Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 192 daysμ=192 days and standard deviation sigma equals 12 daysσ=12 days. what is the probability that a randomly selected pregnancy lasts less than 188188 days? the probability that a randomly selected pregnancy lasts less than 188188 days is approximately

0.5237

Step-by-step explanation:

Mean :

Standard deviation :

The formula to calculate the z-score is given by :-

For x = 188 days ,

For x = 107 miles per day ,

The P-value =

Hence, The probability that a randomly selected pregnancy lasts less than 188 days is approximately 0.5237.

b

step-by-step explanation:

-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