Answer:
189 components must be sampled.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.99}{2} = 0.005](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.99%7D%7B2%7D%20%3D%200.005)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 2.575.
Now, find the margin of error M as such
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Assume that component lifetimes are normally distributed with population standard deviation of 16 hours.
This means that ![\sigma = 16](https://tex.z-dn.net/?f=%5Csigma%20%3D%2016)
How many components must be sampled so that a 99% confidence interval will have margin of error of 3 hours?
n components must be sampled.
n is found when M = 3. So
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![3 = 2.575\frac{16}{\sqrt{n}}](https://tex.z-dn.net/?f=3%20%3D%202.575%5Cfrac%7B16%7D%7B%5Csqrt%7Bn%7D%7D)
![3\sqrt{n} = 2.575*16](https://tex.z-dn.net/?f=3%5Csqrt%7Bn%7D%20%3D%202.575%2A16)
![\sqrt{n} = \frac{2.575*16}{3}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.575%2A16%7D%7B3%7D)
![(\sqrt{n})^2 = (\frac{2.575*16}{3})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B2.575%2A16%7D%7B3%7D%29%5E2)
![n = 188.6](https://tex.z-dn.net/?f=n%20%3D%20188.6)
Rounding up:
189 components must be sampled.
Answer:
162.1%
Step-by-step explanation:
Given that:
Size of previous living space = 1450 sq ft
Size of new living space = 3800 sq ft
% increase in living space ;
Increase in living space = 3800 - 1450 = 2350 sq feet
(Increase / previous size) * 100%
(2350 / 1450) * 100%
1.6206 * 100%
= 162.06%
= 162.1%
Answer:
Step-by-step explanation:
1: 7/6
2: 3/2
3: 0.000288
4: six tenths
<u>Step-by-step explanation:</u>
transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis
(???, 0): 0 = - ln (x - 4)
![\frac{0}{-1} = \frac{-ln (x - 4)}{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B0%7D%7B-1%7D%20%3D%20%5Cfrac%7B-ln%20%28x%20-%204%29%7D%7B-1%7D)
0 = ln (x - 4)
![e^{0} = e^{ln (x - 4)}](https://tex.z-dn.net/?f=e%5E%7B0%7D%20%3D%20e%5E%7Bln%20%28x%20-%204%29%7D)
1 = x - 4
<u> +4 </u> <u> +4 </u>
5 = x
(5, 0)
(???, 1): 1 = - ln (x - 4)
![\frac{0}{-1} = \frac{-ln (x - 4)}{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B0%7D%7B-1%7D%20%3D%20%5Cfrac%7B-ln%20%28x%20-%204%29%7D%7B-1%7D)
1 = ln (x - 4)
![e^{1} = e^{ln (x - 4)}](https://tex.z-dn.net/?f=e%5E%7B1%7D%20%3D%20e%5E%7Bln%20%28x%20-%204%29%7D)
e = x - 4
<u> +4 </u> <u> +4 </u>
e + 4 = x
6.72 = x
(6.72, 1)
Domain: x - 4 > 0
<u> +4 </u> <u>+4 </u>
x > 4
(4, ∞)
Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that
No vertical asymptotes
*************************************************************************
transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis
Domain: there is no restriction on x so domain is all real number
(-∞, ∞)
Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.
(-∞, 0)
Y-intercept is when x = 0:
f(x) = - 3ˣ⁺⁵
= - 3⁰⁺⁵
= - 3⁵
= -243
Horizontal Asymptote: y = 0 <em>(explanation above)</em>
Answer: 55.041
Step-by-step explanation: