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Setler79 [48]
3 years ago
7

The diameter of a circle is 16 kilometers. Find the area of the circle.

Mathematics
1 answer:
Elden [556K]3 years ago
4 0

The diameter of a circle is 16 kilometers. Then the area of the circle is 201.062 \mathrm{km}^{2}

<u>Solution:</u>

Given that , the diameter of a circle is 16 kilometers  

We have to find the area of the circle .

Radius is half of diameter.

\begin{array}{l}{\text { diameter }=2 \times \text { radius } \rightarrow \text { radius }=\frac{\text {diameter}}{2}} \\\\ {\rightarrow \text { radius }=\frac{16}{2} \rightarrow \text { radius }=8 \text { kilometers. }}\end{array}

<em><u>The area of circle is given as:</u></em>

\begin{array}{l}{\text { area of a circle }=\pi \times r^2 \\ {\text { Area of a circle }=\pi \times 8^{2}=\pi \times 64=64 \times 3.14=201.062}\end{array}

Hence, the area of the circle is 201.062 km^2

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6 0
3 years ago
If a sample of 58 runners were taken from a population of 302 runners, could refer to the mean of how many runners' times?
IceJOKER [234]
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4 0
3 years ago
Read 2 more answers
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
3 years ago
What is the extraneous solution?
8_murik_8 [283]
<h3>Answer:</h3>

  x = 1

<h3>Step-by-step explanation:</h3>

<em>The only solution is an extraneous solution</em>, which is to say the equation has no solution.

The rational expression reduces to -1 (for x≠1), which makes the equation ...

  1 = 1/x

The only solution to this is x=1, which is specifically disallowed.

___

If you subtract the right side, the equation becomes ...

  ((-x+1)(x) +2(x)(x -1) -(x -1))/(x -1) = 0

  (-x^2 +x +2x^2 -2x -x +1)/(x -1) = 0 . . . . eliminate parentheses in the numerator

  (x^2 -2x +1)/(x -1) = 0 . . . . . . . . . . . . . . . collect terms

  (x -1)^2/(x -1) = 0 . . . . . . . . . . . . . . . . . . . factor

This is undefined for the only value of x that could possibly be a solution, x=1.

8 0
3 years ago
What is tan angle QTR if line PR is tangent to circle T at Q, and TS = 1 cm?
lions [1.4K]
Tangent = opposite / adjacent

In the diagram:
TQ is the short leg or the opposite
QR is the long leg or the adjacent
TR is the hypotenuse. TS is part of the hypotenuse.

Tan(QTR) = TQ / QR

TS = 1 ; TQ = TS  so, TQ = 1

Tan(QTR) = 1/QR

I think the answer is C.RQ
3 0
3 years ago
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