If you're using a few larger intervals, then your histogram looks more stocky. If you imagine drawing one, it's because you're adding more values into the same category which can make the difference between two intervals much more noticeable. If you're using smaller intervals, however, you can much more accurately assess the difference between two different intervals. For that reason, the transition between one and another interval would look much more 'fluid'.
Answer:
end of third year = 1,200*(1.02)^3 = 1,273.45 plus
end of second year = 1,200*(1.02)^2 = 1,248.48 plus
end of first year = 1,200*(1.02) = 1,224.00
Total = 3,745.9
Step-by-step explanation:
For part A: you will get 3 linear factors (as the degree of the polynomial is 3). perform the division using (x-1) as your known factor and you will get (x-1)(2x²+11x+15). you can then factor the (2x²+11x+15) to get 2x^3 + 9x^2 + 4x - 15 = (x-1)(2x+5)(x+3)
for part B: since 2x+5 will provide the greatest value (assuming x>0) of the 3 factors, then 2x+5=13. solve to get x=4. if x is 4, then the dimensions are 3'x13'x7' [just sub 4 into the x's for each factor]
for part C: as to the graphing calculator, I don't have one. However, if you solve each linear factor for when it is 0, those values will be the x-intercepts. So your graph should cross the x-asix at 1, -5/2, and -3
