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.
So whats the question my guy??
It would be 27/8 = 3 3/8 next 18/5 = 3 3/5 after that 19/6 = 3 1/6 lastly 35/4 = 8 3/4
Hope this helped!
(f×g)(-1) is -14.
Step-by-step explanation:
- Step 1: Calculate (f×g)(x) by using the formula (f×g)(x) = f(x) × g(x)
⇒ (f×g)(x) = (x + 3)(4x - 3)
⇒ (f×g)(x) = 4x² - 3x + 12x - 9
⇒ (f×g)(x) = 4x² + 9x - 9
- Step 2: Find (f×g)(-1) by substituting -1 in the place of x
⇒ (f×g)(-1) = 4(-1)² + 9(-1) - 9
⇒ (f×g)(-1) = 4 - 9 - 9 =-14
