uppose 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 will be accepted. Assume that the competitor's bid is a random variable that is uniformly distributed between and . a. Suppose you bid . What is the probability that your bid will be accepted (to 2 decimals)

The required parameters are missing :

Probable question with parameters :

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 $10,000 will be accepted. Assume that the competitor’s bid x is a random variable that is uniformly distributed

between $10,000 and $15,000. a. Suppose you bid $12,000. What is the probability that your bid will be accepted?

0.4

Step-by-step explanation:

Competitor's bid = 10,000 to 15000

Take the interval as:

a = 15,000 ; b = 20,000

f(y) =1/ interval difference = 1/(20000-15000) = 1/5000 = 0.0002

With a bid of 17000:

P(x < 17000 = ∫f(y) dy

∫0.0002dy = (0.0002x) at x = 17000 to x = 15000

0.0002(17000) - 0.0002(15000)

3.4 - 3

= 0.4

You can plug in your variables, of the variables you have are different.

the answer i believe is 27 units.

step-by-step explanation:

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.