Answer:
108 ft
Correct me, if I am wrong :)
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Given:
To find:
The value of 
.
Solution:
We have,
Using properties of log, we get
![\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Clog_a%5Cdfrac%7Bm%7D%7Bn%7D%3D%5Clog_am-%5Clog_an%5Cright%5D)
![[\log x^n=n\log x]](https://tex.z-dn.net/?f=%5B%5Clog%20x%5En%3Dn%5Clog%20x%5D)
Substitute and .
Therefore, the value of is 
.
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.
The answer is y=34
Explanation:
Y=145, I’m not sure about X
