5 4 3 2 1 0 -2 -1 0 1 5 -2 b = m = Inequality =

What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.

Paul [167]

Answer:

1.60 $ft^{2}$

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = $\pi$ $r^{2}$

So that,

for the inner part of the pipe,

r = $\frac{diameter}{2}$

= $\frac{5}{2}$

= 2.5 feet

Area of the inner part of the pipe = $\pi$ $r^{2}$

= 3.14 x $(2.5)^{2}$

= 19.625

Are of the inner part of the pipe is 19.63 $ft^{2}$ .

Total diameter of the pipe = 5 + 0.2

= 5.2 feet

r = $\frac{5.2}{2}$

= 2.6 feet

Area of the pipe = $\pi$ $r^{2}$

= 3.14 x $(2.6)^{2}$

= 21.2264

Area of the pipe is 21.23 $ft^{2}$ .

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

= 21.2264 - 19.625

= 1.6014

The cross sectional area of the cement pipe is 1.60 $ft^{2}$ .

6 0

3 years ago

The price of an item yesterday was $120. Today, the price fell to $78

marusya05 [52]

Answer:

The answer is decreasing the money

8 0

2 years ago

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The volume of a shipping container is 448.656 cubic inches, and it can hold 24 smaller boxes. What's the volume of each of the s

madreJ [45]

Answer:

The volume of each of the smaller boxes = 18.694 inches³

Step-by-step explanation:

The volume of Shipping container = 448.656 inches³

The shipping container hold small boxes of numbers = 24

So, The volume of each of small boxes = $\frac{448.656}{24}$ inches³

Or, The volume of each of small boxes = 18.694 inches³

Hence The volume of each of 24 small containers is 18.694 inches³ Answer

5 0

4 years ago

There are 190 students taking band, chorus, or both. If there are 180 students taking band and 60 students in both band and chor

Fynjy0 [20]

The answer would be 10. I know this because I got it right on my assignment. Have a great day!

6 0

4 years ago

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Let a "binary code" be the set of all binary words, each consisting of 7 bits (i.e., 0 or 1 digits). For example, 0110110 is a c

Dmitry_Shevchenko [17]

Answer:

a) 128 codewords

b) 35 codewords

c) 29 codewords

Step-by-step explanation:

a) Each 7 bits consist of 0 or 1 digits. Therefore the first bit is two choices (0 or 1), the second bit is also two choices (0 or 1), continues this way till the last bit.

So total number of different code words in 7 bits is 2×2×2×2×2×2×2 = 2⁷ = 128

There are 128 different codewords.

b) A code word contains exactly four 1's this means that it has four 1's and three 0's . Therefore, in 7 bits, we have four of the same kind and three of the same kind. Hence, total number of code words containing exactly four 1's =7!/(4!*3!) = 35 codewords

c) number of code words containing at most two 1's = codewords containing zero 1's + words containing one 1's + words containing two 1's

Now codewords containing zero 1's = 0000000 so 1 word

Codewords containing one 1's = 1000000,0100000,0010000,0001000,0000100,0000010,0000001. That's seven words

Codewords containing two 1's means word containing two 1's and five 0's. So out of seven, two are of one kind and five are of another kind

Therefore, the total number of such words=7!/(2!*5!)=21

Hence, codewords having at most two 1's = 21+7+1 =29 codewords

8 0

3 years ago

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