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Evgen [1.6K]
3 years ago
15

In the diagram find the area of DEF

Mathematics
1 answer:
Agata [3.3K]3 years ago
4 0

Answer:

Im sorry Im not sure

Step-by-step explanation:

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Answer:

See below.

Step-by-step explanation:

Congruent sides:

SU and FO

UN and OG

SN and FG

Congruent angles:

<S and <F

<U and <O

<N and <G

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vaieri [72.5K]

Answer: Choice A.   1/64

Work Shown:

f(x) = 4^x

f(-3) = 4^(-3)

f(-3) = 1/(4^3) ... see note below

f(-3) = 1/64

Note: I'm using the rule that a^(-b) = 1/(a^b)

4 0
3 years ago
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Suppose that two teams play a series of games that ends when one of them has won ???? games. Also suppose that each game played
Musya8 [376]

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.

6 0
2 years ago
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