Answer:
4
Step-by-step explanation:
The easiest way to solve this is just to put it in your calculator.
But if you want to go deeper, then ln(x)^4 = 4*ln(x)
Since x in this case = e then ln (e)^1 = 1.
ln(e)^4 = 4*ln(e) = 4
Two forces act on the fish, and these are:
1) Its weight mg, downwards and
2) pressure in the string, T, upwards which is 22 N when the string breaks.
In the result of these two forces, the fish had an acceleration of 2.5 m/s^2 in
the upward way.
So, T - mg = m*2.5
=> m = T / (2.5 + g)
=> m > 22 / (2.5 + 9.8) ( when the string snaps )
=> m > 1.79 kg.
As a result, the mass of the fish is more than 1.79 kg as the line has been broken.
Solution :
Given :

1. Cumulative distribution function




![$=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D-%5Cfrac%7Bx%5E2%7D%7B2%7D%5Cright%5D_6%5Ex%24)
![$=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right] $](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%5Cleft%5B%20%5Cleft%28%20%5Cfrac%7Bx%5E3%7D%7B3%7D%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%5Cleft%29%20-%20%5Cleft%28%20%5Cfrac%7B6%5E3%7D%7B3%7D%20-%20%5Cfrac%7B6%5E2%7D%7B2%7D%20%5Cright%29%20%5Cright%5D%20%20%24)
![$=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20-%2054%20%5Cright%5D%24)
2. Mean 


![$=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%20%5Cleft%5B%5Cfrac%7Bx%5E4%7D%7B4%7D%20-%20%5Cfrac%7Bx%5E3%7D%7B3%7D%20%5Cright%5D_6%5E%7B12%7D%20%5C%20dx%24)

![$=\frac{1}{450} [4608 - 252]$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%20%5B4608%20-%20252%5D%24)
= 17.2857
20 (4)^3-2
20(64)-2
1280-2
1278