Answer:
L' = (-36, 0)
Step-by-step explanation:
If the point (x, y) is dilated by a scale factor of k about the center (0, 0), then its image is the point (kx, ky)
∵ A dilation has a center (0, 0)
∵ The point L is (-4, 0)
∴ x = -4 and y = 0
∵ The scale factor of dilation is 9
∴ k = 9
→ By using the rule above
∵ kx = 9(-4) = -36
∵ ky = 9(0) = 0
∵ The image of the point L is (kx, ky)
∴ L' = (kx, ky)
∴ L' = (-36, 0)
Answer:
the two values that are a distance of 5 from 0 are -5 and 5.
Step-by-step explanation:
if you look at a number line, you will see that the number 0 is in the middle. if you count 5 from each side, it's -5 and 5. any other number would have a greater distance than 5 from 0. hope this helps.
Answer:
YAY!
Step-by-step explanation:
Happy Friday! :)!
Thanks so much for the free POINTS!
Ur the best!
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.
