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Ostrovityanka [42]
3 years ago
9

The solution set for x^2 -x -56=0 is A.{7,8} b. {-7} c. {8} d. {-7,8} e. {7.-8}

Mathematics
1 answer:
podryga [215]3 years ago
5 0
I believe the answer is E
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What are the real zeros of f(x) = x^3 +2x^2-5x-6
netineya [11]

First of all Horner
First we try with -3
1. 2. -5. -6.
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Now we have this equation
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Aliun [14]

Step-by-step explanation:

see this bro May this help you

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Please solve this question
Vesnalui [34]

Hello,

\boxed{P(X = k) =( {}^{n} _{k}) \times p {}^{k} \times (1 - p) {}^{n - k}   }

P(X= k) = ( {}^{12}  _{6})  \times 0.51 {}^{6}  \times (1 - 0.51) {}^{12 - 6}

We have :

( {}^{n}  _{k})  =  \frac{n!}{k!(n - k)!}

( {}^{12}  _{6})  =  \frac{12!}{6!(12 - 6)!}  =  \frac{12!}{6!6!}  =  \frac{12 \times 11 \times 10 \times ... \times 1}{2(6 \times 5 \times 4 \times... \times 1) }  =  \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 }{6 \times 5 \times 4 \times 3 \times 2 \times 1}  = 924

P(X = k) = 924 \times 0.51 {}^{6}  \times 0.49 {}^{6}  = 0.2250

8 0
2 years ago
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