It’s a little complicated but here’s how it works:
Imagine a table with the intervals
0:4 , 4:6 , 6:7 , 7:10 , 10:13 (10 year intervals)
Then we have different rows
Class width: 4 , 2 , 1 , 3 , 3
Freq density: 0.2 , 0.5 , 1.2 , 0.7 , 0.3
So now calculate frequency where freq = class width * density
Freq: 0.8 , 1 , 3.6 , 2.1 , 0.9
So to find median find cumulative frequency
(Add all freq)
Cfreq = 8.4 now divide by 2 = 4.2
So find the interval where 4.2 lies.
0.8 + 1 = 1.8 + 3.6 = 5.6
So 4.2 (median) will lie in that interval 60-70 years.
Answer:
The first one
Step-by-step explanation:
For box of 100 gloves, each glove = 44.95/100 =
$0.4495
For box of 50 gloves, each glove = 23.50/50 = $0.47
<span>
Thus the box of 100
gloves is good to buy than the box of 50 gloves.</span>
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
