The height of women ages 20-29 is normally distributed, with a mean of 64.9 inches. assume sigmaequals2.3 inches. are you more likely to randomly select 1 woman with a height less than 66 inches or are you more likely to select a sample of 27 women with a mean height less than 66 inches? explain.

To select a sample of 27 women with a mean height less than 66 inches is more likely than to randomly select 1 woman with a height less than 66 inches

Step-by-step explanation:

1) The probability of randomly selecting 1 woman with a height less than 66 inches is

P(z<z(66)) where z(66) is the z-score of the woman whose height is 66 inches.

z score can be calculated using the formula

z(66)= where

Then z(66)= ≈ 0.48

and P(z<0.48) = 0.6844

2) The probability of selecting a sample of 27 women with a mean height less than 66 inches can be calculated using the equation

t= where

t= ≈ 2.49

looking t-table P(t<2.49)≈0.9903

Since 0.9903>0.6844 we can conclude that to select a sample of 27 women with a mean height less than 66 inches is more likely than to randomly select 1 woman with a height less than 66 inches

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15