The probability is 0.36 that the amount spent on a randomly selected child will be between what two value equidistant from the mean?

Assuming a standard normal distribution, the positive value of interest will be that corresponding to the 50 + 36/2 = 68th percentile. A suitable probability calculator will give that value as 0.46770.

The values that are within ±0.46770 standard deviations from the mean will have a probability of 0.36.

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this