Answer:
one thousandnths, sorry i cant rlly spell it
Step-by-step explanation:
Answer:
C. ± 2.326 years.
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution to solve this question.
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so ![z = 2.326/tex]Now, find the width of the interval[tex]W = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=z%20%3D%202.326%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3ENow%3C%2Fstrong%3E%2C%20find%20the%20width%20of%20the%20interval%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which 
is the standard deviation of the population and n is the size of the sample.
In this question:
So
The correct answer is:
C. ± 2.326 years.
Answer:
-f(3x - 1) + 2 = -18x² + 12x + 1
Step-by-step explanation:
Step 1: Find f(3x - 1)
f(3x - 1) = 2(3x - 1)² - 1
f(3x - 1) = 2(9x² - 6x + 1) - 1
f(3x - 1) = 18x² - 12x + 2 - 1
f(3x - 1) = 18x² - 12x + 1
Step 2: Plug in f(3x - 1)
-(18x² - 12x + 1) + 2
Step 3: Evaluate
-18x² + 12x - 1 + 2
-f(3x - 1) + 2 = -18x² + 12x + 1
Answer: 2,100 = 0.02(42) + b
2,100 = 50(42) + b
Step-by-step explanation: good day man
