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Yanka [14]
3 years ago
5

You roll a die 4 times. What is the probability of rolling a 3 on all 4 times?

Mathematics
1 answer:
hram777 [196]3 years ago
4 0

Answer:

to find a probability,we need to follow the rule

P/n

p=1

n=6

=1/6

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Answer:

8 dollars

Step-by-step explanation:

20/100 = x/10

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x=2

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4 years ago
Read 2 more answers
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
Eric’s bicycle tires have a 29-inch diameter and Maggie’s bicycle tires have a 26-inch diameter. How much farther do Eric’s tire
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Answer:

94.2 inches, or 7.85 feet

Step-by-step explanation:

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7 0
3 years ago
Which of these is a solution to the equation 2( x – 5) = 9 – 3 x + 6 + 8 + 3 x + 7? x = 80 x = 20 x = 40 x = 10
bija089 [108]

The solution to the equation is x = 20.

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

The equation is 2( x – 5) = 9 – 3 x + 6 + 8 + 3 x + 7

2x - 10 = 30

2x = 30+10

2x = 40

x = 40/2

x = 20

5 0
3 years ago
The local zoo had two water tanks that are leaking. One tank contains 12 gallons of water and is leaking at a constant rate of 3
fiasKO [112]
Once they're empty, I guess.
Did you put it the wrong way around?
Because it would be after four hours in this case.
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3 years ago
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