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Tcecarenko [31]
3 years ago
8

Let f(t)= -3t+3. Find t such that f(t) = 6

Mathematics
2 answers:
hammer [34]3 years ago
5 0

Answer:

-1

Step-by-step explanation:

-3t+3=6

-3t=3

t=-1

levacccp [35]3 years ago
5 0
T = -1. You just plug in 6 for f(t) and proceed to solve. Picture of work is attached.

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The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in t
andreyandreev [35.5K]

Answer:

The test statistic is c. 2.00

The p-value is a. 0.0456

At the 5% level, you b. reject the null hypothesis

Step-by-step explanation:

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

This means that the mean and the alternate hypothesis are:

Null: H_{0} = 3

Alternate: H_{a} = 3

The test-statistic is given by:

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.

Sample of 100 customers.

This means that n = 100

3 tested at the null hypothesis

This means that \mu = 3

The average length of time it took the customers in the sample to check out was 3.1 minutes.

This means that X = 3.1

The population standard deviation is known at 0.5 minutes.

This means that \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

The test statistic is z = 2.

The p-value is

Mean different than 3, so the pvalue is 2 multiplied by 1 subtracted by the pvalue of Z when z = 2.

z = 2 has a pvalue of 0.9772

2*(1 - 0.9772) = 2*0.0228 = 0.0456

At the 5% level

0.0456 < 0.05, which means that the null hypothesis is rejected.

7 0
3 years ago
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