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sertanlavr [38]
3 years ago
14

Can someone help Find the surface area

Mathematics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

1861.39 cm²

Step-by-step explanation:

Given:

d = 15 cm

radius (r) = ½(15) = 7.5 cm

Height = 32 cm

Surface area of a cylinder = 2πrh + 2πr²

Plug in the values

Surface area = 2*π*7.5*32 + 2*π*7.5²

= 1861.39 cm²

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-1x + 5 = 10 solution please
FinnZ [79.3K]

Answer:i think its -5 because if you subtract 5 from 10 then its five hnd a negative times a negative is a positive

Step-by-step explanation:

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The animal department wants to estimate baboon population. So they paint 50 baboons with a mark. These baboons were then release
Svet_ta [14]

Answer:

  600 baboons

Step-by-step explanation:

Half the number of marked baboons showed up in the sample, so we expect the sample size to be about half of the total population. 300 is half of 600.

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Many assumptions are involved in such an experiment and analysis. Not the least of which are the assumptions that the marked baboons end up evenly distributed through the population, and that the sample is a fair sample of the entire population.

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2 years ago
PLZZZZ HELLPPP MEEE GET ANSWER
nadya68 [22]

Answer:

<em>See above photograph</em>

Step-by-step explanation:

This is how your graph will look like with an x-intercept of [−5, 0] (will not see, according to the scale, but on my device, you can), and a y-intercept of [0, 50]. The way to figure out all the coordinates upon this scale is to simply create a y-x value chart. Plug in each input value [x-value] to get your output value [y-value]:

<u>x|y</u>

0|50

1|60

2|70

3|80

4|90

5|100

6|110

7|120

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I am joyous to assist you anytime.

4 0
3 years ago
What is 2 5/7 as a decimal?
julsineya [31]
25.7 because that is the answer
8 0
3 years ago
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The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6.
Wittaler [7]

Answer:

a) There is a 10.75% probability of observing less than 60 hits in an hour.

b) The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) There is a 24% probability of observing between 80 and 90 hits an hour

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem, we have that

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so \mu = 75, \sigma = 8.6.

a) What’s the probability of observing less than 60 hits in an hour? Use the normal approximation

This is the pvalue of Z when X = 60. So

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

Z = \frac{60 - 75}{8.6}

Z = -1.74

Z = -1.74 has a pvalue of 0.1075. This means that there is a 10.75% probability of observing less than 60 hits in an hour.

b) What’s the 99th percentile of the distribution of the number of hits?

What is the value of X when Z has a pvalue of 0.99.

Z = 2.35 has a pvalue of 0.99

So

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

2.35 = \frac{X - 75}{8.6}

X - 75 = 20.21

X = 95.21

The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) What’s the probability of observing between 80 and 90 hits an hour?

This is the pvalue of the zscore of X = 90 subtracted by the pvalue of the zscore of X = 80.

For X = 90

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

Z = \frac{90 - 75}{8.6}

Z = 1.74

Z = 1.74 has a pvalue of 0.95907

For X = 80

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

Z = \frac{80 - 75}{8.6}

Z = 0.58

Z = 0.58 has a pvalue of 0.71904

So

There is a 0.95907 - 0.71904 = 0.24003 = 24% probability of observing between 80 and 90 hits an hour

6 0
3 years ago
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