Answer:
9 (m+5)-3(m-2)=8m+31
9m+45 - (3m + 6)= 8m+31
9m+45-3m-6=8m+31
45-6+31=8m-9m+3m
70=2m
m=70/2
Therefore m = 35
Hope u understood.............
Answer:
I believe that A would be the correct answer im not sure but ill explain from the best of my ability
Step-by-step explanation:
OK SO FOR THE EXPLANATION I BELIEVE THAT IT IS A BECAUSE USUALLY ON THE OTHER QUESTIONS I HAVE HAD IT WOULD USUALLY SPLIT THE FIRST AMOUNT IN HALF AND THAT IS MY EXPLANATION GOOD LUCK
Answer:
The answer is 4.19
Step-by-step explanation:
12.57 divided by 3 = 4.19
Hope this helps!! :)
Answer:
A) The sampling distribution for a sample size n=50 has a mean of 18.5 weeks and a standard deviation of 0.849.
B) P = 0.7616
C) P = 0.4441
Step-by-step explanation:
We assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks.
A) We take a sample of size n=50.
The mean of the sampling distribution is equal to the population mean:
![\mu_s=\mu=18.5](https://tex.z-dn.net/?f=%5Cmu_s%3D%5Cmu%3D18.5)
The standard deviation of the sampling distribution is:
![\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{50}}=0.849](https://tex.z-dn.net/?f=%5Csigma_s%3D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B6%7D%7B%5Csqrt%7B50%7D%7D%3D0.849)
B) We have to calculate the probability that the sampling distribution gives a value between one week from the mean. That is between 17.5 and 19.5 weeks.
We can calculate this with the z-scores:
![z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{17.5-18.5}{6/\sqrt{50}}=\dfrac{-1}{0.8485}=-1.179\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{19.5-18.5}{6/\sqrt{50}}=\dfrac{1}{0.8485}=1.179](https://tex.z-dn.net/?f=z_1%3D%5Cdfrac%7BX_1-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B17.5-18.5%7D%7B6%2F%5Csqrt%7B50%7D%7D%3D%5Cdfrac%7B-1%7D%7B0.8485%7D%3D-1.179%5C%5C%5C%5C%5C%5Cz_2%3D%5Cdfrac%7BX_2-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B19.5-18.5%7D%7B6%2F%5Csqrt%7B50%7D%7D%3D%5Cdfrac%7B1%7D%7B0.8485%7D%3D1.179)
The probability it then:
![P(|X_s-\mu_s|](https://tex.z-dn.net/?f=P%28%7CX_s-%5Cmu_s%7C%3C1%29%3DP%28%7Cz%7C%3C1.179%29%3D0.7616)
C) For half a week (between 18 and 19 weeks), we recalculate the z-scores and the probabilities:
![z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{18-18.5}{6/\sqrt{50}}=\dfrac{-0.5}{0.8485}=-0.589](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B18-18.5%7D%7B6%2F%5Csqrt%7B50%7D%7D%3D%5Cdfrac%7B-0.5%7D%7B0.8485%7D%3D-0.589)
![P(|X_s-\mu_s|](https://tex.z-dn.net/?f=P%28%7CX_s-%5Cmu_s%7C%3C0.5%29%3DP%28%7Cz%7C%3C0.589%29%3D0.4441)