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3 years ago

Please answer the following questions only 5. 7. 8. 10. 11. and 12.

Mathematics

1 answer:

mojhsa [17]3 years ago

5 0

5) -20b+15

7) -30+9n

8)-48a+36

$10) -{55}{24} n -\frac{185}{24}$

$11) - \frac{15}{15} n + \frac{9}{15}$

$12) - \frac{1}{3} n + \frac{25}{21}$

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Y and x have a proportional relationship, and y = 24 when x = 16.

cluponka [151]

Good evening ,

Answer:

B. 24

Step-by-step explanation:

y and x have a proportional relationship, and y = 24 when x = 16

then

y÷x = 24÷16 = 1.5

Since y=36 then

36÷x = 1.5

then

x = 36÷1.5 = 24.

4 0

4 years ago

Read 2 more answers

The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches

ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

$H_0: \mu = 28.76$

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

$H_1: \mu > 28.76$

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.

28.76 is tested at the null hypothesis:

This means that $\mu = 28.76$

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that $\sigma = 7.5, X = 33, n = 32$ .

Value of the test statistic:

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

$z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}$

$z = 3.2$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level $\alpha = 0.05$ .