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Tasya [4]
3 years ago
5

What is 75% of the area of a circle with a circumference of 10 units? Round the solution to the nearest square unit.

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
5 0

Answer:

75% of the area of the circle is 6 square units.

Step-by-step explanation:

Given:

Circumference of a circle = 10 units

Now we have to find the radius of the circle, then we have to find the area of the circle.

Circumference of a circle = 2*π*r            [π = 3.14]

2*3.14*r = 10

6.28r = 10

Dividing both side by 6.28, we get

r = 10/6.28

r = 1.59

Now let's find the area of the circle.

The area of the circle = π*r^2

= 3.14*1.59*1.59

The area of the circle = 7.94, which is 100 %

Now we have to find the 75% of the area of the circle.

75% = 0.75

75% of the area of the circle = 0.75*7.94

= 5.95

To round off to the nearest whole number, we get 6 square units.

Zolol [24]3 years ago
3 0
Circumference = 2 pi r = 10

r = 10 / 2 pi

75% of the error  = 0.75 * pi *  (10 / 2pi)^2

                           =  5.968 square unts

6 to nearest square unit
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In science class, Paul estimates the volume of a sample to be 37 mL. The actual volume of the sample is 39 mL. Find the percent
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3 years ago
A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two
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Answer:

the probability of no defects in 10 feet of steel = 0.1353

Step-by-step explanation:

GIven that:

A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two.

Let consider β to be the average value for defecting

So;

β = 2

Assuming Y to be the random variable which signifies the anticipated number of defects in a 10-foot segment of this roll.

Thus, y follows a poisson distribution as number of defect is infinite with the average value of β = 2

i.e

Y \sim P( \beta = 2)

the probability mass function can be represented as follows:

\mathtt{P(y) = \dfrac{e^{- \beta} \ \beta^ \ y}{y!}}

where;

y =  0,1,2,3 ...

Hence,  the probability of no defects in 10 feet of steel

y = 0

\mathtt{P(y =0) = \dfrac{e^{- 2} \ 2^ \ 0}{0!}}

\mathtt{P(y =0) = \dfrac{0.1353  \times 1}{1}}

P(y =0) = 0.1353

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