Complete Question
Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150
Answer:
Normal sampling distribution can not be used
Step-by-step explanation:
From the question we are told that
The null hypothesis is 
The alternative hypothesis is 
The sample size is n= 150
Generally in order to use normal sampling distribution
The value 
So
Given that 
normal sampling distribution can not be used
So here are the answers to the given questions above:
1. Since the equation is already given, what we are going to do is to solve for x first.
<span>x + (2x – 10) + 1/2 (2x – 10) = 145
x + (2x -10) + x - 5 = 145
x + 2x + x - 10 - 5 = 145
4x = 145 +10 + 5
4x = 160 <<divide both sides by 4 and we get
x = 40.
Therefore, the amount of white paint is 40ml.
The amount of blue paint is 70ml and the green paint is 35ml.
So the difference between the a</span><span>mounts of white paint and blue paint Vivian combined is 30ml.
2. T</span>he equation of the line that passes through the points (-2, 1) and (1, 10) would be: <span>3x - y = -7, the last option. In order to check, just plug in the given ordered pairs.</span>
Answer:
the slope is -9x
Step-by-step explanation:
whatever has x next to it is the slope
Answer: it is definitely c. y=3x+2
Step-by-step explanation: solve the first equation and then plug in
Notice the picture,
recall your SOH, CAH, TOA
you have,
opposite side, 165
adjacent side, 50
and the angle
that means, we'll need Mrs. tangent
thus
![\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50} \\ \quad \\ tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right] \\ \quad \\ \theta=tan^{-1}\left[ \cfrac{165}{50}\right] \\ \uparrow \\ \textit{angle of elevation}\iff\textit{angle of depression}](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5Cimplies%20tan%28%5Ctheta%29%3D%5Ccfrac%7B165%7D%7B50%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%20tan%5E%7B-1%7D%5Cleft%5B%20tan%28%5Ctheta%29%20%5Cright%5D%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%20%5Cright%5D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ctheta%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%5Cright%5D%0A%5C%5C%20%5Cuparrow%20%20%5C%5C%0A%20%5Ctextit%7Bangle%20of%20elevation%7D%5Ciff%5Ctextit%7Bangle%20of%20depression%7D)
