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astra-53 [7]
3 years ago
14

For a certain group of students, the probability of randomly selecting a student who takes an art class is 41%, the probability

of randomly selecting a student who takes a music class is 29%, and the probability of randomly selecting a student who takes both an art and a music class is 18%.
What is the probability of randomly selecting a student who takes an art or music class?

I got 52%. Is this correct?
Mathematics
1 answer:
Lera25 [3.4K]3 years ago
4 0
41/88 = 0.47
29/88 = 0.33
0.47 + 0.33 = 0.8 = 80%
I hope this is right?? I’m not very confident in that answer :/
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Solve for R using cross multiplication
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3x-(2x-1)=7x-(3*5x)+(-x+24)
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\boxed{\boxed{\sf x=\frac{23}{10}}\: \sf or \:\boxed{x=2.3}}

_________________

\boxed{\sf Step\: By\:Step:- }

\sf 3x-\left(2x-1\right)=7x-\left(3\times \:5x\right)+\left(-x+24\right)

<u>Remove the parentheses:</u>

\to\sf 3x-\left(2x-1\right)=7x-3\times \:5x-x+24

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\to\sf 3x-\left(2x-1\right)=6x-3\times \:5x+24

<u>Multiply 3 and 5x = 15x:-</u>

\to\sf 3x-\left(2x-1\right)=6x-15x+24

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\sf ^*6x-15x=-9x

\to\sf 3x-\left(2x-1\right)=-9x+24

<u>Expand: 3x-(2x-1)= x+1</u>

\to\sf x+1=-9x+24

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\to\sf x+1-1=-9x+24-1

\to\sf x=-9x+23

<u>Add 9x to both sides:</u>

\to\sf x+9x=-9x+23+9x

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<u>Divide both sides by 10:</u>

\to\sf \cfrac{10x}{10}=\cfrac{23}{10}

\to\sf x=\cfrac{23}{10}

<u>________________________________</u>

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