Answer:
The proportion of children that have an index of at least 110 is 0.0478.
Step-by-step explanation:
The given distribution has a mean of 90 and a standard deviation of 12.
Therefore mean,
= 90 and standard deviation,
= 12.
It is given to find the proportion of children having an index of at least 110.
We can take the variable to be analysed to be x = 110.
Therefore we have to find p(x < 110), which is left tailed.
Using the formula for z which is p( Z <
) we get p(Z <
= 1.67).
So we have to find p(Z ≥ 1.67) = 1 - p(Z < 1.67)
Using the Z - table we can calculate p(Z < 1.67) = 0.9522.
Therefore p(Z ≥ 1.67) = 1 - 0.9522 = 0.0478
Therefore the proportion of children that have an index of at least 110 is 0.0478
P = 2(L + W)
P = 400
L = W + 40
400 = 2(W + 40 + W)
400 = 2(2W + 40)
400 = 4W + 80
400 - 80 = 4W
320 = 4W
320/4 = W
80 = W.....so the width (W) = 80 yards
L = W + 40
L = 80 + 40
L = 120...and the length (L) = 120 yards
Answer:
Step-by-step explanation:
Given,

we know,
(f-g)(x)=f(x)-g(x)
So, here we get
(f-g)(x)
=f(x)-g(x)
=
=
=
=
So, the answer is

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