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kumpel [21]
3 years ago
10

What are the zeros of the polynomial function? f(x)=x2−4x−60 Enter your answers in the boxes. The zeros of f(x) are and.

Mathematics
1 answer:
slamgirl [31]3 years ago
4 0

Answer:

x = -6, 10

Step-by-step explanation:

Hi there!

f(x)=x^2-4x-60

Let f(x) = 0:

0=x^2-4x-60

Factor the equation:

0=x^2-10x+6x-60\\0=x(x-10)+6(x-10)\\0=(x+6)(x-10)

The zero-product property tells us that when two terms are multiplied and they equal zero, one of the terms must be zero.

Therefore, either (x+6) or (x110) is equal to 0:

x+6 = 0 ⇒ x = -6

x-10 = 0 ⇒ x = 10

Therefore, the zeros of f(x) are -6 and 10.

I hope this helps!

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1. A gambler makes 100 column bets at roulette. The chance of winning on any one play is 12/38. The gambler can either win $ 5 o
lapo4ka [179]

Answer:

a) Expected amount of the gambler's win = $0.209

b) SD = 2.26

c)P (X >1) = P(z >0.35) = 0.36317

Step-by-step explanation:

The probability of winning, p = 12/38 =6/19

Probability of losing, q = 1 -p = 1-6/19

q = 13/19

Win amount = $5

Loss amount = $2

a) Expected total amount of win = ((6/19)*5) - ((13/19)*2)

Expected total amount of win = 1.579 - 1.369

Expected amount of win, E(X) = $0.209

b) Standard Deviation for the total amount of the gambler's win

SD = \sqrt{E(X^{2}) - (E(X))^{2} }

E(X²) = (6/19)*5²  -  (13/19)*2²

E(X²) = 5.158

SD = \sqrt{5.158 - 0.209^2}

SD = 2.26

c)  probability that, in total, the gambler wins at least $1.

P(X >1)

z = \frac{X - \mu}{\sigma}

μ = E(x) = 0.209

z = (1-0.209)/2.26

z = 0.35

P( X >1) = P(z >0.35)

P(z >0.35) = 1 - P(z <0.35)

P(z >0.35) = 1 - 0.63683

P(z >0.35) = 0.36317

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3 years ago
How do you write 26.17 in word form
olga_2 [115]
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Answer:

Step-by-step explanation:

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Answer:

1st time: 50% chance or 8/16 chance

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Selec the correct answer.
sp2606 [1]
Domain= [0,18], Range = [0,31.50]
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