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wlad13 [49]
2 years ago
7

A small publishing company is planning to publish a new book. The production costs will include one-time fixed costs (such as ed

iting) and variable costs (such as printing). There are two production methods it could use. With one method, the one-time fixed costs will total $82,910, and the variable costs will be $10.75 per book. With the other method, the one-time fixed costs will total $22,110 , and the variable costs will be $23.25 per book. For how many books produced will the costs from the two methods be the same?
Mathematics
1 answer:
MatroZZZ [7]2 years ago
8 0

Answer:

When 4864 books are sold

Step-by-step explanation:

Create an equation where x is the number of books produced:

The first method will be represented by 10.75x + 82,910

The second method will be represented by 23.25x + 22,110

Set these equal to each other and solve for x:

10.75x + 82,910 = 23.25x + 22,110

82,910 = 12.5x + 22,110

60,800 = 12.5x

4864 = x

So, the costs will be the same for both methods when 4864 books are sold

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In a study of parents' perceptions of their children's size, researchers asked parents to estimate their youngest child's height
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Answer:

Test statistic t=6.505

P-value=0

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that parents tend to underestimate their youngest child's size.

Then, the null and alternative hypothesis are:

H_0: \mu=0\\\\H_a:\mu> 0

The significance level is assumed to be 0.05.

The sample has a size n=39.

The sample mean is M=7.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=7.2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{7.2}{\sqrt{39}}=1.153

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.5-0}{1.153}=\dfrac{7.5}{1.153}=6.505

The degrees of freedom for this sample size are:

df=n-1=39-1=38

This test is a right-tailed test, with 38 degrees of freedom and t=6.505, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>6.505)=0

As the P-value (0) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that parents tend to underestimate their youngest child's size.

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