Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $9,500 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $9,500 and $14,700. Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?

Probability (bid accepted) = 0.48

Step-by-step explanation:

Probability density is given byF(y)= 1/(b-a)

a=9500

b= 14700

F(y)= 1/(14700-9500) =1/5200=0.00019

Probability (bid accepted)= (12000-9500)÷1/5200

P( bid accepted) = 2500×0.00019=0.475 approximately 0.48

555.99.

step-by-step explanation:

hope this

first the equation needs to be simplified. so the equation would now look like:   52-24+11

then you multiply.

and when you multiply, you will get the answer of 39.