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crimeas [40]
3 years ago
10

A Nashville-area radio station plays songs from a specific, fixed set of artists. The station has no DJ; instead, a computer ran

domly selects which songs to play. The songs themselves are picked randomly, and the same song may be played many times in a row. In the set of songs, 45% are sung by a female singer, 45% are sung by a male singer, and 10% are instrumental with no vocals. What is the probability that a particular set of 3 songs contains exactly 2 female singers and 1 male singer? (Hint: Be aware that there are multiple ways to achieve this pattern of songs.)
A. 279/8000
B. 2187/8000
C. 729/2000
D. 9/10
Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0
There are 3 ways to get 2 female and 1 male singer.
These 3 ways are 
<span>FFM, FMF and MFF </span>
<span>Each of the three ways has probability 45% * 45% * 45% = 9.1125% </span>
<span>As there are three ways multiply by 3 </span>
<span> = 27.3375% 
hope it helps</span>
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Answer:

1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

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Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

We also know that: a^{-m}=\frac{1}{a^m}

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

We know that: a^{-m}=\frac{1}{a^m}

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

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