All categories

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

You might be interested in

If the sum of interior angle is 3780 degree.find the number of sides

timurjin [86]

The sum of the interior angles of a polygon is.

Sum = 180(n-2) where n is the number of sides

Thus 180(n-2) = 3780

n-2=3780/180 =21

⇒ n = 21+2 = 23

number of sides is 23

5 0

3 years ago

Read 2 more answers

"if a 40-pound weight stretches the spring 4 inches, ,find the distance that a 45-pound weight stretchs the spring"

shepuryov [24]

40=4x then 45=4.5x solving "x" as 1 in per 10 pounds.

3 0

3 years ago

Please hurry I will give brainless answer if correct

Natali [406]

Answer:

we have to be able to read it first tho. we can't see what the problem is so we need to see it please and thank you...

7 0

2 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago

Subtract 5 and 3 1/3

Ostrovityanka [42]

Answer:

-2 2/3

Step-by-step explanation:

I am pretty sure

5 0

3 years ago

Read 2 more answers