Suppose the GPAs of all VVC students have an approximately normal distribution with a mean of 2.98 and a standard deviation of 0.31. Find the probability that the mean GPA of a random sample of 15 VVC students is

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560

graph #4 (last graph) depict the function   f(x)= -m(x-d)^1/5 i

step-by-step explanation:

we know that

m=slope

but in this function slope is -m that means slope is decreasing and it is decreasing by 1/5 points down the x-axis.

which is shown in graph four

a decreasing slope down x-axis by 1/5 points.

hence 4th graph is the right option for this.

hey there! : )

4y < -4

we need to isolate y. we can do this by simply dividing both sides by 4.

4y ÷ 4 > -4 ÷ 4

simplify.

y > -1

5 - y > 0

once again, we need to isolate y, so we can start off by subtracting 5 from both sides.

5 - y - 5 > 0 - 5

simplify.

-y > -5

divide both sides by -1.

** remember that when dividing any side by a negative, you must switch the sign! **

-y ÷ -1 < -5 ÷ -1

simplify.

y < 5

~hope i ! ~

use a calculator or a graph