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dalvyx [7]
3 years ago
6

Factorise 8x^2+20XY-12y

Mathematics
1 answer:
Julli [10]3 years ago
4 0

Answer:

4(2x^2+5xy-3y)

Step-by-step explanation:

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Exhibit 9-2 The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the cu
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Answer:

At a .05 level of significance, it can be concluded that the mean of the population is significantly more than 3 minutes.

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

At the null hypothesis, we test if the mean is of at most 3 minutes, that is:

H_0: \mu \leq 3

At the alternative hypothesis, we test if the mean is of more than 3 minutes, that is:

H_1: \mu > 3

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.

3 is tested at the null hypothesis:

This means that \mu = 3

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

This means that n = 100, X = 3.1, \sigma = 0.5

Value of the test statistic:

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

z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}

z = 2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 3.1, which is 1 subtracted by the p-value of z = 2.

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

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The p-value of the test is of 0.0228 < 0.05, meaning that the is significant evidence to conclude that the mean of the population is significantly more than 3 minutes.

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