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OverLord2011 [107]
2 years ago
15

Evaluate (c − b)2 + a2  for a = –5, b = –2, and c = –4.

Mathematics
1 answer:
frosja888 [35]2 years ago
7 0

Answer:

29

Step-by-step explanation:

(c − b)^2 + a^2

Let a = –5, b = –2, and c = –4.

(-4--2)^2 + (-5)^2

(-4+2)^2 + (-5)^2

(-2)^2+ (-5)^2

4 +25

29

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Answer:

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Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

\mu = 0 - 5 = -5

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

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

Z = \frac{1 - (-5)}{\sqrt{13}}

Z = \frac{6}{\sqrt{13}}

Z = 1.66

Z = 1.66 has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

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