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erastova [34]
3 years ago
8

I think I got this wrong I need help

Mathematics
1 answer:
JulijaS [17]3 years ago
8 0

Answer:

I think 3/4 will be the answer if you let

(-5 2) =(x1 y1)

(3 8) =( x2 y2)

y1-y2/x2-x2

8-2/3+5

6/8

3/4

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ANSWEEER FAST PLEASE PLEASEEE
IrinaVladis [17]

Answer:

E

Step-by-step explanation:

A Slope of 5/2 means you have to go up 5 and over 2 every time a point is on the line.

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3 years ago
Which values are solutions to the inequality below? √x > 25?
dezoksy [38]

Answer:

C and D

Step-by-step explanation:

So basically when u square -5 or +5, u get 25 either way

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3 years ago
Select Equal or Not Equal to correctly classify each statement.
emmainna [20.7K]
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kkurt [141]
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Problem written out
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3 years ago
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The time for a professor to grade an exam is normally distributed with a mean of 16.3 minutes and a standard deviation of 4.2 mi
dem82 [27]

Answer:

62.17% probability that a randomly selected exam will require more than 15 minutes to grade

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 16.3, \sigma = 4.2

What is the probability that a randomly selected exam will require more than 15 minutes to grade

This is 1 subtracted by the pvalue of Z when X = 15. So

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

Z = \frac{15 - 16.3}{4.2}

Z = -0.31

Z = -0.31 has a pvalue of 0.3783.

1 - 0.3783 = 0.6217

62.17% probability that a randomly selected exam will require more than 15 minutes to grade

8 0
3 years ago
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