All categories

3 years ago

Choose the correct answer below.

1 answer:

4 0

A z score is a standardized score, measuring the number of standard deviations a given score is away from the mean. If the original scores are normal, the z scores will have a standard normal distribution.

Choice b

You might be interested in

What is the value of x in the solution to the system of linear equations? {y=3x+2 y=x-4 a –7 b –3 c 1 d 5

lyudmila [28]

Since y must equal both 3x+2 and x-4, we deduce that 3x+2 and x-4 must equal each other. So, we have

$3x+2=x-4 \iff 2x+2=-4 \iff 2x=-6 \iff x=-3$

5 0

3 years ago

Read 2 more answers

The Butlers bought a $332,000 house. They made a down payment of 42,000 and took out a mortgage for the rest. Over the course of

Zarrin [17]

Answer:

Step-by-step explanation:

(a)Mortgage = 332000 - 42000 = 290000

15 * 12 = 180 months

Total monthly payments = 180 * 2447.19 =440494.2

Total = 440494.2 + 42000 = 482494.2

(b) Interest = 482494.2 -290000

= 192494.2

4 0

3 years ago

Read 2 more answers

Your friend says that 14.5% is same as the fraction 29/200. explain whether your friend is correct or incorrect by comparing the

OlgaM077 [116]

14.5% = 0.145

Step-by-step explanation:

The term 'percent' (%) actually means per 100. So when you say 14.5% it means 14.5 per 100. So when you put that into an equation it actually means:

So when you solve it, 14.5% is actually 0.145 and not 14.5.

Your friend is then incorrect because:

14.5% ≠ 14.5

3 0

3 years ago

Read 2 more answers

Which of the following are true of line graphs?

ASHA 777 [7]

Specific data is show I believe

6 0

2 years ago

The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

ankoles [38]

Answer:

2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

$T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}$

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

$D_x=T_x\sqrt{T_x}$

The mean distance of planet Y can be found using the given relation ...

$T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x$

The mean distance of planet Y is increased from that of planet X by the factor ...

2√2

8 0

2 years ago