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Komok [63]
4 years ago
7

What is the mesure of fsm

Mathematics
1 answer:
krok68 [10]4 years ago
4 0

Answer:

The Functional Strength Measurement (FSM) is a norm‐referenced test for functional strength. It consists of eight items including muscle power (overarm and underarm throwing, standing long jump, chest pass) and muscle endurance (lateral step‐up, sit to stand, lifting a box and stair climbing.) ;.; next time put a question with fsm...

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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
If a/2 = b/3, then b/a = ?<br><br> A. 3/2<br><br> B. 2/3
Paul [167]
It would be A because b+a
4 0
3 years ago
(Answer with Explanation. I will mark BRAINLIEST! 100 pts.)
katovenus [111]

Answer:

lemme answer it,I'll put the answer at the comment section

8 0
3 years ago
4.
tresset_1 [31]

Answer:

The answer is C. 28.8 in

Step-by-step explanation:

Given a=16 and b=24,

c = 28.84441 = 8√13

∠α = 33.69° = 33°41'24" = 0.588 rad

∠β = 56.31° = 56°18'36" = 0.98279 rad

h = 13.3128

area = 192

perimeter = 68.84441

inradius = 5.57779

circumradius = 14.42221 = 4√13

5 0
3 years ago
Cabrina and Dabney are attending a conference. After the​ conference, Cabrina drives home to Boise at an average speed of 7575 m
kifflom [539]

Answer:

Cabrina's time = 5 hours  

Dabney's time = 6.5 hours

Step-by-step explanation:

The data of the exercise are:

Cabrina speed = 75 mph

Daney speed = 60 mph

The sum of their distance is 765 miles, therefore

Cabrina's distance Dabney's distance = 765

Cabrina's distance = x

Dabney's distance = 765-x

The sum of their times is 11.5 hours, therefore

Cabrina's time Dabneys time = 11.5

Cabrina's time = x / 75

Dabney's time = (765-x) / 60

x / 75 (765-x) / 60 = 11.5

Lowest common denominator of 75 and 60 is 300, therefore we have:

330 * x / 75 300 * (765-x) / 60 = 300 * 11.5

4 * x 5 * (- x 765) = 3450

4 * x - 5 * x 3825 = 3450

x = 375

Replacing:

Cabrina's time = 375/75 = 5 hours took

Dabney's time = (765-375) / 60 = 6.5 hours took

7 0
3 years ago
Read 2 more answers
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