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Nadya [2.5K]
3 years ago
10

Josh likes to play an outdoor beanbag toss game in his spare time. On any toss, he has about a 30% chance of getting a bag into

the hole. As a challenge one day, he decides to keep tossing beanbags until he makes it. Starting on line 5 of the random number table, how many times does Josh have to toss the beanbag in order to make it in the hole if we let success = 0–2 and failure = 3–9?
Mathematics
1 answer:
MArishka [77]3 years ago
5 0

Answer:

B

Step-by-step explanation:

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Answer:

I believe he slept for 9 hours of the day

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3 years ago
A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
VMariaS [17]

Answer:

Q = arctan(7.1739) = 82.06

Step-by-step explanation:

Given:

- The mass of the person m = 20.3 kg

- The distance traveled up the ladder s = 1.1 m

- The gravitational constant g = 9.8 m/s^2

- The coefficient of static friction u_s = 0.23

- Total length of the ladder

Find:

The minimum angle θ, that would allow the person to climb without ladder slipping

Solution:

- Taking moments about point of ladder and wall contact A to be zero:

                   -F_n,b*cos(Q)*4 - F_f*sin(Q)*4+ m*g*cos(Q)*2.6 = 0

- Taking Sum of vertical forces to be zero:

                    F_n,b - m*g = 0

                    F_n,b = m*g

- The frictional force F_f is given by:

                   F_f = u_s*F_n,b = u_s*m*g

- Plug the values back in:

                  - m*g*cos(Q)*4 + u_s*m*g*sin(Q)*4 - m*g*cos(Q)*2.6 = 0

Simplify:

                  4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)

                        6.6*cos(Q) = 4*u_s*sin(Q)

                               tan(Q) = 6.6 / 4*u_s

- Plug in the values:

                               tan(Q) = 6.6 / 4*0.23

                                    Q = arctan(7.1739) = 82.06

                   

                     

4 0
3 years ago
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