Frieda is 67 inches tall and weighs 135 pounds. Women her age have a mean height of 65 inches with a standard deviation of 2.5 inches and a mean weight of 125 pounds with a standard deviation of 10 pounds. In relative terms, it is correct to say that:.
A) For this group of women, weight has greater variation than height.
B) The variation coefficient exceeds 10 percent for both height and weight.
C) Frieda is taller and thinner than women in her age group.
D) Frieda's height is more unusual than her weight.

C is correct

Step-by-step explanation:

To answer this question, we shall be examining the options one after the other to ascertain their level of correctness. We proceed as follows;

Option A is incorrect because she is taller than the mean but her weight is greater than the mean so she is heavier not thinner than most women in her age group.

Option B is incorrect as we do not know about the variation in the distribution.

Option D is incorrect as we would need to calculate the variation coefficient.

Option C is correct

We can see that Frieda's weight is 1 Standard Deviation above the mean while her height is less than 1 Standard Deviation away from the mean. This means her height is closer to the mean than her weight. As a result, we would say that her weight is definitely more unusual than her height because her weight is more standard deviations away from the mean than is her height.

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.