Find the surface area of the cylinder leave your answer in terms of pie

Mathematics

1 answer:

Sonbull [250]3 years ago

5 0

Answer:

54π cm²

Step-by-step explanation:

It's pi, or π, but definitely not "pie."

The surface area has two parts: the vertical side and the two ends.

The formula for the area of the vertical sides is A = πr²h, where r is the radius and h is the height. Here, we have:

A = π(3 cm)²(5 cm) = 45π cm²

Each of the ends has the following area: A = πr², or A = 9π cm²

Thus, the total surface area of the cylinder is 45π cm² + 9π cm², or 54π cm²

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irina1246 [14]

Answer: The answer is (d) Accurately predicting who will be promoted at work.

Step-by-step explanation: We are given four options and asked in which of these four, we can use binomial distribution.

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In the given four options, (a), (c) and (d) are incorrect, because the number of outcomes are not fixed there and hence we cannot use binomial distribution.

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

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Step-by-step explanation:

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

Normal probability distribution

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.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$