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Yuliya22 [10]
3 years ago
11

Find CD? Plssss help

Mathematics
2 answers:
matrenka [14]3 years ago
4 0

Answer:

CD = 10

Step-by-step explanation:

tresset_1 [31]3 years ago
3 0

Answer:

CD=10 is the answer to your question

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If a line has 4 points how many distinct line segments can be formed using the set of points
Harrizon [31]

Answer: The will be 5 distinct lines



Step-by-step explanation:


7 0
3 years ago
The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds
Feliz [49]

Answer:

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.

Step-by-step explanation:

When the distribution is normal, we use 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, we have that:

\mu = 200, \sigma = 50

Find the probability that he weighs between 170 and 220 pounds.

This is the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 170.

X = 220

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

Z = \frac{220 - 200}{50}

Z = 0.4

Z = 0.4 has a pvalue of 0.6554

X = 170

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

Z = \frac{170 - 200}{50}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.6554 - 0.2743 = 0.3811

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.

6 0
4 years ago
Choose the answer that shows exponential decay?*
Vlad1618 [11]
I’m not sure what that is
6 0
3 years ago
Help, please! sorry I have so many of these
Mademuasel [1]
Median = 70.

If you arrange all the numbers in ascending order, 70 will be the middle number.
7 0
3 years ago
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