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givi [52]
4 years ago
6

Perform the following calculation. Express your answer in scientific notation. (2×105) + (6×105) Write your answer in scientific

notation using the format: 4.2x10^5 with no spaces
Mathematics
2 answers:
ZanzabumX [31]4 years ago
7 0
Answer:

8.4x10^1
Thanks
nirvana33 [79]4 years ago
6 0
8.4x10^1 I think that should be the answer
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The answer is B. 44.52

Step-by-step explanation:

Just added them all together =P

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Can you help me with the Anabella question pls
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Hey! I need help on this question as well . I know it is b or c. But I’m not sure which one . Thank you.
Airida [17]
P = 2(l + w)

p/2 = l + w    divide both sides by 2

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5 0
3 years ago
1)
lisabon 2012 [21]
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3 years ago
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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
3 years ago
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