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ss7ja [257]
3 years ago
11

A book publisher counted the words on each page of a series of books. They found that the number of words per page followed a no

rmal distribution with a mean of 265 words and a standard deviation of 25 words.
What percentage of the pages contain fewer than 215 words?
Mathematics
2 answers:
vagabundo [1.1K]3 years ago
7 0

Answer:

The answer is 2.5%

Step-by-step explanation:

Took test on edmentum and made 100%.

galben [10]3 years ago
5 0

Answer:

b. 2.5%

Step-by-step explanation:

Plato

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15 points !<br> Find m &lt; RQP
Phoenix [80]

Answer:

<RQP = 70°

Step-by-step explanation:

<RQP = 1/2 (140)

<RQP = 70°

4 0
2 years ago
Rewrite 6√2 in the simplest form
mylen [45]
6<span>√2 is already in it's simplest form as a radical.


</span>
5 0
3 years ago
In a recently administered iq test, the scores were distributed normally, with mean 100 and standard deviation 15. what proporti
iragen [17]

To solve for proportion we make use of the z statistic. The procedure is to solve for the value of the z score and then locate for the proportion using the standard distribution tables. The formula for z score is:

z = (X – μ) / σ

where X is the sample value, μ is the mean value and σ is the standard deviation

 

when X = 70

z1 = (70 – 100) / 15 = -2

Using the standard distribution tables, proportion is P1 = 0.0228

 

when X = 130

z2 = (130 – 100) /15 = 2

Using the standard distribution tables, proportion is P2 = 0.9772

 

Therefore the proportion between X of 70 and 130 is:

P (70<X<130) = P2 – P1

P (70<X<130) = 0.9772 - 0.0228

P (70<X<130) = 0.9544

 

Therefore 0.9544 or 95.44% of the test takers scored between 70 and 130.

6 0
3 years ago
At a Psychology final exam, the scores are normally distributed with a mean 73 points and a standard deviation of 10.6 points. T
USPshnik [31]

Answer:

The score that separates the lower 5% of the class from the rest of the class is 55.6.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this question:

\mu = 73, \sigma = 10.6

Find the score that separates the lower 5% of the class from the rest of the class.

This score is the 5th percentile, which is X when Z has a pvalue of 0.05. So it is X when Z = -1.645.

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

-1.645 = \frac{X - 73}{10.6}

X - 73 = -1.645*10.6

X = 55.6

The score that separates the lower 5% of the class from the rest of the class is 55.6.

3 0
3 years ago
What is the LCM 18,10,12? yeah
Karo-lina-s [1.5K]

Answer:

The LCM of 10, 12, and 18 is 180.

HOPE THIS HELPED!!!!!!!!!!!!XDDDDDDDD

4 0
2 years ago
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