to answer this question u have to tell me the size of the bear or its not solvabal
Answer:
about 5.8cm
Step-by-step explanation:
Answer:
0.2389
Step-by-step explanation:
The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0.1 mm.
![\mu=1.3mm\\\sigma=0.1mm](https://tex.z-dn.net/?f=%5Cmu%3D1.3mm%5C%5C%5Csigma%3D0.1mm)
A random sample of 200 wafers is drawn
we are supposed to find What is the probability that the sample mean warpage exceeds 1.305 mm
We will use central limit theorem
According to central limit theorem:
![\sigma^2_{\bar{x}}=\frac{\sigma^2}{n}](https://tex.z-dn.net/?f=%5Csigma%5E2_%7B%5Cbar%7Bx%7D%7D%3D%5Cfrac%7B%5Csigma%5E2%7D%7Bn%7D)
![\sigma^2_{\bar{x}}=\frac{0.1^2}{200}](https://tex.z-dn.net/?f=%5Csigma%5E2_%7B%5Cbar%7Bx%7D%7D%3D%5Cfrac%7B0.1%5E2%7D%7B200%7D)
![\sigma_{\bar{x}=\sqrt{\frac{0.1^2}{200}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%7Bx%7D%3D%5Csqrt%7B%5Cfrac%7B0.1%5E2%7D%7B200%7D%7D)
![\sigma_{\bar{x}=\sqrt{0.00005}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%7Bx%7D%3D%5Csqrt%7B0.00005%7D)
Now we are supposed to find What is the probability that the sample mean warpage exceeds 1.305 mm i.e.P(x>1.305)
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
![z=\frac{1.305-1.3}{\sqrt{0.00005}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B1.305-1.3%7D%7B%5Csqrt%7B0.00005%7D%7D)
![z=0.71](https://tex.z-dn.net/?f=z%3D0.71)
Refer the z table
P(z<0.71)=0.7611
![P(\bar{x}>1.305)=1-P(z](https://tex.z-dn.net/?f=P%28%5Cbar%7Bx%7D%3E1.305%29%3D1-P%28z%3C0.71%29%20%3D1-0.7611%3D0.2389)
Hence the probability that the sample mean warpage exceeds 1.305 mm is 0.2389
The answer is:
x = 4hrs
Explanation:
2.5(12.75x + 24.50) = 188.75
31.875x + 61.25 = 188.75
31.875x = 188.75 - 61.25
31.875x = 127.50
x = 127.50/31.874
x = 4
I hope this helps!!!
Answer:
y= -2x - 1
Step-by-step explanation:
that is the answer sorry if wrong