To solve for proportion we make use of the z statistic. The procedure is to solve for the value of the z score and then locate for the proportion using the standard distribution tables. The formula for z score is:

z = (X – μ) / σ

where X is the sample value, μ is the mean value and σ is the standard deviation

when X = 70

z1 = (70 – 100) / 15 = -2

Using the standard distribution tables, proportion is P1 = 0.0228

when X = 130

z2 = (130 – 100) /15 = 2

Using the standard distribution tables, proportion is P2 = 0.9772

Therefore the proportion between X of 70 and 130 is:

P (70<X<130) = P2 – P1

P (70<X<130) = 0.9772 - 0.0228

P (70<X<130) = 0.9544

Therefore 0.9544 or 95.44% of the test takers scored between 70 and 130.

6 0

3 years ago

Simplify the expression. Quantity cosecant of x to the power of two times secant of x to the power of two divided by quantity se

Art [367]

Answer:

D. 1

Step-by-step explanation:

We have the expression, $\frac{\csc^{2}x\sec^{2}x}{\sec^{2}x+\csc^{2}x}$

We get, eliminating the cosecant function,

$\frac{\sec^{2}x}{\frac{\sec^{2}x}{\csc^{2}x}+1}$

As, sinx is reciprocal of cosecx and cosx is reciprocal of secx,

i.e. $\frac{\sec^{2}x}{\frac{\sin^{2}x}{\cos^{2}x}+1}$

i.e. $\frac{1}{\cos^{2}x}\times \frac{\cos^{2}x}{\sin^{2}x+\cos^{2}x}$

Since, we know that, $\sin^{2}x+\cos^{2}x=1$

Thus,

$\frac{1}{\cos^{2}x}\times \frac{\cos^{2}x}{\sin^{2}x+\cos^{2}x}=1$

So, after simplifying, we get that the result is 1.

Hence, option D is correct.

3 0

3 years ago

How to do permutations?

PIT_PIT [208]

Permutations are written as nPx where n is the number of total choices possible and x is the number of choices that will be used. This is calculated as nPx = n! / (n-x)!.
Permutations represent the number of ways we can choose x objects from n possibilities where the order of selection matters.

7 0

3 years ago