Please Help Me With This Please!

The state of mishgan recored an all time high snowball of 355.9 inches during the winter of 1978-79 its lowest snowfall of 81.3

Rama09 [41]

Answer:

<u>274.6 inches</u> more snowfall Michigan received during the winter of 1978-79 than in 1930-31.

Step-by-step explanation:

Given:

The state of Michigan recorded an all time high snowfall of 355.9 inches during the winter of 1978-79.

Its lowest snowfall of 81.3 inches was recorded during the winter of 1930-31.

Now, to find how much more snowfall did Michigan received during the winter of 1978-79 than in 1930-31.

The snowfall in 1978-79 = 355.9 inches.

And, the snowfall in 1930-31 = 81.3 inches.

So, to get the how much more snowfall in 1978-79 than 1930-31 we subtract:

Snowfall in 1978-79 - Snowfall in 1930-31.

$355.9\ inches-81.3\ inches = 274.6\ inches.$

Therefore, 274.6 inches more snowfall Michigan received during the winter of 1978-79 than in 1930-31.

8 0

3 years ago

In a class with 28 students, there are 19 boys. What is the ratio of girls to boys? 9:19 19:28 19:9 28:9

leva [86]

Answer:

9:19

Step-by-step explanation:

Total number of students = 28

Number of boys = 19

Number of girls = Total number of students - Number of boys = 28-19 = 9

Therefore, ratio of girls to boys = Number of girls/Number of boys = 9/19 = 9:19

5 0

3 years ago

Read 2 more answers

You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard devia

Arturiano [62]

Answer:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (3)

$n=(\frac{2.58(54.5)}{5})^2 =790.846 \approx 791$

So the answer for this case would be n=791 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$