Answer:
24
Step-by-step explanation:
A=LxW
6x4=24
The question is not complete so I will answer it with an example and a few assumptions. Follow the steps to find the answer to your question.
Important to note:
Your question is a z-scores problem
Assume that for the population of unemployed individuals the population standard deviation is 4 weeks.
Thus, we need to find the z-value.
The z-value is the sample mean decreased by the population mean, divided by the standard deviation that we assumed. So, we have:
![z = \frac{\bar x - \mu}{\sigma / \sqrt{n} } = \frac{\pm 1}{4/\sqrt{60} } \approx \pm 1.94](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B%5Cbar%20x%20-%20%5Cmu%7D%7B%5Csigma%20%2F%20%5Csqrt%7Bn%7D%20%7D%20%20%3D%20%5Cfrac%7B%5Cpm%201%7D%7B4%2F%5Csqrt%7B60%7D%20%7D%20%20%5Capprox%20%5Cpm%201.94)
![P = P(-1 < \bar x - \mu < 1) = P(-1.94 < z < 1.94) = 1 - 2P(z < -1.94)](https://tex.z-dn.net/?f=P%20%3D%20P%28-1%20%3C%20%5Cbar%20x%20-%20%5Cmu%20%3C%201%29%20%3D%20P%28-1.94%20%3C%20z%20%3C%201.94%29%20%3D%201%20-%202P%28z%20%3C%20-1.94%29)
Using any standard negative z-scores table, we can find that:![P(z < -1.94) = 0.0262](https://tex.z-dn.net/?f=P%28z%20%3C%20-1.94%29%20%3D%200.0262)
Thus, we get:
![P(| \bar x - \mu | < 1) = 1 - 2 \times 0.0262 = 1 - 0.0524 = 0.9476](https://tex.z-dn.net/?f=P%28%7C%20%5Cbar%20x%20-%20%5Cmu%20%7C%20%3C%201%29%20%3D%201%20-%202%20%5Ctimes%200.0262%20%3D%201%20-%200.0524%20%3D%200.9476)
Therefore, the probability that a simple random sample of 60 unemployed individuals will provide a sample mean within 1 week of the population mean is 0.9476
Answer:
0.9476
Answer:
200 I think just kidding or you ca
Answer:
17
Step-by-step explanation:
Given:
e = 15
f = 2
To Find:
e - -f
Solution:
e - -f
= e + f
= 15 + 2
Therefore, 17
PLZ MARK ME AS BRAINLIEST!!!