Answer:
The horizontal shift is 6 units left (so -6).
Step-by-step explanation:
The horizontal shift is located inside the absolute value bars. While F(x) adds 6 to x, so you may think it's 6 units right, <em>horizontal</em> shifts are backwards as opposed to vertical shifts: we go 6 units left.
(a + 1)(a² + 2a + 1)
= a(a² + 2a + 1) + 1(a² + 2a + 1)
= a³ + 2a² + a + a² + 2a + 1
= a³ + 3a² + 3a + 1
Answer: b
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
Answer:its the same thing
Step-by-step explanation: