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timama [110]
3 years ago
6

Luis solves the following system of equations by elimination. What is the value of s in the solution of the system? 11 33

Mathematics
2 answers:
-Dominant- [34]3 years ago
5 0

Answer:

Step-by-step explanation:

ok so this is what you would do

5s+3t=30

-1(2s+3t=-3)

which will become

5s+3t=30

-2s-3t=3

which turns into

3s = 33, s=11

devlian [24]3 years ago
3 0

Answer:

Its C on edegneuity

Step-by-step explanation:

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6 0
3 years ago
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
dangina [55]

Answer:

A.0.4477

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 between 14 and 19 minutes to​ grade?

This probability is the pvalue of Z when X = 19 subtracted by the pvalue of Z when X = 14. So

X = 19

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

Z = \frac{19 - 16.3}{4.2}

Z = 0.64

Z = 0.64 has a pvalue of 0.7389.

X = 14

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

Z = \frac{14 - 16.3}{4.2}

Z = -0.55

Z = -0.55 has a pvalue of 0.2912

0.7389 - 0.2912 = 0.4477

So the correct answer is:

A.0.4477

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3449995000 is what I got, I did 3.45x10^9 and then used that number and subtracted the answer from 0.5x10^4 to get the final answer, if that makes any sense.
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2 years ago
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Norma-Jean [14]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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