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Masteriza [31]
4 years ago
12

Find the perimeter of the figure to the nearest hundredth

Mathematics
1 answer:
Ghella [55]4 years ago
3 0

Answer: 33.13 in

Step-by-step explanation:

First, let's start with the perimeter of the semicircles.

Since there are two, it is easier to do because the perimeter of the two halves will add to one full circle; so we can just calculate without having to halve anything.

The formula for the perimeter (aka circumference) of a circle is 2πr

Plug the radius into the equation to get 2π4, or 8π

8π ≈ 25.13

The perimeter of the rectangle is 6(4) = 24.

We now have to subtract the diameters of the circles, as they are not on the outside of the figure and won't be counted in the total perimeter. From the 24, we subtract the two diameters, or 16 in.

24 - 16 = 8

The total perimeter then comes out to be 25.13 + 8, or 33.13 in

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Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t
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(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

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The sample space for the event of rolling two dice is presented below

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6)  =\dfrac{20}{36}

The probability distribution of this event is given as follows.

\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|

First, we determine the expected Value of this event.

Expected Value

=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning  =\dfrac{20}{36}

If the game is played 100 times

Number of times expected to win

=\dfrac{20}{36} \times 100\\=56$ times

Therefore, number of times expected to loose

= 100-56

=44 times

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