3. A=bh/2
2A=bh
h=2A/b.
8. Dy-Cx=E
Dy=E+Cx
y=(E+Cx)/D
The answer is nine because 2(4)=8 3(2)=6 7=7, os 8-6+7= 2+7 =9
it is b.9 because u have to multiply (w×h×l)/3
If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show
Let 
. Compute the inverse:
![f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}](https://tex.z-dn.net/?f=f%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%3D%20%5Csqrt%7B1%20%2B%20f%5E%7B-1%7D%28x%29%5E3%7D%20%3D%20x%20%5Cimplies%20f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%5E2-1%7D)
and we immediately notice that ![f^{-1}(x+1)=\sqrt[3]{x^2+2x}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%2B1%29%3D%5Csqrt%5B3%5D%7Bx%5E2%2B2x%7D)
.
So, we can write the given integral as
Splitting up terms and replacing 
in the first integral, we get
A) the z score for 95% is 1.96
Multiply by the deviation:
1.96 x 0.005 = 0.0098
Now add and then subtract that from the mean:
4.035 - 0.0098 = 4.0252
4.035 + 0.0098 = 4.0448
The interval is (4.0252, 4.0448)
B) 4.035 +/- 1.96 x sqrt( 0.005/sqrt(25))
= 4.035 +/-0.00196
Answer: ( 4.033, 4.037)
C) the conclusion is that both 4.020 and 4.055 are out of the range.
4.020 is below the lowest range and 5.055 is higher than highest range.
