Height and weight are two measurements used to track a child's development. the world health organization measureschild development by comparing the weights of children who are the same height and the same gender. in 2009, weightsfor all 80 cm girls in the reference population had a mean u = 10.2 kg and standard deviation 0 = 0.8 kg. weights arenormally distributed. x ~ n(10.2, 0.8). calculate the z-scores that correspond to the following weights and interpret them. a. 11 kgb. 7.9 kgc 122 kg

Let x be the random variable that represents the weights of all 80 cm girls.

Given :

Formula to calculate z-score :

a) Z-score for x= 11 kg

i.e. z=1

Interpretation : A child who weighs 11 kg is 1 standard deviations above the mean weight 10.2 kg.

b) Z-score for x= 7.9 kg

i.e. z=-2.875

Interpretation : A child who weighs 11 kg is 2.875 standard deviation deviation below the mean weight 10.2 kg.

c) Z-score for x= 12.2 kg

i.e. z=2.5

Interpretation : A child who weighs 11 kg is 2.5 standard deviations above the mean weight 10.2 kg.

it is not that hard all u have to do is plot the graph

step-by-step explanation:

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb