62.1 divided by 58.6

The probability, P, that a person responds to an advertisement can be modeled by the exponential function P = 1 – e^-0.047t. In

MA_775_DIABLO [31]

To solve this equation, start by noticing that you want to find t when P is 75%, or 0.75. Plug in 0.75 for P. Please see the attached screenshot for a step-by-step. The correct answer is 29.496 days.

EDIT 2: I actually was correct! You check the answer by substituting 29.496 for t in the original equation set equal to 0.75. It is 0.75. I had forgotten the negative sign when I checked.

5 0

3 years ago

Here is the question

artcher [175]

3.17 is the answer and u can’t round to the nearest 10 cause it’s not necessary

5 0

3 years ago

Working for a car company, you have been assigned to find the average miles per gallon (mpg) for acertain model of car. you take

Orlov [11]

Answer:

a) The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b) Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c) 37 cars would have to be sampled.

Step-by-step explanation:

Question a:

We have the sample standard deviation, and thus, the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.1448

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.1448\frac{6.2}{\sqrt{15}} = 3.4$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 26.7 - 3.4 = 23.3 mpg.

The upper end of the interval is the sample mean added to M. So it is 26.7 + 3.4 = 30.1 mpg.

The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b. What would happen to the interval if you increased the confidence level from 95% to 99%? Explain

Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c. The lead engineer is not happy with the interval you constructed and would like to keep the width of the whole interval to be less than 4 mpg wide. How many cars would you have to sample to create the interval the engineer is requesting?

Width is twice the margin of error, so a margin of error of 2 would be need. To solve this, we have to consider the population standard deviation as $\sigma = 6.2$ , and then use the z-distribution.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How many cars would you have to sample to create the interval the engineer is requesting?

This is n for which M = 2. So

$M = z\frac{\sigma}{\sqrt{n}}$

$2 = 1.96\frac{6.2}{\sqrt{n}}$

$2\sqrt{n} = 1.96*6.2$

$\sqrt{n} = \frac{1.96*6.2}{2}$

$(\sqrt{n})^2 = (\frac{1.96*6.2}{2})^2$

$n = 36.9$

Rounding up:

37 cars would have to be sampled.

8 0

3 years ago

A town's population has been growing linearly. In 2003, the population was 65000, and the population has been growing by 1700 pe

Virty [35]

Since X represents the number of years and the population consistently increases by 1700 each year the equation would be 65000 + 1700X

7 0

4 years ago

School meeting was attended only by sophomores, juniors, and seniors. 5/12 of those who attended were juniors, and 1/3 were seni

Veronika [31]

If $\frac{9}{12}$ of students were juniors and seniors that leaves the 36 sophomores to be $\frac{3}{12}$ or $\frac{1}{4}$ of attendees. I got $\frac{9}{12}$ by converting $\frac{1}{3} to \frac{4}{12} by multiplying it by 4$ and adding it to $\frac{5}{12}$ .

Now if 36 is $\frac{3}{12}$ of the equation that means theirs at least 36 juniors and seniors.

Lets start with seniors. there should be just over 36 of them there. Lets start by multiplying 36 by $\frac{3}{12}$ to figure out how many $\frac{1}{12}$ is.
The equation would look like this 36*(3/12)
start with 3/12. this equals .25 or $\frac{1}{4}$ .
Now multiply 36 by .25. this equals 9.

We have 9 students per $\frac{1}{12}$ of the attendees.

To get the amount of seniors add 9 to 36 $\frac{4}{12}$ because we already stated $\frac{3}{12}$ is 36.
36+9=45

Now to find the juniors add another 9 to create $\frac{5}{12}$ .
45+9=54

In total you have:
36 sophomores
54 juniors
45 seniors

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3 0

3 years ago