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

i think the answer is a,c,e.

i hope i love yaaa..bruno mars from california.

step-by-step explanation:

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.