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____ [38]
3 years ago
9

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.)
Mathematics
2 answers:
k0ka [10]3 years ago
7 0

The verified answer is correct, but not in a fraction. The answer is B, 2187/8000. Check for yourself right here

https://www.calculatorsoup.com/calculators/math/percent-to-fraction-calculator.php

<em>Type in 27.3375 to get your answer.</em>

<em />

Hope I helped!

vesna_86 [32]3 years ago
4 0
The answer is 27.3375%.

To calculate this, a multiplication rule is used. The multiplication rule calculates the probability that both of two events will occur. In this method, the possibilities of each event are multiplied.

So, we have three events occurring simultaneously:

1. set contains female singer: 45% = 0.45

2. set contains female singer: 45% = 0.45

3. set contains female singer: 45% = 0.45


Also, it should be taken into consideration that there are three set combinations:

female-female-male,

female-male-female,

male-female-female


So, the probability for one set of the song is:

0.45 × 0.45 × 0.45 = 0.091125

Therefore, the probability is multiple ways to achieve this pattern of songs):

3 × 0.091125 = 0.273375 = <span>27.3375%.</span>

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10.) state whether the given side lengths can form a triangle
Verizon [17]

Answer:

A. 5,8,12, side lengths can form a triangle.

B. 20, 18, 2, side lengths can not form a triangle.

C. 15, 26 , 9, side lengths can not form a triangle.

D. 4.75, 12.25, 16.25, side lengths can form a triangle.

Step-by-step explanation:

We are given three sides in options A, B , C and D.

We need to check if the given sides would form a triangle or not.

Note: Sum of two sides is always greater than third sides in a triangle.

So, we need to check the sum of two sides of given sides length to check if sum is greater than third side length.

A. 5,8,12.

5+8 is greater than 12.

5+12 is greater than 8.

12+8 is greater than 5.

Therefore A. 5,8,12, side lengths can form a triangle.

B. 20, 18, 2

18+2 is not greater than 20.

Therefore B. 20, 18, 2, side lengths can not form a triangle.

C. 15, 26 , 9

15+9 is less than 26.

Therefore C. 15, 26 , 9, side lengths can not form a triangle.

D. 4.75, 12.25, 16.25.

4.75+12.25 is greater than 16.25.

4.75+16.25 is greater than 12.25.

12.25+16.25 is greater than 4.75.

Therefore D. 4.75, 12.25, 16.25, side lengths can form a triangle.


3 0
3 years ago
Twenty four over the quantity of 3 x minus two. Find f(−2).
TEA [102]
Answer:

f(-2) = -3

Explanation:

Not sure if those last couple numbers are answer choices, but I'm going to infer that they might be.

Twenty four over the quantity of 3 x minus two in standard form is:

f(x) = <span>24<span>3x−2</span></span>

Since the number in the ( ) = x, plug in for x using f(-2)

<span>24<span>3∗<span>(−2)</span>−2</span></span>

P E {M D} {A S} for where to solve first

Multiply the 3 and (-2):

<span>24<span>−6−2</span></span>

Add -6 and -2:

<span>24<span>−8</span></span>

Divide the rest:

f(-2) = -3

3 0
4 years ago
Simplify the following expression:<br><br> (66)4
iren [92.7K]

Step-by-step explanation:

Given

( {6}^{6})^{4}  \\  =  {6}^{6 \times 4}  \\  =  {6}^{24}  \\  = 4.7383813e18

3 0
3 years ago
I NEED HELP ASAP!!!!! In one version of a trail mix, there are 3 cups of peanuts mixed with 2 cups of raisins. In another versio
wolverine [178]

Answer:

What are the answers

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What is 0.68 written as a slimplifed fraction
netineya [11]
0.68 = 68/100

68/100 simplest form is 17/25

hope this helps
5 0
4 years ago
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