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andreev551 [17]
3 years ago
14

what is the approximate volume of a cylinder with a radius of 4 cm and a height of 5 cm drag and drop the answer into the box

Mathematics
1 answer:
liraira [26]3 years ago
8 0
The formula for volume of a cylinder is  V= <span>π</span>r2h. The two means to square it, not multiply by two. So you would have 3.14 x 16 x 5. The sixteen in the equation comes from squaring 4. The answer is 251.2.
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Convert 2 Liters to gallons. Use the fact that 1 gal.= 3.8L.
iris [78.8K]

we know that

1 gal.= 3.8L

we can also write as

3.8L=1 gal

Firstly, we will find for 1L

so, we can divide both sides by 3.8

\frac{3.8}{3.8} L=\frac{1}{3.8}gal

1 L=\frac{1}{3.8}gal

now, we will multiply both sides by 22

22*1 L=22*\frac{1}{3.8}gal

22L=5.78947 gal

22L=5.8 gal

so, option-b.............Answer

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Find FGH~ HOP. Find the area of X​
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First, you put the triangles in a ratio
Then, you ask yourself "what do I have to divide seven by to get 5" so you do 7.5/5 to get that answer (which is to divided by 1.5)
Then, on the other side, you divide 22.5 by 1.5 as well

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3 years ago
The perimeter of a rectangle is 50 cm and the length is 4 cm longer than the width. Write and solve the system of equations usin
laila [671]
I hope this helps you

3 0
3 years ago
Choose the best description of the following number 18
Mrac [35]

Answer:

4.243

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5 0
3 years ago
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
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