The dimensions of a rectangular prism are 1 mm, 6 mm, and 11 mm. Find the volume and the surface area.

1 answer:

8 0

The volume is the product of the side lengths which is 1*6*11= 66 mm^3. The surface area is twice the sum of all the possible products of the side lengths, or 2lw+2wh+2hl which is 2(6*11+6*1+11*1) = 166 mm^2 if my mental math is decent.

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denis23 [38]

Answer:

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Step-by-step explanation:

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55 ÷ 5

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Find constants A and B such that (x + 7)/(x^2 - x - 2) = A/(x - 2) + B/(x + 1) for all x such that x≠ -1 and x≠ 2 Give your answ

Yanka [14]

For this case we have the following expression:
(x + 7) / (x ^ 2 - x - 2) = A / (x - 2) + B / (x + 1)
Rewriting we have:
(x + 7) / ((x-2) (x + 1)) = A / (x - 2) + B / (x + 1)
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A = 3
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mr_godi [17]

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A population has a mean of 180 and a standard deviation of 24. A sample of 100 observations will be taken. The probability that

jeka94

Answer:

The probability that the mean from that sample will be between 183 and 186 is 0.0994 = 9.94%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

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

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .