4000 is 1/10 of what number

-202.5÷ 13.5<br><br> I need the work shown, please.

s344n2d4d5 [400]

Please find attached photograph for your answer.

5 0

2 years ago

Line/ has a ___ slope

Masja [62]

A line that would appear like this ( / ) on a graph would have a positive slope because the coordinates that the line passes through are increasing from left to right.

3 0

3 years ago

Solve 2x+2>10 i need to know what this is?

mrs_skeptik [129]

Answer:

x>4

Step-by-step explanation:

$2x + 2 > 10 \\ 2x > 10 - 2 \\ 2x > 8 \\ x > 4$

Sorry if there is any mistakes

6 0

3 years ago

The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean

Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .