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ioda
3 years ago
5

How to find average.

Mathematics
2 answers:
larisa [96]3 years ago
6 0
Add up all of the numbers and then divide by the amount of numbers you have, basically the sum divided by the amount. :)
Ray Of Light [21]3 years ago
3 0

Answer:

Add up all the numbers then divided by how many numbers there is and the you'll get your answer.

Example:

70+80+75+85+90=400

400 divided by 5=80

So 80 is my answer

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4. Six numbers are shown.
Reptile [31]

Step-by-step explanation:

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7 0
3 years ago
40,-20,10,-5,__,__,__<br> What are the next three numbers.
valkas [14]

Answer:

2.5, -1.25, 0.625

Step-by-step explanation:

4 0
3 years ago
Which statements are true of the graph of h(x) = ^3 square root of x-4 ? Check all that apply.
kolezko [41]

Answer:

A,B,E

Step-by-step explanation:


4 0
3 years ago
Read 2 more answers
A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 33.9 wee
MArishka [77]

Answer:

P(35.3 < M < 35.4) = 0.0040.

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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 33.9, \sigma = 6.7, n = 119, s = \frac{6.7}{\sqrt{119}} = 0.6142

Find the probability that a single randomly selected value is between 35.3 and 35.4

This is the pvalue of Z when X = 35.4 subtracted by the pvalue of Z when X = 35.3. So

X = 35.4

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{35.4 - 33.9}{0.6142}

Z = 2.44

Z = 2.44 has a pvalue of 0.9927

X = 35.3

Z = \frac{X - \mu}{s}

Z = \frac{35.3 - 33.9}{0.6142}

Z = 2.28

Z = 2.28 has a pvalue of 0.9887

0.9927 - 0.9887 = 0.0040

So the answer is:

P(35.3 < M < 35.4) = 0.0040.

4 0
3 years ago
20/23 as a decimal rounded to the nearest tenth
AlladinOne [14]

Answer:

0.9

Step-by-step explanation:

20/23 = 0.869565217, which rounded to the nearest tenth is 0.9

6 0
2 years ago
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