Three equals some number plus five.

A = kr² choices: direct inverse joint combined

Lina20 [59]

Answer:

Direct.

Step-by-step explanation:

A varies directly as the square of r.

6 0

2 years ago

Read 2 more answers

A professor gives a statistics exam. The exam has 100 possible points. The scores for the students in the second classroom are a

Vesna [10]

Answer:

N = 9

M = 87.11

Step-by-step explanation:

According to the situation, the data provided and the solution are as follows

The scores for the student in the classroom is

= 88 88 92 88 88 72 96 88 84

The different student's number of scores is 9

The solution of new n and M is shown below:-

Sample size (n) = 9

Sample mean (m) is

$= \frac{Sum\ of\ all\ data\ values}{Sample\ size}$

= $\frac{88 + 88 + 92 + 88 + 88 + 72 + 96 + 88 + 84}{9}$

$= \frac{784}{9}$

= 87.11

3 0

2 years ago

An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you th

aliya0001 [1]

Answer:

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .