If

How do i turn 2/5 into a percentage

likoan [24]

Numerator divided by the denominator then multiplied by 100.

(2/5) * 100

7 0

3 years ago

Combine like terms.

miss Akunina [59]

The answer is -13z because there is no reasons to square it, it cannot be positive, and there has to be a z in the answer.

8 0

3 years ago

6:30-33D<br> Points are getting up .

Klio2033 [76]

What is the question pls I don’t understand

7 0

3 years ago

Does this equation have:

Leni [432]

Answer:

C

Step-by-step explanation:

$\sqrt{3x} + 8 = x - 2$

4 0

2 years ago

All 423 Wisconsin public schools were all given a rating by the Wisconsin Department ofPublic Instruction based on several varia

Blizzard [7]

Answer:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779$

From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.

Step-by-step explanation:

For this case from all the population we know that the population mean and deviation are:

$\mu = 71.5,\sigma = 4.87$

And we take a random sample of size n =40 and we got a sample mean calculated with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}= \frac{\sum_{i=1}^{40} X_i}{40}=70.9$

And we want to test if this value is unusually low.

Since the sample size is large n>30 we can use the central limit theorem who says that the distribution for the sample mean is given by:

$\bar X \sim N (\mu , \frac{\sigma}{\sqrt{n}})$

And on this case if we replace the values that we have we got:

$\bar X \sim N (\mu_{\bar X}=71.5,\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{4.87}{\sqrt{40}}=0.77)$

For this case we can calculate how many deviations above or below is our calculated value from the sample of size 40, using the z score given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779$

From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.

3 0

3 years ago