Let x and y be independent continuous random variables that are uniformly distributed on (0,1). let h=(x+2)y. find the probability p(lnh≥z) where z is a given number that satisfies ez< 2. your answer should be a function of z. hint: condition on x. p(lnh≥z1)= ? let x be a standard normal random variable, and let fx(x) be its cdf. consider the random variable z=fx(x). find the pdf fz(z) of z. note that fz(z) takes values in (0,1) .fz(z)= ?

Looks like there are two distinct problems here.

Conditioning on involves interpreting this probability for all possible fixed values of :

We have , and over this domain . Since is fixed, we can omit the conditioning notation to leave us with

has CDF

Then we can write our integral as

Using the method of distribution functions, we have

all three statements are true

step-by-step explanation:

a shipping fee of 5% is charged online with every item purchased online

first let's just take each option and calculate them based on this, to see if the answers matched

keith buys a 40 dollar book

40 x 0.05 = 2

therefore there will be a two dollar fee on keith's book

40 + 2 =42

the first statement is true

a chair costs $80

80 x 0.05 = 4

80 + 4 = 84

the second statement is true

george buys a 20 dollar calculator

20 x 0.05 = 1

20 + 1 = 21

the third statements are true

all three statements are true

114 plus 113 times 2 and 114 times 3