The heights of a group of trees is normally distributed with a mean of 14.3 feet. if the z score for a 20-foot y’all tree is 1.9, what is the standard deviation?

The standard deviation of the heights of the group of trees is 3 feet

Step-by-step explanation:

For a normal distribution, the z-score corresponding to a particular observed value is calculated as;

z-score = (observed value - mean) / (standard deviation)

The mean is given as 14.3

The z-score is 1.9

The observed value is 20 (the given height)

We can make the standard deviation to be the subject of the formula in the above equation;

standard deviation = (observed value - mean) / ( z-score)

We then substitute the given values into this equation;

standard deviation =

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.