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3 years ago

Can someone solve this for me (:

Mathematics

1 answer:

mario62 [17]3 years ago

5 0

I have no idea what the answer to that is or how to solve it but i LOVE UR PROFILE PICTURE

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Five percent of computer parts produced by a certain supplier are defective. what is the probability that a sample of 16 parts c

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It is 3 in 320 chance first it's 3/16 then divide by 5%

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3 years ago

Solve:⬆️ plzzz<br> 6th grade math

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Step-by-step explanation:

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2 years ago

F(x) = (x - 4)(x+2) ?

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Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

John wants to find the center of a wall so he can hang a picture. He measures the wall and determines it is 65.25" wide. 65.25"

antiseptic1488 [7]

Option B is correct.

John wants to find the center of a wall so he can hang a picture. He measures the wall and determines it is 65.25" wide.

Here, 65.25" is Quantitative, continuous

There are two types of quantitative data or numeric data: continuous and discrete.

As a general rule, counts are discrete and measurements are continuous. A continuous data can be recorded at many different points (length, size, width, time, temperature, etc.)

So, option B is the answer.

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2 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$