Y = x^2 + 10x - 171
y = (x - 9)(x + 19)

x - 9= 0 x + 19 = 0
x = 9 x = -19

Answer B covers all requirements... the factored form is
y= (x + 19)(x - 9) 
and the zeros are -19 and 9

4 0

3 years ago

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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%

Final answer:-

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

P(E) = 0.067 = 6.71%

3 0

3 years ago

Can can someone help me with this I don’t know

Zarrin [17]

Answer:

29500

Step-by-step explanation:

subtract the two numbers

7 0

3 years ago

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