The lengths of salamanders have a normal distribution with mean 15cm, and standard deviation 2cm. What length of salamander would place it at the 80th percentile of salamander lengths?

16.68 cm

Step-by-step explanation:

Given that :

Mean , m = 15

Standard deviation, s = 2

Length of salamander that would Place it at 80% of salamander length :

P(Z ≤ x) = 0.8

Zscore equivalent to 0.8 = 0.842

Using the relation :

Zscore = (x - m) /s

0.842 = (x - 15) / 2

1.684 = x - 15

Add 15 to both sides

1.684 + 15 = x - 15 + 15

16.684 = x

Hence, x = 16.68 cm

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b