3 years ago

Need help answering this math problem.

Mathematics

1 answer:

Fed [463]3 years ago

4 0

The second bubble is the answer

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A population is estimated to have a standard deviation of 9. We want to estimate the population mean within 2, with a 99% level

cupoosta [38]

Answer:

The sample required is $n = 135$

Step-by-step explanation:

From the question we are told that

The standard deviation is $\sigma = 9$

The margin of error is $E = 2$

Given that the confidence level is 99% then the level of significance is mathematically evaluated as

$\alpha = 100-99$

$\alpha = 1 \%$

$\alpha = 0.01$

Next we will obtain the critical value $\frac{\alpha }{2}$ from the normal distribution table(reference math dot armstrong dot edu) , the value is

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.05 }{2} } = 2.58$

The sample size is mathematically represented as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

substituting values

$n = [ \frac{ 2.58 * 9 }{2} ]^2$

$n = 135$

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3 years ago

Suppose that x and y vary inversely and x = 1 when y =5. Write a function that models the inverse variation Graph the function a

Vaselesa [24]

The answer is a function

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3 years ago

A publisher requires 2⁄3 of a page of advertisements for every 5 pages in a magazine. If a magazine has 98 pages, to the nearest

rjkz [21]

Let x be the number of pages of advertisements.

Write a proportion from the table:

$\dfrac{2}{3}$ of a page of advertisment - 5 pages in a magazine

x pages of advertisment - 98 pages in a magazine,

then

$\dfrac{\dfrac{2}{3}}{x}=\dfrac{5}{98}.$

Find x:

$x=\dfrac{\dfrac{2}{3}\cdot 98}{5}=\dfrac{196}{15}=13\dfrac{1}{13}.$

Answer: 13 whole pages and $\dfrac{1}{13}$ of 14th page are advertisments, so 14 pages contain advertisments.

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A waiter had fifty-seven customers. If twenty-three left, how many customers would the waiter still have?

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Answer:

Step-by-step explanation:

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What is the measure of angle x?

blondinia [14]

Answer:

Step-by-step explanation:

Because they are alternate, duh

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2 years ago

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