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slega [8]
3 years ago
5

Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each, and the store so

ld no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p(125–p)100p(125–p)
B. 150p(250–p)150p(250–p)
C. 300p(375–p)300p(375–p)
D. 400p(500–p)400p(500–p)
E. 500p(625–p)
Mathematics
1 answer:
Artemon [7]3 years ago
5 0
Answer is b 150p(250-p)150p(250-p
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(x^3 - 158) divided by (x-8)
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Answer:


Step-by-step explanation:

I'll do this with synthetic division. It is the easiest to show in this editor.

8 || 1  +   0   +   0    -   158

             8        64  +   512  

============================

     1       8       64     354


The results of the synthetic division are

x^2 + 8x + 64 remainder 358 / (x - 8)

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company A offers a semimonthly salary of $2468, company B offers a biweekly salary of $2421. Which company offers the greater an
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Find the PERIMETER of the following shape.
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6 0
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How many angles are supplementary to the angle of 100 degrees?
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Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of
zhannawk [14.2K]

Answer:

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

H_{0}: p = 0.29

We want to test the hypothesis that this is true, so the alternate hypothesis is:

H_{a}: p \neq 0.29

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that \mu = 0.29, \sigma = \sqrt{0.29*0.71}

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that n = 100, X = 0.21

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}

z = -1.76

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

5 0
3 years ago
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