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masya89 [10]
3 years ago
10

Brianna works at an electronics store as a salesperson. Brianna earns a 4% commission on the total dollar amount of all phone sa

les she makes, and earns a 4% commission on all computer sales. How much money would Brianna earn in commission on a day that she sold $1900 worth of phones and $1600 worth of computers? How much money would Brianna earn in commission on a day that she sold $xx worth of phones and $yy worth of computers?
Mathematics
2 answers:
svetoff [14.1K]3 years ago
8 0

Answer:

P(x)=0.025x+100

Step-by-step explanation:

The 0.025 is 2.5% as a decimal. Then, the +100 is for the base pay amount Hailey receives.

Fofino [41]3 years ago
7 0

Answer:

Lacy would earn a commission of $104.

Lacy would earn 0.06x dollars on phones sale and 0.02y dollars on computer sales.

Step-by-step explanation:

Hopefully this helped, if not HMU and I will try my best to get u a better aswer!

Have a great day! :)

And if u need help HMU and I got you, no points needed:)

Andf thanks  for the 25 points! :D

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Answer:

The value of the test statistic is z = 1.39.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

Step-by-step explanation:

Using traditional methods it takes 92 hours to receive an advanced flying license.

This means that at the null hypothesis, it is tested if the mean is of 92, that is:

H_0: \mu = 92

Test if there is evidence that the technique lengthens the training time

At the alternative hypothesis, it is tested if the mean is more than 92, that is:

H_1: \mu > 92

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.

92 is tested at the null hypothesis:

This means that \mu = 92

A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.

This means that n = 70, X = 93, \sigma = \sqrt{36} = 6

Value of the test statistic:

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

z = \frac{93 - 92}{\frac{6}{\sqrt{70}}}

z = 1.39

The value of the test statistic is z = 1.39.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.

Looking at the z-table, z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

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