Here are some of my notes from my 7th grade math class!
Answer:
c) .22
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{1-0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.95%7D%7B2%7D%20%3D%200.025)
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 = 1.96](https://tex.z-dn.net/?f=z%20%3D%201.96)
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%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:
![\sigma = 2.5, n = 500](https://tex.z-dn.net/?f=%5Csigma%20%3D%202.5%2C%20n%20%3D%20500)
Then
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![M = 1.96*\frac{2.5}{\sqrt{500}}](https://tex.z-dn.net/?f=M%20%3D%201.96%2A%5Cfrac%7B2.5%7D%7B%5Csqrt%7B500%7D%7D)
![M = 0.22](https://tex.z-dn.net/?f=M%20%3D%200.22)
Answer:
Step-by-step explanation:
2(6x² - 3) = 11x² - x
2*6x² - 2*3 = 11x² - x
12x² - 6 = 11x² -x
Subtract 11x² from both sides
12x² - 11x² - 6 = -x
x² - 6 = -x
x² + x - 6= 0
Sum =1
Product = -6
Factors = 3 , (-2) { 3*(-2) = -6 & 3 +(-2) = 1}
x² + 3x - 2x - 6 = 0
x(x + 3) - 2(x + 3)= 0
(x +3)(x - 2) = 0
I think it’s the second one
Sector = theta/360° × pi×radius×radius
40/360×22/7×10×10
4/36×22/7×100
1/9×22/7×100
22/63×100
2200/63=34.92 sq feets (approximately)
Option C is the answer . Don't get confused with decimals as maybe the person who provided these options used pi =3.14 instead of what i used i.e. =22/7.