150, 155, 163, 168, 172, 177, 186, 190, 205
Answer:
33
Step-by-step explanation:
it is 33 if you get it wrong than dab
The expected length of code for one encoded symbol is

where
is the probability of picking the letter
, and
is the length of code needed to encode
.
is given to us, and we have

so that we expect a contribution of

bits to the code per encoded letter. For a string of length
, we would then expect
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have

so that the variance for the length such a string is

"squared" bits per encoded letter. For a string of length
, we would get
.
Answer-
The exponential model best fits the data set.
Solution-
x = input variable = number of practice throws
y = output variable = number of free throws
Using Excel, Linear, Quadratic and Exponential regression model were generated.
The best fit equation and co-efficient of determination R² are as follows,
Linear Regression
Quadratic Regression
Exponential Regression
The value of co-efficient of determination R² ranges from 0 to 1, the more closer its value to 1 the better the regression model is.
Now,
Therefore, the Exponential Regression model must be followed.