Answer:
40 % of the snacks were carrot sticks.
Step-by-step explanation:
Total number of carrot sticks in lunch = 18
Total number of apple slices in lunch = 27
So, Total Snacks in the Lunch = Carrot Sticks + Apple Slices
= 18 + 27 = 45
Now,
=
⇒Percentage of Carrot Sticks = 40%
Hence, 40 % of the snacks were carrot sticks.
Answer:
I think it is 6/7
Step-by-step explanation:
I am really bad at explaining but I'm not completely sure
Answer: Only the second sequence is arithmetic.
Work:
An arithmetic sequence is a sequence in which there is a common value added or subtracted between each term in the sequence.
The first option, -2, 4, -6, 8,... is not arithmetic because the terms switch back and forth between positive and negative signs meaning that there is no common value that can be added between each term.
The second option, -8, -6, -4, -2,... is arithmetic because there is a common difference between each term which is +2.
The third option, 2, 4, 8, 16,... is not arithmetic but is, in fact, geometric. There is no common value that is added or subtracted between each term.
Answer:
(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0
(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0
Step-by-step explanation:
(a) when i = 2, the expected number of played games will be:
E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] = 4p²-4p+2-6p²+6p = -2p²+2p+2.
If p = 1/2, then:
d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.
(b) when i = 3;
E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]
Simplification and rearrangement lead to:
E(X) = 6p⁴-12p³+3p²+3p+3
if p = 1/2, then:
d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10
Therefore, E(X) is maximized.
$21 for 3 pounds of steak the rate would be 7 because 21/3= 7
