F(x) = 2^x; h(x) = x^3 + x + 8
Table
x f(x) = 2^x h(x) = x^3 + x + 8
0 2^0 = 1 0 + 0 + 8 = 8
1 2^1 = 2 1^3 + 1 + 8 = 10
2 2^2 = 4 2^3 + 2 + 8 = 8 + 2 + 8 = 18
3 2^3 = 8 3^3 + 3 + 8 = 27 + 3 + 8 = 38
4 2^2 = 16 4^3 + 4 + 8 = 76
10 2^10 = 1024 10^3 +10 + 8 = 1018
9 2^9 = 512 9^3 + 9 + 8 = 729 + 9 + 8 = 746
Answer: an approximate value of 10
Each coin has a head and a tail, therefore when you toss two coins, you have 4 possible outcomes. You have two heads in only one of these outcomes, while the other three have at least one tail.
The expected value of the game is the price paid/gained times the probability of loss/victory:
E = (1 / 4) · (-6) + (3 / 4) · (2)
= -3 / 2 + 3 / 2
= 0
Bob expects to tie with Will.
Answer:
81 m/s
Step-by-step explanation:
If the box is sliding at 27 m per second, and it's been sliding for 3 seconds, then you should multiply 3 × 27 and your answer would be 81 m/s
The change of the cost changes when the pounds of ham changes. When the pounds of ham is 0, it costs 0 dollars. When the pounds of ham is 3, it costs 12 dollars. The cost changes consistently with the pounds of ham. They change together. Every time the pounds of ham goes up by 3, the cost goes up by 12.
$3 went to the men and and $2 to the bellboy. That's $5 in total, which is what the bellboy was supposed to give the men. So the $1 is with one of the men.
