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sp2606 [1]
3 years ago
15

Please don't waste my add a random answer if you don't know it don't answer.

Mathematics
1 answer:
love history [14]3 years ago
5 0
So basically are doing keep change flip of ((x^2)/(x-1))/((3x^2)/(x+4))
keep ((x^2)/(x-1)) change the division symbol (/) to a multiplication symbol (*) and flip the denominator ((3x^2)/(x+4)) to ((x+4)/(3x^2))
now we are multiplying ((x^2)/(x-1)) by ((x+4)/(3x^2)). 
((x^2)/(x-1))*((x+4)/(3x^2))
multiply the numerators and then multiply the denominators to get 
((x^3)+(4x^2))/((3x^3)-(3x^2))
let's now extract x^2 from the numerator and 3x^2 from the denominator to get ((x^2)(x+4))/((3x^2)(x-1)
now we can simplify it down to (x/3)((x+4)/(x-1))
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What is the slope of the line represented by the equation 10×+5y=4<br>a.2<br>b.1/2<br>c.-1/2<br>d.-2
VikaD [51]
Equation: 10x + 5y = 4
5y = -10x + 4

Divide the equation by 5, 
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Now Compare it with, y = mx + c
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6 0
3 years ago
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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