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viktelen [127]
4 years ago
5

The alphabet game costs $.25 to play. Before the game, 26 slips of paper with a different letter of the alphabet on it are put i

nto a bag. A player draws one slip from the bag. If the player draws a vowel (A, E, I, O, or U), he or she wins $1. If a player plays the alphabet game 2 times in a row, replacing the slip of paper after each turn, what is the probability that they win twice? Write as a fraction.
Mathematics
1 answer:
iragen [17]4 years ago
3 0

Answer: 25/676

Step-by-step explanation:

Number of possible outcomes = 26

In other to win, one must draw must be either (A, E, I, O or U)

Therefore required drws to win = 5

First draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Second draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Therefore,

P(winning twice) = (5/26) × (5/26) = 25/676

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Find d for the arithmetic series with S17=-170 and a1=2
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So, we know the sum of the first 17 terms is -170, thus S₁₇ = -170, and we also know the first term is 2, well

\bf \textit{ sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
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\end{cases}
\\\\\\
-170=\cfrac{17(2+a_{17})}{2}\implies \cfrac{-170}{17}=\cfrac{(2+a_{17})}{2}
\\\\\\
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well, since the 17th term is that much, let's check what "d" is then anyway,

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=17\\
a_{17}=-22\\
a_1=2
\end{cases}
\\\\\\
-22=2+(17-1)d\implies -22=2+16d\implies -24=16d
\\\\\\
\cfrac{-24}{16}=d\implies -\cfrac{3}{2}=d
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