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Darya [45]
1 year ago
14

id="TexFormula1" title=" - 10a - 9n + 16 - 30 + 3n + 15a + 50 + a = " alt=" - 10a - 9n + 16 - 30 + 3n + 15a + 50 + a = " align="absmiddle" class="latex-formula">I'm having trouble comebineing those like terms
Mathematics
1 answer:
Rina8888 [55]1 year ago
4 0

Problem:

Solution:

let the expression

-10a -9n + 16-30 + 3n + 15 a + 50 + a =

Putting together similar terms we have

(-10a +15a+a) + (-9n+3n ) + (16-30+50) =

this is equivalent to:

(6a) + (-6n) + (36)

this is equivalent to

6a -6n+36

then, the correct answer is:

6a -6n+36

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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
Reptile [31]

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%

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Simplify. 1+32⋅2−5 A −48 B14 C15 D27
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y is cost

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