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svetoff [14.1K]
2 years ago
10

How do you this .. I don’t know how to do it

Mathematics
1 answer:
IRISSAK [1]2 years ago
7 0
I would help but your missing parts of the problem
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There are 3 different roads from city a to city b and 2 different roads from city b to city
Nat2105 [25]

3\cdot2=6

7 0
3 years ago
sandra brings cheese and crakers for lunch every 6 days, and lily brings cheese and crakers every 8 days. if they both brought c
Evgesh-ka [11]

Answer:

24 Days

Step-by-step explanation:

6 0
3 years ago
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.

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

1 - 0.9993 = 0.0007

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.

5 0
2 years ago
Can someone help me with this question please?
zheka24 [161]
(1,0) or (2,5) or (3,10)
5-0=5, 10-5=5, and 15-10=5
7 0
3 years ago
In a school, 130 students are sixth graders. If 52% of the students in the school are sixth graders, how many students are there
emmasim [6.3K]

Answer:

250

Step-by-step explanation:

52%-> 130

100%-> 250

3 0
3 years ago
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