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.

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

the answer would be 150 books

step-by-step explanation:

if you divide 300 by 2 you will get 150