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Galina-37 [17]
3 years ago
11

Hey can you please help me posted picture of question :)

Mathematics
2 answers:
otez555 [7]3 years ago
4 0
Answer:
The solutions are -3 and -6

Explanation:
First we would need to put the equation in standard form which is:
ax² + bx + c = 0
This can be done as follows:
x² + 18 = -9x
x² + 9x + 18 = 0
By comparison, we would find that:
a = 1
b = 9
c = 18

Now, to get the roots, we would use the quadratic formula shown in the attached image.

By substitution, we would find that:
either x = \frac{-9+ \sqrt{(9)^2-4(1)(18)} }{2(1)} = -3

or x = \frac{-9- \sqrt{(9)^2-4(1)(18)} }{2(1)} = -6

Hope this helps :)

soldi70 [24.7K]3 years ago
4 0
The solutioms for the equation are a and b or -3 & -6
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murzikaleks [220]

Answer:

a) 0.73684

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Step-by-step explanation:

part a)

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222

Hence,

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684

Part b

P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}

Cases for denominator when:

P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)

= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}

Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}

Hence,

P ( B = 12 / A+B+C = 21) = \frac{\frac{1}{144} }{\frac{1}{96} }\\\\= \frac{2}{3}

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