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cestrela7 [59]
4 years ago
6

Find the volume of a cylinder with radius 8cm and height 10 cm, correct to 2 decimal places.

Mathematics
1 answer:
serious [3.7K]4 years ago
6 0

Answer:

2010.62cm^3

Step-by-step explanation:

The formula for the area of a circle is : \pi *radius^2

To work this out you would first have to work out the area of the circle. You can do this by multiplying pi by 8 squared, this gives you 201.06. This is because in order to calculate the volume you must calculate the area of one face of the shape.

The next step is to multiply 201.06 by 10, this gives you 2010.62.

1) Multiply pi by 8 squared.

\pi*8^2=201.06

2) Multiply 201.06 by 10.

201.06*10=2010.62cm^3

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Answer:

1767 cm³

Step-by-step explanation:

We can use volume V = 4/3 π r³

We have diameter of 15 cm. To get radius, we need to divide diameter by 2.

r = 7.5 cm

So now plug in r = 7.5 then calculate V:

V = 4/3 π ( 7.5 )³

V =  4/3 π * 421.875 ≈ 1767.145 ≈ 1767

Hope this helps.

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3 years ago
Rafael counted a total of 40 white cars and yellow cars. There were 9 times as many white cars as yellow cars. How many whit car
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3 0
3 years ago
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

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

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

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

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

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

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
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