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ivolga24 [154]
2 years ago
6

If p represents the original price, which of the following expressions could be used to calculate the sale price of a shirt that

is on sale for 15% off? Select all that apply.
A: p+15
B:p+0.15p
C: p - 0.15p
D: 0.85p
Mathematics
1 answer:
dezoksy [38]2 years ago
7 0

Answer:

C & D

Step-by-step explanation:

You can do these two ways. You can multiply the cost of the the shirt by 15% (.15 as a decimal.) and then subtract that answer from the original price of the shirt to get the sales price of the shirt.

You can also multiply the price of the shirt by 85% (.85 as a decimal) and that will give you the sales price without having to subtract. 15% + 85% = 100%, so 85% represents the sales price of the shirt.

Anyway, since you multiply, you can eliminate choice A & B.

Since the direction in the question says to select all that apply, I would check both C & D.

C is the equation of the first way I talked about- Price - 15% * price

C       p - 0.15p

D is the equation of the second way I talked about. Price times 85%

D            0.85p

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(-y+5•3)+7.2y-9)=6.2y+n
kramer
(-y+5x3)+(7.2y-9)=6.2y+n
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since you're adding the two parentheses, you don't need to have them there
-y+15+7.2y-9=6.2y+n
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3 years ago
A market research team thinks that their new ad campaign is better than their old one. They used to be able to sell to 50% of th
andreev551 [17]

Answer:

a. The test statistic is 2 and we conclude that the new ad campaign is not signficantly better.

Step-by-step explanation:

They used to be able to sell to 50% of those who saw their ads. Test if the new campaign is better.

At the null hypothesis, we test is it is the same, that is, the proportion is the same.

H_0: p = 0.5

At the alternate hypothesis, we test if it is significantly better, that is, the proportion is above 50%.

H_1: p > 0.5

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.

0.5 is tested at the null hypothesis:

This means that \mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5

They take a random sample of 100 potential buyers and find that they convinced 60 of these people to buy their product.

This means that n = 100, X = \frac{60}{100} = 0.6

Test statistic:

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

z = \frac{0.6 - 0.5}{\frac{0.5}{\sqrt{100}}}

z = 2

The test statistic is 2.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.6, which is 1 subtracted the by p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

The p-value of the test is 0.0228 > 0.01, which means that we cannot conclude that the new ad campaign is signficantly better, so the correct answer is given by option A.

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