Assume the given distributions are normal. Cucumbers grown on a certain farm have weights with a standard deviation of 2 ounces. What is the mean weight if 85% of the cucumbers weigh less than 16 ounces?Assume the given distributions are normal. Group of answer choices14.4014.3014.8813.92

13.92

Step-by-step explanation:

We have that the critical z-score associated with 85% to the left is 1.04, we know that by table.

So we have to:

m + z * (sd) = 16

where m is the mean, z is the critical z-scor and sd is the standard deviation, if we replace we are left with:

m + 1.04 * (2) = 16

m = 16 - 1.04 * (2)

m = 13.92

Therefore, the average weight if 85% of cucumbers weigh less than 16 ounces is 13.92

ege

step-by-step explanation:

egge

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3