Answer: There are 75 books.
Price of each book = $4.
Step-by-step explanation:
Let x = Number of books in the box.
Then as per given,
Cost of x books = $300
Cost of one book = 
Books left after giving 15 of them = x-15
Selling price of (x-15) books= $330
Selling price of one book = 
Profit on each book= $1.50
Profit = selling price - cost price
![\Rightarrow 1.50=\dfrac{330}{x-15}-\dfrac{300}{x}\\\\\Rightarrow\ 1.50=\dfrac{330(x)-300(x-15)}{x(x-15)}\\\\\Rightarrow\ 1.50=\dfrac{330x-300x+4500}{x^2-15x}\\\\\Rightarrow\ 1.50(x^2-15x)=30x+4500\\\\\Rightarrow\ 1.50x^2-22.5x=30x+4500\\\\\Rightarrow\ 1.50x^2-52.5x-4500=0\\\\\Rightarrow\ 1.50x^2-52.5x-4500=0\\\\\Rightarrow\ x^2-25x-3000=0\ \ [\text{divide by 1.5}]](https://tex.z-dn.net/?f=%5CRightarrow%201.50%3D%5Cdfrac%7B330%7D%7Bx-15%7D-%5Cdfrac%7B300%7D%7Bx%7D%5C%5C%5C%5C%5CRightarrow%5C%201.50%3D%5Cdfrac%7B330%28x%29-300%28x-15%29%7D%7Bx%28x-15%29%7D%5C%5C%5C%5C%5CRightarrow%5C%201.50%3D%5Cdfrac%7B330x-300x%2B4500%7D%7Bx%5E2-15x%7D%5C%5C%5C%5C%5CRightarrow%5C%201.50%28x%5E2-15x%29%3D30x%2B4500%5C%5C%5C%5C%5CRightarrow%5C%201.50x%5E2-22.5x%3D30x%2B4500%5C%5C%5C%5C%5CRightarrow%5C%201.50x%5E2-52.5x-4500%3D0%5C%5C%5C%5C%5CRightarrow%5C%201.50x%5E2-52.5x-4500%3D0%5C%5C%5C%5C%5CRightarrow%5C%20x%5E2-25x-3000%3D0%5C%20%5C%20%5B%5Ctext%7Bdivide%20by%201.5%7D%5D)

Number of books cannot be negative.
So, there are 75 books.
Price of each book = 
So price of each book = $4.
Answer:5
dhhddhhhhddkskkStep-by-step explanation:
It’s c you add the exponents
Since you have to distribute both numbers, you'll end up with x^2-3x+4x-12 then simplify and it is x^2+x-12
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:

And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221