Answer:
Step-by-step explanation:
![\ln \dfrac{4y^5}{x^2}\\\\=\ln(4y^5) - \ln(x^2)~~~~~~~~~~~;\left[ \log_b\left( \dfrac mn \right) = \log_b m - \llog_b n \right]\\\\=\ln 4 + \ln y^5 - 2\ln x~~~~~~~~~~~~;[\log_b m^n = n \log_b m ~\text{and}~\log_b(mn) = \log_b m + \log_b n ]\\\\=\ln 4 + 5 \ln y -2 \ln x\\\\=\ln 4 -2 \ln x +5 \ln y](https://tex.z-dn.net/?f=%5Cln%20%5Cdfrac%7B4y%5E5%7D%7Bx%5E2%7D%5C%5C%5C%5C%3D%5Cln%284y%5E5%29%20-%20%5Cln%28x%5E2%29~~~~~~~~~~~%3B%5Cleft%5B%20%5Clog_b%5Cleft%28%20%5Cdfrac%20mn%20%5Cright%29%20%20%3D%20%5Clog_b%20m%20-%20%5Cllog_b%20n%20%5Cright%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%20%5Cln%20y%5E5%20-%202%5Cln%20x~~~~~~~~~~~~%3B%5B%5Clog_b%20m%5En%20%3D%20n%20%5Clog_b%20m%20~%5Ctext%7Band%7D~%5Clog_b%28mn%29%20%3D%20%5Clog_b%20m%20%2B%20%5Clog_b%20n%20%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%205%20%5Cln%20y%20-2%20%5Cln%20x%5C%5C%5C%5C%3D%5Cln%204%20-2%20%5Cln%20x%20%2B5%20%5Cln%20y)
Your answer should be 0.3000
2,12, 1 ,24,3,8,4,3,6,
THese number has 24 as a multiple so they are actually factors
~JZ
Hope it helps
Bo, 4.23 is not bigger than 4.3
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.
