What is your conclusion?
Answer:
See the explanation
Step-by-step explanation:
(a)
H0: Consultant with more experience has the higher population mean service rating.
H1: Consultant with more experience doesn't have the higher population mean service rating.
(b)
t = 1.9923 (see the attached image)
(c)
The degrees of freedom for the test statistic,
df = 16
The P-value of the one tailed t- test with 16 degrees of freedom is,
P−value = tdist(X,df,tails)
P-value = tdist(1.9923,16,1)
P-value = 0.032
(d)
Since, P-value 0.032 is less than the significance level 0.05, there is an enough evidence to reject the null hypothesis.
Hence, there is a sufficient evidence to conclude that Consultant with more experience doesn't have the higher population mean service rating.
Hope this helps!
At a nationwide travel agency, newly hired employees are classified as beginners (B). Every six months the performance of each agent is reviewed. Past records indicate that transitions through the ranks to intermediate (I) and qualified (Q) are according to the following Markov chain, where F indicates workers that were fired BIQ F B 45.4 0.15 Q 0 0 1 0 F 0 0 0 1 (a) What fraction eventually become qualified? (b) What is the expected time until a beginner is fired or becomes qualified?
Answer
The answer and procedures of the exercise are attached in the following archives.
The information complete of the exercise is attached in the sheet.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
If other factors are held constant, what happens to a confidence interval if the sample variance increases? The range of t scores increases and the width of the interval increases. The range of t scores decreases and the width of the interval decreases. The standard error increases and the width of the interval increases. The standard error decreases and the width of the interval decreases.
Answer:
The standard error increases and the width of the interval increases
Step-by-step explanation:
The variance is equal to the square of the standard deviation. An increase in variance will increase the standard deviation.
The width of the confidence interval can be defined as:
[tex]W=UL-LL=2z\sigma/\sqrt{n}[/tex]
The width is proportional to the standard deviation, so when the variance increases, the standard deviation and also the width of the interval increases.
Of the 50 students in an undergraduate statistics class, 60% send email and/or text messages during any given lecture. They have a 50/50 chance of being caught by their professor, who is not amused by such unprofessional conduct. The eagle-eyed professor never charges an innocent student. What is the probability a student sends an email and/or text message during lecture and gets caught?
Answer:
0.3 or 30%
Step-by-step explanation:
Since no innocent student will ever be caught, the probability that a student sends an email and/or text message during a lecture AND gets caught is given by the product of the probability of a student sending a message (60%) by the probability of the professor catching them (50%) :
[tex]P = 0.5*0.6 = 0.3[/tex]
The probability is 0.3 or 30%.
The number of rescue calls received by a rescue squad in a city follows a Poisson distribution with an average of 2.83 rescues every eight hours. What is the probability that the squad will have at most 2 calls in an hour? Round answer to 4 decimal places.
Answer:
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
In this problem, we have that:
2.83 rescues every eight hours.
What is the probability that the squad will have at most 2 calls in an hour?
This is
[tex]P = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
We have 2.83 rescues every 8 hours. So for an hour, we have [tex]\mu = \frac{2.83}{8} = 0.354[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.354}*(0.354)^{0}}{(0)!} = 0.7019[/tex]
[tex]P(X = 1) = \frac{e^{-0.354}*(0.354)^{1}}{(1)!} = 0.2485[/tex]
[tex]P(X = 2) = \frac{e^{-0.354}*(0.354)^{2}}{(2)!} = 0.0440[/tex]
So
[tex]P = P(X = 0) + P(X = 1) + P(X = 2) = 0.7019 + 0.2485 + 0.0440 = 0.9944[/tex]
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
Final answer:
The probability that the rescue squad will receive at most 2 calls in an hour is 0.9951
when the average is 2.83 calls every eight hours.
Explanation:
The number of rescue calls that a rescue squad receives is given to be a Poisson distribution with an average of 2.83 rescues every eight hours. To calculate the probability of receiving at most 2 calls in one hour, we first find the average number of rescues per hour by dividing the given average by eight, since there are eight hours in the period mentioned. This gives us an average rate ([tex]λ[/tex]) of 2.83 rescues / 8 hours = 0.35375 rescues per hour. The Poisson probability formula is:
[tex]P(X=k) = (e^{-λ} * λ^k) / k![/tex]
To find the probability of at most 2 calls, we sum the probabilities of 0, 1, and 2 calls:
[tex]P(X=0) = e^{-0.35375} * (0.353750)^0 / 0! = 0.7026 \\\\P(X=1) = e^{-0.35375} * (0.353751)^1 / 1! = 0.2485 \\\\P(X=2) = e^{-0.35375} * (0.353752)^2 / 2! = 0.0440[/tex]
Thus, the desired probability is the sum of these probabilities:
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) ≈ 0.7026 + 0.2485 + 0.0440 ≈ 0.9951
After rounding to four decimal places we have that the probability that the squad will have at most 2 calls in an hour is 0.9951.
The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random sample of 50 intermediate level executives is selected. What is the probability that the mean annual salary of the sample is between $71000 and $73500?A 0.079B. 0.500C. 0.487D. 0.306
Answer: D. 0.306
Step-by-step explanation:
Assuming a normal distribution for the annual salary for intermediate level executives, the formula for normal distribution is expressed as
z = (x - u)/s
Where
x = annual salary for intermediate level executives
u = mean annual salary
s = standard deviation
From the information given,
u = $74000
s = $2500
We want to find the probability that the mean annual salary of the sample is between $71000 and $73500. It is expressed as
P(71000 lesser than or equal to x lesser than or equal to 73500)
For x = 71000,
z = (71000 - 74000)/2500 = - 1.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.1151
For x = 73500,
z = (73500 - 74000)/2500 = - 0.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.4207
P(71000 lesser than or equal to x lesser than or equal to 73500) is
0.4207 - 0.1151 = 0.306
20)
The projected value of an investment is modeled by the exponential function V(t) = 30,000(1.125) where V(t) is the total value
after t years. What does 1.125 represent in the function?
The growth factor of the investment
The initial value of the investment
The projected value of the investment
D)
The value of the investment after t years:
The 1.125 in the exponential function V(t) = 30,000(1.125)^t represents the growth factor of the investment, indicating an annual growth of 12.5%.
Explanation:In the exponential function V(t) = 30,000(1.125)^t, the 1.125 represents the growth factor of the investment. This means each year, the value of the investment grows by 12.5%. It's the rate at which the initial value of the investment, 30,000 in this case, increases on an annual basis. Therefore, the exponential function shows us how the value of the investment changes over time with this constant growth factored in.
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Write and simplify the integral that gives the arc length of the following curve on the given interval b. If necessary, use technology to evaluate or approximate the integral. y:31n x, for 2sxs5 a. The integral that gives the arc length of the curve is L dx &The arc length of the curve is approximately (Round to three decimal places as needed.)
Answer:
a. [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex]
b. [tex]\displaystyle AL = 4.10322[/tex]
General Formulas and Concepts:
Calculus
Differentiation
DerivativesDerivative NotationDerivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Integration
IntegralsDefinite IntegralsIntegration Constant CIntegration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
U-Substitution
Arc Length Formula [Rectangular]: [tex]\displaystyle AL = \int\limits^b_a {\sqrt{1+ [f'(x)]^2}} \, dx[/tex]
Step-by-step explanation:
Step 1: Define
Identify
y = 3ln(x)
Interval [2, 5]
Step 2: Find Arc Length
[Function] Differentiate [Logarithmic Differentiation]: [tex]\displaystyle \frac{dy}{dx} = \frac{3}{x}[/tex]Substitute in variables [Arc Length Formula - Rectangular]: [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ [\frac{3}{x}]^2}} \, dx[/tex][Integrand] Simplify: [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex][Integral] Evaluate: [tex]\displaystyle AL = 4.10322[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Applications of Integration
Book: College Calculus 10e
The arc length of the curve [tex]\( y = 3 \ln x \)[/tex] over the interval [2, 5] is approximately 12.092 units, obtained through numerical methods such as Simpson's rule or a CAS like Wolfram Alpha.
a. Writing and Simplifying the Integral:
The arc length L of the curve y = f(x) over the interval [a, b] is given by the following definite integral:
L = [tex]\int_a^b \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} dx[/tex]
First, find the derivative dy/dx of the given function:
y' = [tex]\dfrac{d}{dx} (3 \ln x) = \dfrac{3}{x}[/tex]
Now, plug f(x) = 3 \ln x and [tex]f'(x) = \dfrac{3}{x}[/tex] into the arc length formula for the interval [2, 5]:
L = [tex]\int_2^5 \sqrt{1 + \left( \dfrac{3}{x} \right)^2} dx[/tex]
L = [tex]\int_2^5 \sqrt{1 + \dfrac{9}{x^2}} dx[/tex]
b. Evaluating or Approximating the Integral:
This integral cannot be solved analytically using standard techniques. Therefore, we can use numerical methods like Simpson's rule or a CAS (Computer Algebra System) to approximate the value.
Using Simpson's rule with 10 subintervals, we get an approximate arc length of:
L ≈ 12.092
Alternatively, using a CAS like Wolfram Alpha, we can obtain a more precise numerical value:
L ≈ 12.091971077
Therefore, the arc length of the curve is approximately 12.092 units.
In Pennsylvania the average IQ score is 101.5. The variable is normally distributed, and the population standard deviation is 15. A school superintendent claims that the students in her school district have an IQ higher than the average of 101.5. She selects a random sample of 30 students and finds the mean of the test scores is 106.4. Test the claim at ???? = 0.05.
Answer:
We conclude that the Pennsylvania school district have an IQ higher than the average of 101.5
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 101.5
Sample mean, [tex]\bar{x}[/tex] = 106.4
Sample size, n = 30
Alpha, α = 0.05
Population standard deviation, σ = 15
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 101.5\\H_A: \mu > 101.5[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.96[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that the Pennsylvania school district have an IQ higher than the average of 101.5
An analyst is forecasting net income for Excellence Corporation for the next fiscal year. Her low-end estimate of net income is $250,000, and her high-end estimate is $350,000. Prior research allows her to assume that net income follows a continuous uniform distribution. The probability that net income will be greater than or equal to $337,500 is
Answer:
[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]
Step-by-step explanation:
probability distribution fucntion is given as
[tex] F_x = P(X \leq x)[/tex]
[tex] = \frac{x -a}{b -a} a<x< b[/tex]
where a indicate lower end estimate = $250,000
b indicate high end estimate = $350,000
probability greater than $337500
[tex] P(X \geq 337,500) = 1- P(X < 337500)[/tex]
[tex] = 1 - \frac{x -a}{b -a}[/tex]
[tex] = 1 - \frac{337500 - 250000}{350000 - 250000}[/tex]
[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]
Calculate the probability of net income being greater than or equal to $337,500 for Excellence Corporation following a continuous uniform distribution. Probability will be equal to 12.5%.
The probability that net income will be greater than or equal to $337,500 is 0.25. To calculate this probability for a continuous uniform distribution, we need to find the proportion of the area under the distribution curve that corresponds to values greater than $337,500.
Step-by-step calculation:
Calculate the total range of net income: $350,000 - $250,000 = $100,000
Calculate the proportion of the area for values greater than $337,500: ($350,000 - $337,500) / $100,000 = 0.125
Since the distribution is continuous and uniform, the probability is equal to the proportion calculated: 0.125 = 12.5%
In a certain city, projections for the next year indicate there is a 20% chance that electronics jobs will increase by 1300, a 50% chance that they will increase by 300, and a 30% chance they will decrease by 900. Find the expected change in the number of electronics jobs in that city in the next year. The expected change is an increase of nothing jobs.
Answer:
110
Step-by-step explanation:
%20 increase = [tex]\frac{20}{100} 1300 = 260\\[/tex]
%50 decrease = [tex]\frac{50}{100}300 = 150[/tex]
%30 no change = [tex]\frac{30}{100}0 = 0[/tex]
Expected change = Increase - Decrease + No Change
Expected change = 260 - 150 + 0
Expected change = 110
The blood type of successive children born to the same parents are independent and have fixedprobabilities that depend on the genetic makeup of the parents. Each child born to a particularset of parents has probability 0.15 of having blood type O. If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?
Answer:
There is a 9.75% probability that exactly 2 of them have type O blood.
Step-by-step explanation:
For each children, there is only two possible outcomes. Either they have blood type O, or they do not. This means that we use the binomial probability distribution to solve this problem.
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 combinatios 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.
In this problem we have that:
4 children, so [tex]n = 4[/tex]
Probability 0.15 of having blood type O, so [tex]p = 0.15[/tex]
If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{4,2}.(0.15)^{2}.(0.85)^{2} = 0.0975[/tex]
There is a 9.75% probability that exactly 2 of them have type O blood.
A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats and dogs. Which of the following statements must be true? Indicate all such statements.
Answer:
The three statements are true.
Step-by-step explanation:
The question is incomplete:
A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats and dogs. Which of the following statements must be true? Indicate all such statements.
A. There are 54 clients who own dogs.
B. There are 34 clients who own dogs but not cats.
C. There are 16 clients who own cats but not dogs.
A. There are 54 clients who own dogs.
TRUE. Of the 70 clients, only 36 own cats. There are left 34 clients that own only dogs. If we add the 20 clients that own both cats and dogs, we have 34+20=54 clients who own dogs.
B. There are 34 clients who own dogs but not cats.
TRUE. Of the 70 clients, only 36 own cats. Then, there are left 70-36=34 clients that own only dogs.
C. There are 16 clients who own cats but not dogs.
TRUE. Out of the 36 clients that own cats (only of with dogs), there are 20 that own both. Therefore, there are 36-20=16 clients that own only cats.
Extreme cold and hot temperatures are known to affect the operation of electronic components. Winter is approaching and you are interested in the mean minimum temperature that will damage an iPod. As a research project, 9 iPods were tested to see what minimum temperature damaged the electronic component. The damaging temperature averaged 5 ºF with a standard deviation of 3 ºF. Find a 95% confidence interval for the true mean minimum temperature that will damage an iPod. (Disclaimer: Apple recommends that to "keep iPod comfy," the iPod’s battery should be kept between temperatures of 32 ºF and 95 ºF.)If the question requires a test of significance, your solution should clearly show the four steps to the test. In addition, you should show that you have checked that the conditions for inference are met.Step 1: State the null and alternative hypothesis.Step 2: Calculate the test statistic.Step 3: Find the p-value.Step 4: State your conclusion in the context of the problem. (Do not just say "Reject H0" or "Do not reject H0.")If the question requires a confidence interval, be sure to make a statement about the confidence interval in context of the problem.
Answer:
A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F
Test statistic(Z) = 2.31
P-value = 0.0104
Step-by-step explanation:
Step 1
Null hypothesis: The average damaging temperature of nine iPods is 5°F
Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F
Step 2
Mean=5°F, Sd=3, df=n-1=9-1=8
The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306
Confidence Interval(CI) = (mean + or - t×sd/√n)
CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F
CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F
Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31
Step 3
Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896
P-value = 1 - 0.9896 = 0.0104
Step 4
Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F
The GT brewery has hired you as an analyst to understand its market position. It is particularly concerned about its major competitor, the ML Brewery, and which brewery has the ‘lead’ market share. Recent history has suggested that which brewery has the ‘lead market share’ can be modeled as a Markov Chain using three states: GT, ML, and Other Company (OC). Data on the lead market share is taken monthly and you have constructed the following one-step transition probability matrix from past data:
GT ML OC
GT 0.75 0.15 0.10
ML 0.30 0.60 0.10
OC 0.20 0.10 0.70
a. The current state of the lead market share in November is GT·The GT Brewery is considering launching a new brand in March only if it has the lead market share in February. Determine the probability that GT Brewery will launch this new brand. Please show any equations or matrices used to calculate this probability and briefly explain how you obtained your answer.
b. Provide the steady-state equations and calculate the steady-state probabilities for the Markov Chain.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
An air-water vapor mixture gas enters an insulated mixing chamber at 300 K, 1 atm, 60 % relative humidity with a mass flow rate of 1 kg/s. A second steam of moist air enters at 380 K, 1 bar, 80% relative humidity at a mass flow rate of 0.5 kg/s. Assume the mixing takes place completely and any condensed liquid or vapor mixture exit at 1 bar. Determine: (a) The humidity ratio for both entering streams. (b) The humidity ratio and the temperature of the mixed stream exiting the control volume. (c) The amount of liquid condensate that leaves the control volume. (d) Challenge: Determine the rate of entropy production, in kW/K.
Answer:
i don't know sorry hope this helps
Step-by-step explanation:
Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and released to mix into the population. After they have had an opportunity to mix, a random sample of 10 of these animals is selected. Let X = the number of tagged animals in the second sample. Assuming there is a total of 25 animals of this type in the region, what are E(X) and Var(X)?
Answer:
[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]
Step-by-step explanation:
Previous concepts
The hypergeometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:
[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]
Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes
The expected value and variance for this distribution are given by:
[tex]E(X)= n\frac{M}{N}[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}[/tex]
What is the distribution of X?
For this case the random variable X follows a hypergometric distribution.
Compute the values for E(X) and Var(X)
For this case n=10, M=5, N=25, so then we can replace into the formulas like this:
[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]
What is the probability that none of the animals in the second sample are tagged?
So for this case we want this probability:
[tex]P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565[/tex]
What is the probability that all of the animals in the second sample are tagged?
So for this case we want this probability:
[tex]P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474[/tex]
Given is a finite set of spherical planets, all of the same radius and no two intersecting. On the surface of each planet consider the set of points not visible from any other planet. Prove that the total area of these sets is equal to the surface area of one planet.
Answer:
Step-by-step explanation:
To answer this question, First, we define a direction in space, which defines
the North Pole of each planet.
We also assume that the direction is not at 90° perpendicular to the axis of any two planet.
Assume we now define an order on the set of planet by saying that planet A ≥ B, when removing all the other planet from space, the north pole B, is visible from A.
If we refer to the direction of the planet in the diagram, we discover that,
1, A ≥ A
2, If A ≥ B and B≥ A,
Then A = B
3, If A ≥ B, and B ≥ C , then A ≥ C
4, Either A ≥ B or B ≥ A
It should also be noted that the array of order above has a unique maximal element (M). This is the only planet whose North Pole is not visible from another.
If we now consider a sphere of the same radius as the planets. Remove
from it all North Poles defined by directions that are perpendicular
to the axis of two of the planets. This is a set of area zero.
For every other point on this sphere, there exists a direction in
space that makes it the North Pole, and for that direction, there
exists a unique North Pole on one of the planets which is not visible
from the others. Hence the total area of invisible points is equal
to the area of this sphere, which in turn is the area of one of the planets.
In a survey, 31% of a randomly selected sample of n = 935 American adults said that they do not get enough sleep each night. The margin of error for the 95% confidence level was reported to be 3.2%. (a) Use the survey information to create a 95% confidence interval for the percentage that feels they don't get enough sleep every night_________%.(b) Do you think this sample could be used to estimate the percentage of college students who think they don't get enough sleep each night? Why or why not? A. Yes, the population the sample comes from is representative of college students, fulfilling the Fundamental Rule for Using Data For Inference B. No, the population the sample comes from may not be representative of college students, violating the Fundamental Rule for Using Data For Inference. O C. Yes, the population the sample comes from is representative of college students, violating the Fundamental Rule for Using Data For Inference. D. No, the population the sample comes from may not be representative of college students, fulfilling the Fundamental Rule for Using Data For Inference
Answer:
a) [0.278; 0.342]
b) No, the population the sample comes from may not be representative of college students, violating the Fundamental Rule for Using Data For Inference.
Step-by-step explanation:
Hello!
a)
The study variable is
X: Number of American adults that said that they do not get enough sleep each night.
The sample is n= 935 American Adults
Sample proportion 'p= 0.31
The margin of error of the 95% CI d= 0.032
The formula for a CI for the population proportion is:
['p ± [tex]Z_{1-\alpha /2}[/tex]*[tex]\sqrt{\frac{'p(1-'p)}{n} }[/tex]]
The basic structure of the CI formula is "Point estimator" ± "Margin of error"
So with the given data, you can calculate the CI.
[0.31 ± 0.032]
[0.278; 0.342]
So with a confidence level of 95%, you'd expect that the interval [0.278; 0.342] will contain the population proportion of the American adults that don't get enough sleep at night.
b)
The answer is no. The sample is of "American adults" there is no information on their education level, therefore it is possible that all of them have a college education or that none of the have. Since no further information about their education then you cannot be certain that this sample will be representative of the American College Students.
Then the correct option is B
I hope you have a SUPER day!
The following 5 questions are based on this information: An economist claims that average weekly food expenditure of households in City 1 is more than that of households in City 2. She surveys 35 households in City 1 and obtains an average weekly food expenditure of $164. A sample of 30 households in City 2 yields an average weekly expenditure of $159. Historical data reveals that the population standard deviation for City 1 and City 2 are $12.50 and $9.25, respectively.Col1 City 1 x1(bar)=164 σ(1)=12.5 n(1)=35Col2 City 2 x2 (bar) =159 σ(2) =9.25 n2=30Let μ(1) be the mean weekly food expenditure for City 1 and μ(2) be that for City 2.1. To test the economist’s claim, the competing hypotheses should be formulated asSelect one:a. H0:μ1-μ2>0 versus Ha:μ1-μ2≤0b. H0:μ1-μ2≥0 versus Ha:μ1-μ2<02.The standard error of x(1)bar- x(2) bar isSelect one:a. 0.82b. 2.70c. 12.5d. 9.253.The value of the test statistics isSelect one:a. 0.40b. 1.85c. 0.54d. 27.784. The p-value of the test is
Select one:
a. 0.34
b. 0.03
c. 0.29
d. 0.08
5.At α=0.05,
Select one:
a. We can reject H(0) in favor of H(a)
b. We cannot reject H(0)
c. We can conclude that average weekly food expenditures in City 1 is less than that of City 2
Answer:
Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]
[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]
b) 2.70
[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]
b. 1.85
[tex]p_v =P(Z>1.85)=0.032[/tex]
b. 0.03
a. We can reject H(0) in favor of H(a)
Step-by-step explanation:
Data given and notation
[tex]\bar X_{1}=164[/tex] represent the mean for the sample 1
[tex]\bar X_{2}=159[/tex] represent the mean for the sample 2
[tex]\sigma_{1}=12.5[/tex] represent the population standard deviation for the sample 1
[tex]s_{2}=9.25[/tex] represent the population standard deviation for the sample B2
[tex]n_{1}=35[/tex] sample size selected 1
[tex]n_{2}=30[/tex] sample size selected 2
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]
We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:
[tex]z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^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".
Standard error
The standard error on this case is given by:
[tex]SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}[/tex]
Replacing the values that we have we got:
[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]
b. 2.70
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]
P-value
Since is a one side right tailed test the p value would be:
[tex]p_v =P(Z>1.85)=0.032[/tex]
b. 0.03
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
a. We can reject H(0) in favor of H(a)
Find equations of both the tangent lines to the ellipse x2 + 9y2 = 81 that pass through the point (27, 3). y = (smaller slope) y = (larger slope)
Answer:
The tangent line equations are:
- The one that has the smaller slope [tex]\bold{y =3}[/tex]
- The one that has the larger slope [tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]
Step-by-step explanation:
In order to find the tangent lines we need to find first the first derivative since the first derivative evaluated at a point give us the slope of the tangent line.
Working with implicit differentiation.
In order to find the derivative we can work with implicit differentiation and we get
[tex]2x+18y \cfrac{dy}{dx}= 0[/tex]
Notice that only when we are finding the derivative of an expression that has y we multiply by dy/dx due chain rule.
Solving for the first derivative we get
[tex]18y\cfrac{dy}{dx}=-2x[/tex]
[tex]\cfrac{dy}{dx}=-\cfrac{x}{9y}[/tex]
So the equation of the first derivative at the point (x,y) give us the first equation for the slope.
[tex]m=-\cfrac{x}{9y}[/tex]
Slope of a line using definition.
We can also find the slope of the line that will pass the point (x,y) trough the point (27,3) using the definition:
[tex]m = \cfrac{y_2-y_1}{x_2-x_1}[/tex]
Replacing the point we get
[tex]m=\cfrac{y-3}{x-27}[/tex]
That is another equation for the slope.
Finding the value of the slopes.
We can now set both slope equations equal to each other to find the point (x,y) where they intersect and the value of the slopes.
[tex]-\cfrac{x}{9y}= \cfrac{y-3}{x-27}[/tex]
We can work with cross multiplication to get
[tex]-x(x-27)=9y(y-3)[/tex]
And we can distribute and simplify
[tex]-x^2+27x=9y^2-27y[/tex]
Moving everything to the right side
[tex]0=x^2-27x+9y^2-27y\\\\0=x^2+9y^2-27x-27y[/tex]
At this point we can use the ellipse equation so we can replace [tex]x^2+9y^2[/tex] with 81, that will give us
[tex]0=81-27x-27y[/tex]
Then we can divide by 27 and solve for y.
[tex]0=3-x-y\\y= 3-x[/tex]
At this point we can replace on the ellipse equation.
[tex]x^2+9(3-x)^2=81[/tex]
And we can distribute and simplify.
[tex]x^2+9(9-6x+x^2)=81\\x^2+81-54x+9x^2=81\\x^2-54x+9x^2=0\\10x^2-54x=0\\[/tex]
So then we can factor and solve for x.
[tex]2x(5x-27)=0[/tex]
Setting each factor and solve for x we get
[tex]x= 0, \cfrac{27}{5}[/tex]
At x = 0, we can find the y value.
[tex]y= 3-x\\y= 3-0\\y= 3[/tex]
And the slope
[tex]m = -\cfrac{x}{9y}\\m = -\cfrac{0}{9y}\\m = 0[/tex]
So the line equation is
[tex]y-3=0(x-0)\\y=3[/tex]
For the second point x = 27/5 we have:
[tex]y= 3-\cfrac{27}{5}\\y=-\cfrac{12}{5}[/tex]
So slope is:
[tex]m = -\cfrac{\cfrac{27}{5}}{9\left(-\cfrac{12}{5}\right)}[/tex]
[tex]m = \cfrac 14[/tex]
So the line equation is
[tex]y-\left(-\cfrac{12}{5}\right)= \cfrac 14 \left(x -\cfrac{27}{5}\right)\\y = \cfrac 14 x - \cfrac{15}{4}[/tex]
Thus the equations of tangent lines are:
[tex]\bold{y =3}[/tex]
and
[tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]
To find the equations of the tangent lines to the ellipse that passes through a specific point, differentiate the equation of the ellipse to get a formula for the slope at any point on the ellipse. Set this equal to the slope of the line from the point on the ellipse to the specific point, and solve. The resultant solutions give the points of tangency, and the lines through these points and the specific point are the desired tangent lines.
Explanation:The subject of the question is about finding the equations of the tangent lines to the ellipse x^2 + 9y^2 = 81 that pass through a specific point (27, 3). We first differentiate the equation of the ellipse to get a general formula for the slope of the tangent line at any point on the ellipse. The differentiation of x^2 + 9y^2 = 81 with respect to x gives 2x + 18yy' = 0, so y' = -x/ (9y). Now, we put the slopes y' equal to the slopes of line from (x, y) to (27, 3), and solve the resulting system of equations. The solutions give the points of tangency, and the lines through these points and (27, 3) are the desired tangent lines. The exact computations involve solving a quadratic equation and it'll give two slopes for the tangent line that passes through the point (27,3).
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find an equation of the line satisfying the given conditions. write the equation using function notation. Through (-20,5) and (-36, 9) And can you show me the steps?
Answer:
The equation of the line using function notation is
[tex]y=f(x)=-\frac{1}{4}x[/tex]
Step-by-step explanation:
Given points are (-20,5) and (-36, 9)
Now to find the equation of the line passes through these points
Let [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] be the two given points (-20,5) and (-36, 9) respectively.
To find slope
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]m=\frac{9-5}{-36-(-20)}[/tex]
[tex]m=\frac{4}{-36+20}[/tex]
[tex]m=\frac{4}{-16}[/tex]
[tex]m=-\frac{1}{4}[/tex]
Therefore [tex]m=-\frac{1}{4}[/tex]
The equation of the line is of thr form y=mx+c
The point (-20,5) passes through the above line and [tex]m=-\frac{1}{4}[/tex]
[tex]5=-\frac{1}{4}(-20)+c[/tex]
[tex]5=5+c[/tex]
[tex]c=0[/tex]
[tex]y=-\frac{1}{4}x+0[/tex]
Therefore [tex]y=-\frac{1}{4}x[/tex]
Therefore the equation of the line using function notation is
[tex]y=f(x)=-\frac{1}{4}x[/tex]
Choose an American household at random, and let the random variable X be the number of cars, including SUVs and light trucks, the residents own. The table gives the the probability model if we ignore the few households that own more than 8 cars.Number of cars X 0 1 2 3 4 5 6 7 8Probability 0.087 0.323 0.363 0.144 0.053 0.019 0.007 0.002 0.001A housing company builds houses with two?car garages. What percent of households have more cars than the garage can hold?A- 22.7%B- 41.0%C- 59.0%D- 14.4%
Answer:
Option A.
Step-by-step explanation:
The given probability table is
Number of cars X: 0 1 2 3 4 5 6 7 8
Probability :0.087 0.323 0.363 0.144 0.053 0.019 0.007 0.002 0.001
It is given that a housing company builds houses with two car garages.
We need to find the percent of households have more cars than the garage can hold.
[tex]P(X>2)=1-P(X\leq 2)[/tex]
[tex]P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]
Substitute the probability values from the given table.
[tex]P(X>2)=1-[0.087+0.323+0.363][/tex]
[tex]P(X>2)=1-0.773[/tex]
[tex]P(X>2)=0.227[/tex]
It means 22.7% of households have more cars than the garage can hold.
Therefore, the correct option is A.
The percent of households that have more cars (including SUVs and light trucks) than a two-car garage can accommodate is approximately 22.7%.
Explanation:To answer the student's question, let's examine the probability model provided. The random variable X represents the number of cars, including SUVs and light trucks, in a randomly selected American household.
If a housing company builds houses with two-car garages, we want to determine what percent of households have more cars than the garage can hold. To find this, we would add up the probabilities for the households that own more than 2 cars.
The probabilities for owning 3, 4, 5, 6, 7 or 8 cars are 0.144, 0.053, 0.019, 0.007, 0.002, and 0.001, respectively. Adding all these probabilities gives us:
0.144 + 0.053 + 0.019 + 0.007 + 0.002 + 0.001 = 0.226
To convert this to a percentage, we multiply by 100, which gives us 22.6%. Therefore, approximately 22.7% of households have more cars than the garage can hold, which corresponds to answer choice A.
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A farmer uses two types of fertilizers. A 50-lb bag of Fertilizer A contains 8 lb of nitrogen, 2 lb of phosphorus, and 4 lb of potassium. A 50-lb bag of Fertilizer B contains 5 lb each of nitrogen, phosphorus, and potassium. The minimum requirements for a field are 440 lb of nitrogen, 260 lb of phosphorus, and 360 lb of potassium. If a 50-lb bag of Fertilizer A costs $80 and a 50-lb bag of Fertilizer B costs $30, find the amount of each type of fertilizer the farmer should use to minimize his cost C in dollars while still meeting the minimum requirements. (Let x represent the number of bags of Fertilizer A and y represent the number of bags of Fertilizer B.)
Answer:
The amount fertilizer A the farmer should use while still meeting the minimum requirement is approximately seventy-six 50-lb bags which would cost $6080.
The amount of fertilizer B the farmer should use while still meeting the minimum requirement is approximately seventy-one 50- lb bags which would cost $2120
Step-by-step explanation:
Fertilizer A
Let x represent the number of 50- lb bags needed for the minimum requirement
Cost of I 50-lb bag = $80
Cost of x 50-lb bag = $80x
1 50-lb bag contains 14lb nutrient ( 8lb Nitrogen+ 2lb Phosphorus + 4lb Potassium)
x 50-lb bag contains 1060lb nutrient (440lb N + 260lb P + 360lb K) which is the minimum requirement
x = 1060lb ÷ 14lb = 75.71
x = 76 ( to the nearest whole number)
76 50-lb bags would cost 76×$80 = $6080
Fertilizer B
Cost of a 50- lb bag is $30
Let y be the number of bags required
Cost of y bag is $30y
y = 1060÷ 15= 71
Cost = 71× $30 = $2130
Answer:
0 fertilizer A
88 fertilizer B
Step-by-step explanation:
Shown in the attachment
The logistic growth function at right describes the number of people, f(t), who have become ill with influenza t weeks after its initial outbreak in a particular community. f (t )equals StartFraction 107 comma 000 Over 1 plus 4200 e Superscript negative t EndFraction.
a. How many people became ill with the flu when the epidemic began?
b. How many people were ill by the end of the fourth week?
c. What is the limiting size of the population that becomes ill?
Answer:
Part (a): 25 people became ill with the flu when the epidemic began.
Part (b): 1373 people were ill by the end of the fourth week.
Part (c): The limiting size of the population that becomes ill is 107,000.
Step-by-step explanation:
Consider the provided logistic growth function
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-t}}[/tex]
Part (a) How many people became ill with the flu when the epidemic began?
Substitute t = 0 in above function.
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-0}}[/tex]
[tex]f (t )=\frac{ 107,000}{4201}[/tex]
[tex]f (t )=25.47[/tex]
Hence, 25 people became ill with the flu when the epidemic began.
Part (b) How many people were ill by the end of the fourth week?
Substitute t = 4 in above function.
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-4}}[/tex]
[tex]f (t )=1373.10[/tex]
Hence, 1373 people were ill by the end of the fourth week.
Part (c) What is the limiting size of the population that becomes ill?
Substitute t = ∞ in above function.
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-\infty}}[/tex]
[tex]f (t )=\frac{ 107,000}{1 +0}[/tex]
[tex]f (t )=107,000[/tex]
Hence, the limiting size of the population that becomes ill is 107,000.
Using the logistic growth function, the number of sick people at the start of the epidemic and at the end of the fourth week can be found by substituting the appropriate values for time into the function. The limiting size of the potentially ill population is the maximum value of the function, which is 107,000.
Explanation:The logistic growth function provided describes the number of people, represented by f(t), who have become ill with influenza, t weeks after its initial outbreak. The function is given as f(t) = 107000/(1 + 4200 * e-t).
a. At the beginning of the epidemic (i.e., when t=0), the number of ill people, f(0), can be calculated by substituting 0 for t in the equation, resulting in 107,000/(1+4200) people.
b. By the end of the fourth week (i.e., when t=4), the number of ill people, f(4), can be calculated by substituting 4 for t in the equation.
c. The limiting size of the population that becomes ill is represented by the maximum value the growth function can take, which in this case is 107,000 (as seen by the numerator of the function). This means that illness will eventually spread to reach a potential maximum of 107,000 people in this specific community.
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A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is
A. 0.037773 to 0.202227
B. 3.7773% to 20.2227%
C. 59.98786% to 61.01214%
D. 51.7773% to 68.2227%
E. 10.988% to 13.012%
A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of 60 funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is
A. 0.037773 to 0.202227
B. 3.7773% to 20.2227%
C. 59.98786% to 61.01214%
D. 51.7773% to 68.2227%
E. 10.988% to 13.012%
Answer: E. 10.988% to 13.012%
Step-by-step explanation:
Given;
Mean x= 12%
Standard deviation r = 4%
Number of samples tested n = 60
Confidence interval is 95%
Z' = t(0.025)= 1.96
Confidence interval = x +/- Z'(r/√n)
= 12% +/- 1.96(4%/√60)
= 12% +/- 0.01214%
Confidence interval= (10.988% to 13.012%)