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melisa1 [442]
3 years ago
9

Y=3x-5 2y - 3x = 5 Solve by subsitution

Mathematics
2 answers:
m_a_m_a [10]3 years ago
6 0
\left\{\begin{array}{ccc}y=3x-5\\2y-3x=5\end{array}\right\\\\substitute\ y=3x-5\ to\ 2y-3x=5\\\\2(3x-5)-3x=5\\2\cdot3x-2\cdot5-3x=5\\6x-10-3x=5\\3x-10=5\ \ \ /+10\\3x=15\ \ \ \ /:3\\x=5\\\\

\left\{\begin{array}{ccc}x=5\\y=3x-5\end{array}\right\\\left\{\begin{array}{ccc}x=5\\y=3\cdot5-5\end{array}\right\\\left\{\begin{array}{ccc}x=5\\y=15-5\end{array}\right\\\left\{\begin{array}{ccc}x=5\\y=10\end{array}\right
Dovator [93]3 years ago
4 0


y = 3x - 5
2y - 3x = 5

2(3x - 5) - 3x = 5
6x - 10 - 3x = 5
3x - 10 = 5
3x = 15
x = 5

y = 3(5) - 5
y = 15 - 5
y = 10

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Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
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Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0
4 years ago
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3 years ago
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Step-by-step explanation:

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3 years ago
What value can be written on the blank line to make the expressions equivalent?
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Hello!

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