Alan

Do the following:

\(Expected value = \frac{n_3\times3+n_8\times8}{n_3+n_8}\\ a) 6 = \frac{8\times3+(2+n)\times8}{8+2+n}\)

Solve for n.

Let \(S(n)=\frac{n(n+1)(2n+1)}{6}\)

Then you want  \(S(100)-S(20)\)

Like this:

Like this:

Like this:

Answered here: https://web2.0calc.com/questions/even-more-more-more-geometric-series-and-sequences#r3

The line will touch the parabola at just one point where the line is tangent to the parabola, at which point their slopes are equal.

Slope of straight line = 1

Slope of parabola = 2x/3

So the x-coordinate is given by  2x/3 = 1  or x = 3/2

At this point the value of y is y = (1/3)*(3/2)^2  or y = 3/4

So, using the straight line equation:    3/4 = 3/2 + c    or    c = -3/4

Here's one way of approaching this:

q(4) = 3*4 - b  or  q(4) = 12 - b

p(q(4)) = 2*q(4) - 7   or  p(q(4)) = 2*(12-b) - 7  or  p(q(4)) = 24 - 2b - 7   or  p(q(4)) = 17 - 2b

But p(q(4)) = 7, so  17 - 2b = 7   or 17 - 7 = 2b    or  10 = 2b   or  b = 5