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Vika [28.1K]
3 years ago
11

Solve for n. n+4+10=25 Enter your answer in the box. n= __

Mathematics
2 answers:
-BARSIC- [3]3 years ago
8 0

Answer:

11

Step-by-step explanation:

n+4+10=25

n+14=25

n+14=25

  -14  -14

n=11

KatRina [158]3 years ago
4 0
N = 11 that is the answer
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At the 2012 Summer Olympic Games in London, in the Men's Shot Put qualifying round, the distances ranged from 17.58 meters to 21
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Answer:

One standard deviation above the mean means a z-score of 1.0. If we put this into the calculator as normalcdf(1,999) = .159

15.9 percent of the athletes would have a distance greater than 20.56 meters.

Now multiply the number of athletes by the percent.

.159 * 40 = 6.346 = about 6 athletes

c. about 6 athletes

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Alexis loves to work out on the treadmill. On Monday, she ran 1.52 miles. On Tuesday she ran 0.75 miles. On Wednesday she ran 2.
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1.08 Miles

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3 years ago
Lengths of full-term babies in the US are Normally distributed with a mean length of 20.5 inches and a standard deviation of 0.9
mash [69]

Answer:

66.48% of full-term babies are between 19 and 21 inches long at birth

Step-by-step explanation:

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.

Mean length of 20.5 inches and a standard deviation of 0.90 inches.

This means that \mu = 20.5, \sigma = 0.9

What percentage of full-term babies are between 19 and 21 inches long at birth?

The proportion is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19. Then

X = 21

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

Z = \frac{21 - 20.5}{0.9}

Z = 0.56

Z = 0.56 has a p-value of 0.7123

X = 19

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

Z = \frac{19 - 20.5}{0.9}

Z = -1.67

Z = -1.67 has a p-value of 0.0475

0.7123 - 0.0475 = 0.6648

0.6648*100% = 66.48%

66.48% of full-term babies are between 19 and 21 inches long at birth

5 0
3 years ago
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