I received a letter from your competitor that mentions a savings of 15% over your prices. Can you match this offer? My current r

The tension does not approach infinity.
<span>Let's analyze free body diagrams (FBDs) for each mass, considering the direction of motion of m₁ as positive.

For m₁: m₁*g - T = m₁*a

For m₂: T - m₂*g = m₂*a

Assuming a massless cord and pulley without friction, the accelerations are the same.

From the second equation: a = (T - m₂*g) / m₂

Substitute into the first:
m₁*g - T = m₁ * [(T - m₂*g) / m₂]
Rearranging:
m₁*g - T = (m₁*T)/m₂ - m₁*g
2*m₁*g = T * (1 + m₁/m₂)
2*m₁*m₂*g = T * (m₂ + m₁)
T = (2*m₁*m₂*g) / (m₂ + m₁)
Taking the limit as m₁ approaches infinity:
T = 2*m₂*g

This aligns with intuition since the greatest acceleration m₁ can have is -g. The cord then accelerates m₂ upward at g while gravity acts downward, leading to a maximum upward acceleration of 2*g for m₁.</span>

D. 20.64% or C. 32.37.

The range consists of all the valid y values, starting from 5.

Answer:

a) 2.60x10^-7

b) 5.31x10^-5

c) 2.19x10^-3

Step-by-step explanation:

X=the number of hits

Probability can be determined by taking the number of successful outcomes and dividing it by the total number of possible outcomes.

Now

a) P(X=6)=P({1, 1, 1, 1, 1, 1})=6/40*5/39*4/38*3/37*2/36*1/35=2.60x10^-7

b) P(X=5)=P({0, 1, 1, 1, 1, 1})+P({1, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 1, 0}), and each scenario carries the same probability

P(X=5)=6P({1, 1, 1, 1, 1, 0})=6*(6/40*5/39*4/38*3/37*2/36*34/35)=5.31x10^-5

c) P(X=4)=P({0, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 0, 0}), where each case has an identical probability.

We must determine the value of x