The nickel an the quarter doesn't affect it. We list out the possibilities.

p:h d:h

p:h d:t

p:t d:h

p:t d:t

2 out of the 4 combinations work, so the probability is 1/2.

-nx=-b-hx

-nx+hx=-b

(h-n)x=-b

x=-b/(h-n)

if it's supposed to be the sqaure root of 64 the answer is \(\frac{11}{16/8} = \frac{11}{2}\)

if it's 4 times the sqare root of 6 the answer is \(\frac{11}{4/\sqrt6}=\frac{11\sqrt6}{4}\)

(a)

I'm not sure if this is correct but the expected value for each question is (1/10)*0+(9/10)*1=0.9. Since the expected value for one question is 0.9 and there are ten questions. The expected value for the exam is 10*0.9=9.

I was thinking, this is like putting 9 indistinguishable balls into 3 distinguishable boxes where every box must have one ball. Once you put a ball in each box then we have two but 6 balls into 3 boxes. We can use stars (are the balls) and bars (the box dividers). For example this |**||****, would be 2-0-4. So that's the same as 9C3=84.

f+s= total # of trips

x+y= amount of cement they bring when they each go one trip

xf + sy = total number of concrete they bring after all their trips

\(6(\frac{4}{3}(14\div\frac{1}{7}))\div\frac{13}{10}= 6(\frac{4}{3}(14\cdot7))\cdot\frac{10}{13}= 6(\frac{4}{3}(98))\cdot\frac{10}{13}= 6(\frac{392}{3}))\cdot\frac{10}{13}=784\cdot\frac{10}{13}=\frac{7840}{13}\)or \(603 \frac{1}{13}\)

-16+9

17-24

Basically any expression that equals -7.