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Anon25 [30]
2 years ago
11

What is the sum of the roots of the polynomial shown below? f(x) = x^3 + 2x^2 - 11x-12

Mathematics
1 answer:
svp [43]2 years ago
4 0

<u>Answer:  </u>

Sum of the roots of the polynomial x^{3}+2 x^{2}-11 x-12 \text { is }-2

<u>Solution:</u>

The general form of cubic polynomial is a x^{3}+b x^{2}+c x+d=0 ---- (1)

If we have any cubic polynomial a x^{3}+b x^{2}+c x+d=0 having roots \alpha , \beta , \theta

Sum of roots \alpha + \beta + \theta = \frac{-b}{a} ---(2)

From question given that,

x^{3}+2 x^{2}-11 x-12 --- (3)

On comparing equation (1) and (3), we get a = 1, b = 2, c = -11 and d = -12

Hence the sum of roots using eqn 2 is given as,

= \frac{-2}{1}

= -2

Hence the sum of the roots of the polynomial x^{3}+2 x^{2}-11 x-12 \text { is }-2

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The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

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Answer:

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Step-by-step explanation:

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