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olganol [36]
3 years ago
7

I NEED HELP ASAP IM TIMED... picture connected!

Mathematics
1 answer:
cluponka [151]3 years ago
5 0

Answer:

d

Step-by-step explanation:

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D

Step-by-step explanation:

15 + 2x + x + 3 = 3x + 18

hope this helps

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2 years ago
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It’s timed can sb please help me
alex41 [277]

Answer:

D

Step-by-step explanation:

The intersection of the graphs is at (-3,4).

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3 years ago
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patriot [66]

Answer:

1. x = -4

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2 years ago
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The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 1000 lbs
ElenaW [278]

Answer:

0.9466 = 94.66% probability that the weight of a randomly selected steer is between 639 and 1420lbs.

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 = 1000, \sigma = 200

Find the probability that the weight of a randomly selected steer is between 639 and 1420lbs.

This is the pvalue of Z when X = 1420 subtracted by the pvalue of Z when X = 639. So

X = 1420

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

Z = \frac{1420 - 1000}{200}

Z = 2.1

Z = 2.1 has a pvalue of 0.9821

X = 639

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

Z = \frac{639 - 1000}{200}

Z = -1.805

Z = -1.805 has a pvalue of 0.0355

0.9821 - 0.0355 = 0.9466

0.9466 = 94.66% probability that the weight of a randomly selected steer is between 639 and 1420lbs.

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3 years ago
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