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bulgar [2K]
4 years ago
13

A function is graphed on the coordinate plane.

Mathematics
1 answer:
Eduardwww [97]4 years ago
5 0
2........................................
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The three circles are all centered at the center of the board and are of radii 1, 2, and 3, respectively.Darts landing within th
FromTheMoon [43]

Answer:

a) P=0.262

b) P=0.349

c) P=0.215

d) E(x)=12.217

e) P=0.603

Step-by-step explanation:

The question is incomplete.

Complete question: "The three circles are all centered at the center of the board (square of side 6) and are of radii 1, 2, and 3, respectively.Darts landing within the circle of radius 1 score 30 points, those landing outside this circle, butwithin the circle of radius 2, are worth 20 points, and those landing outside the circle of radius2, but within the circle of radius 3, are worth 10 points. Darts that do not land within the circleof radius 3 do not score any points. Assume that each dart that you throw will land on a point uniformly distributed in the square, find the probabilities of the accompanying events"

(a) You score 20 on a throw of the dart.

(b) You score at least 20 on a throw of a dart.

(c) You score 0 on a throw of a dart.

(d) The expected value on a throw of a dart.

(e) Both of your first two throws score at least 10.

(f) Your total score after two throws is 30.

As the probabilities are uniformly distributed within the area of the board, the probabilities are proportional to the area occupied by the segment.

(a) To score 20 in one throw, the probabilities are

P(x=20)=P(x=20\&30)-P(x=30)=\frac{\pi r_2^2}{L^2} -\frac{\pi r_1^2}{L^2}\\\\P(x=20)=\frac{\pi(r_2^2-r_1^2)}{L^2}=\frac{3.14*(2^2-1^2)}{6^2}=\frac{3.14*3}{36} =0.262

(b) To score at least 20 in one throw, the probabilities are:

P(x\geq 20)=P(x=20\&30)=\frac{\pi r_2^2}{L^2}=\frac{3.14*2^2}{6^2} =0.349

(c) To score 0 in one throw, the probabilities are:

P(x=0)=1-P(x>0)=1-\frac{\pi r_3^2}{L^2} =1-\frac{3.14*3^2}{6^2} =1-0.785=0.215

(d) Expected value

E(x)=P(0)*0+P(10)*10+P(20)*20+P(30)*30\\\\E(x)=0+\frac{\pi(r_3^2-r_2^2)}{L^2}*10+ \frac{\pi(r_2^2-r_1^2)}{L^2}*20+\frac{\pi(r_1^2)}{L^2}*30\\\\E(x)=\pi[\frac{(3^2-2^2)}{6^2}*10+\frac{(2^2-1^2)}{6^2}*20+\frac{1^2}{6^2}*30]\\\\E(x)=\pi[1.389+1.667+0.833]=3.889\pi=12.217

(e) Both of the first throws score at least 10:

P(x_1\geq 10; x_2\geq 10)=P(x\geq 10)^2=(\frac{\pi r_3^2}{L^2} )^2=(\frac{3.14*3^2}{6^2} )^2=0.785^2=0.616

(f) Your total score after two throws is 30.

This can happen as:

1- 1st score: 30, 2nd score: 0.

2- 1st score: 0, 2nd score: 30.

3- 1st score: 10, 2nd score: 20.

4- 1st score: 20, 2nd score: 10.

1 and 2 have the same probability, as do 3 and 4, so we can add them.

P(2x=30)=2*P(x_1=30;x_2=0)+2*P(x_1=20;x_2=10)\\\\P(2x=30)=2*P(x_1=30)P(x_2=0)+2*P(x_1=20)P(x_2=10)\\\\P(2x=30)=2*\frac{\pi r_1^2}{L^2}*(1-\frac{\pi r_3^2}{L^2})+2*\frac{\pi(r_2^2-r_1^2)}{L^2}*\frac{\pi(r_3^2-r_2^2)}{L^2}\\\\P(2x=30)=2*\frac{\pi*1^2}{6^2}*(1-\frac{\pi*3^2}{6^2})+2*\frac{\pi(2^2-1^2)}{6^2}*\frac{\pi(3^2-2^2)}{6^2}\\\\P(2x=30)=2*0.872*0.215+2*0.262*0.436=0.375+0.228=0.603

6 0
3 years ago
A town recreation hall needs to build a ramp. The height of the ramp must be 2 ft. The ramp will start 6 ft from the door. To th
zimovet [89]

Answer:

6.32 feets

Step-by-step explanation:

Given that:

Height of ramp = 2 feets

Length of base = 6 feets

Length of ramp = l

Using Pythagoras rule :

Hypotenus = sqrt(opposite² + adjacent²)

Length of ramp = sqrt(height of ramp² + length of base²)

l = sqrt(2² + 6²)

l = sqrt(4 + 36)

l = sqrt(40)

l = 6.3245

Length of ramp = 6.32 feets

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