What is the slope of the line that contains these points. X -4 -3 -2 -1 Y 2 5 8 11
1 answer:
Answer:
m=
Step-by-step explanation:
In order to get your you can do it 2 ways: 1). which is how far up you go (rise) over how far left or right you go (run). 2).
which is pick two values and you subtract your second pair from your first pair. You get the same answer either way. In this case, the slope is 1 over 3 or 1/3.
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The graph of the function f(x)=3(2)^x -4 passes through the points (0,-1), (1,2), (2,8) and (3,20)
<h3>How to determine the graph of the function?</h3>The equation of the function is given as:
f(x)=3(2)^x -4
Set x = 0, 1, 2 and 3
f(0)=3(2)^0 -4 = -1
f(1) = 3(2)^(1) - 4 = 2
f(2) = 3(2)^(2) - 4 = 8
f(3) = 3(2)^(3) - 4 = 20
This means that the graph of the function f(x)=3(2)^x -4 passes through the points (0,-1), (1,2), (2,8) and (3,20)
See attachment for the graph of the function
Read more about exponential graphs at:
#SPJ1
Answer:
No, there is not enough evidence at the 0.02 level to support the manager's claim.
Step-by-step explanation:
We are given that the company's promotional literature states that 68% of the chips fail in the first 1000 hours of their use. Also, a sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use.
And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;
Null Hypothesis, : p = 0.68 {means that the actual percentage that fail is same as the stated percentage}
Alternate Hypothesis, : p 0.68 {means that the actual percentage that fail is different from the stated percentage}
The test statistics we will use here is;
T.S. = ~ N(0,1)
where, p = actual percentage of chip fail = 0.68
= percentage of chip failed in a sample of 900 chips = 0.66
n = sample size = 900
So, Test statistics =
= -1.267
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have sufficient evidence to accept null hypothesis.
Therefore, we conclude that the actual percentage that fail is same as the stated percentage and the manager's claim is not supported.