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noname [10]
3 years ago
12

The probability of passing the math class of Professor Jones is 64%, the probability of passing Professor Smith's physics class

is 32%, and the probability of passing both is 30%. What is the probability of passing one or the other? The probability of passing one or the other is %.

Mathematics
1 answer:
alexgriva [62]3 years ago
5 0

Answer:

The probability of using one or the other is 36%

Step-by-step explanation:

For solving this problem it is easy if we see it in a ven diagram, for this first we are going to name the initial conditions with some variables:

Probability of passing Professor Jones math class = 64% =0,64

P(J) = 0.64

Probabiliry of passing Professor Smith's physics class = 32% =0.32

P(S) = 0.32

Probability of passing both is = 30% = 0.30

P(JnS) = 0.30 (Is is an intersection so it is in the middle of the ven diagram

We need to know which is the probability of pasing one or the other for this we need to take out the probability of passing both for this we have to add the probability of passing  Professor Jones math class with the probabiliry of passing Professor Smith's physics class and substract the probability of passing both for each one:

P(JuS) = (P(J) - P(JnS)) + (P(S) - P(JnS)) = (0.64 - 0.30) + (0.32 - 0.30) = 0.34 + 0.02 = 0.36 = 36%

If you check the ven diagram you can see that if we add all what is in red we will have the probability of passing Professor Jones math class and if we add all what is in blue we wiill have the probability of passing Professor Smith's physics class, and if we add just what is in each corner we will get the same value that is the probabilty of passsing one or the other.

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pshichka [43]

Answer:

\boxed{\sf x=-2}

\boxed{\sf y=-4}

Step-by-step explanation:

<u>First, Let's  solve for x in -2x+2y=-4:</u>

\sf -2x+2y=-4

<u>Subtract 2y from both sides:</u>

\sf -2x+2y-2y=-4-2y

\sf -2x=-4-2y

<u>Divide both sides by -2:</u>

\sf \cfrac{-2x}{-2}=-\cfrac{4}{-2}-\cfrac{2y}{-2}

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<u>Now, we'll substitute x=y+2 to 3x+3y=-18:</u>

\sf 3x+3y=-18

→ let x=2+y

\sf 3\bold{(2+y)}+3y=-18

<u>Simplify:</u>

\sf 6+6y=-18

<u>Now, let's solve for y in 6+6y=-18</u>

\sf 6+6y=-18

<u>Subtract 6 from both sides:</u>

\sf 6+6y-6=-18-6

\sf 6y=-24

<u>Divide both sides by 6:</u>

\sf \cfrac{6y}{6}=\cfrac{-24}{6}

\bold{ y=-4}

<u>Now, substitute y=-4 into x=2+y:</u>

\sf x=2+y

→ let y = -4

\sf x=2+\bold{-4}

\bold{x=-2}

Therefore, x=-2 and y=-4.

<u>_____________________________________</u>

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