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3 years ago

A drainage pipe 66 in. tall measures 25.12 in. around.

1 answer:

8 0

In order to solve this using V= B(h) (The B is area of the base) you need a radius it doesn't give us that but it does give us the circumference.
C= 3.14(2r)
25.12=2r divide each by 2
12.56=r
now use this in the volume formula:
V= 12.56^2(3.14)(66)
V=32,692.86 cubic inches

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Find the magnitude of WX for W(-2, 8, -3) and X(1, 4, -1)

zheka24 [161]

The magnitude of WX is $\sqrt{29}$ ⇒ c

Step-by-step explanation:

The magnitude of XY (its length) where,

$X=(x_{1},y_{1},z_{1})$
$Y=(x_{2},y_{2},z_{2})$

is IXYI = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

∵ W = (-2 , 8 , -3)

∵ X = (1 , 4 , -1)

- To find the magnitude of WX use the formula above

∴ $x_{1}$ = -2 and $x_{2}$ = 1

∴ $y_{1}$ = 8 and $y_{2}$ = 4

∴ $z_{1}$ = -3 and $z_{2]$ = -1

- Substitute these values in the rule above

∵ IWXI = $\sqrt{(1--2)^{2}+(4-8)^{2}+(-1--3)^{2}}=\sqrt{9+16+4}$

∴ IWXI = $\sqrt{29}$

The magnitude of WX is $\sqrt{29}$

Learn more:

You can learn more about the position of an object in brainly.com/question/10940255

#LearnwithBrainly

8 0

3 years ago

Read 2 more answers

1. Solve: 6x - 12 < 6<br> 2. Solve: 2(3x-8) >8

8090 [49]

Answer:

Step-by-step explanation:

1. here you go mate

step 1

6x - 12 < 6 equation

step 2

6x - 12 < 6 simplify

6x-12+12<6+12

step 3

6x<18 Divide both sides

$\frac{6x}{6}$

answer

x<3

2. Here you go mate

step 1

2(3x-8) >8 equation

step 2

2(3x-8) >8 simplify

6x-16+16>8+16

step 3

6x>24 divide both sides

$\frac{6x}{6} >\frac{24}{6}$

answer

x>4

4 0

3 years ago

I don't understand this quadratic functions question please help.

djyliett [7]

I think it would be b

6 0

2 years ago

What's is the absolute value -18 degrees

dybincka [34]

The absolute value would be 18 !

5 0

2 years ago

Read 2 more answers

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

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

$Z = \frac{5.8-6}{0.2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

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

$Z = \frac{5.8-6}{0.2}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

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

$Z = \frac{5-6}{0.2}$

$Z = -5$

$Z = -5$ has a pvalue of 0.

1 - 0 = 1

5 0

3 years ago