Cylinder (A):
-Surface area: adding all areas of all faces of the shape.
*10 x 6= 60m^3
*3.14 x 3^2 = 28.274 m^2. Then We times it by 2 since we have 2 circles. Which equals to 56.55m^2
Total surface area: 60 + 56.55 = 116.55m^2
-Volume: 3.14 x r^2 x h, then substitute.
*3.14 x 3^2 x 10 = 282.74m^3
(B):
-surface area:
*7 x 11= 77
*5 x 11 = 55
*11 x 11 = 121
*7 x 5 = 35/2 = 17.5 > then again we times it by 2 cuz we have 2 triangles. Which equals to 35.
Total surface area: 77 + 55 + 121 + 35 = 288 cm^2
-Volume:
*7 x 5 x 11 = 385 cm^3
Make sure to check the units!!!
Hope this helped :)
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.
Answer:
i don't know this type of work.
Step-by-step explanation:
i'm only in 10th grade
