The lengths of pregnancies in a small rural village are normally distributed with a mean of 264.1 days and a standard deviation of 12.9 days. In what range would you expect to find the middle 95% of most pregnancies

The range of the 95% data (X) =  238.3 days <  X < 289.9 days

Step-by-step explanation:

Given;

mean of the normal distribution, m = 264.1 days

standard deviation, d = 12.9 days

between two standard deviation below and above the mean is 96% of all the data.

two standard deviation below the mean = m - 2d

                                                                   = 264.1 - 2(12.9)

                                                                   = 238.3 days

two standard deviation above the mean = m + 2d

                                                                    = 264.1 + 2(12.9)

                                                                     = 289.9 days

The middle of the 95% of most pregnancies would be found in the following range;

238.3 days <  X < 289.9 days

the equivalent expression are always the hardest ones but you look at the gcf also

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.