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ikadub [295]
3 years ago
13

Verify in the following whether g(x) is a factor of p(x)

Mathematics
1 answer:
notsponge [240]3 years ago
8 0

\red{\large\underline{\sf{Solution-i}}}

Given that,

\rm \longmapsto\: p(x) =  {x}^{4} -  {x}^{3} -  {x}^{2} - x - 2

and

\rm \longmapsto\:g(x) = x - 2

<em>We know, </em>

Factor theorem states that if g(x) = x - a is a factor of polynomial f(x), then remainder f(a) = 0.

So, using Factor theorem, Consider

\rm \longmapsto\:p(2)

\rm \:  =  \:  {2}^{4} -  {2}^{3} -  {2}^{2} - 2 - 2

\rm \:  =  \:  16 - 8 - 4 - 4

\rm \:  =  \:  16 - (8 + 4 +  4)

\rm \:  =  \:  16 - 16

\rm \:  =  \: 0

\rm\implies \:p(2) \:   =  \: 0

<u>So, </u>

\bf\implies \:g(x) \: is \: factor \: of \: p(x)

\red{\large\underline{\sf{Solution-ii}}}

\rm \longmapsto\: p(x) = 2{x}^{3}  +  {x}^{2} - 2x + 1

and

\rm :\longmapsto\: g(x) = x + 1

Now, By using Factor Theorem, Consider

\rm \longmapsto\:p( - 1)

\rm \:  =  \: 2 {( - 1)}^{3} +  {( - 1)}^{2} - 2( - 1) + 1

\rm \:  =  \:  - 2 + 1  + 2 + 1

\rm \:  =  \: 2

\rm\implies \:p( - 1) \:  \ne \: 0

<u>So, </u>

\bf\implies \:g(x) \: is  \: not \: a\: factor \: of \: p(x)

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3 years ago
Evaluate the limit 9n^3 + 5n - 2/2n^3
VikaD [51]

Answer:

9/2    if  n goes to infinity  and that the 2n^3 is under the whole expression

Step-by-step explanation:

Let me clear this .

find  limit  (9n^3  + 5*n  - 2)/ (2n^3)

as n --> infinity

Did I put the parentheses in the right spot?

because if you leave it the way you did, then the whole expression goes to positive infinity as n goes to infinity  But I will do this with parentheses

so

find  limit  (9n^3  + 5*n  - 2)/ (2n^3)

simplify expression

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=  (9/2)  + 0  - 0

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4 0
3 years ago
The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.
mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.96516 probability of a chipmunk living through the year, thus p = 0.96516

Item a:

  • Two is P(X = 2) when n = 2, thus:

P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

  • Six is P(X = 6) when n = 6, then:

P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

  • At least one not living is:

P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

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nordsb [41]

Answer:

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Let the invested amount is x.

<u>Then the interest amount is:</u>

  • x*5*2.25/100 = 675
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Correct choice is A

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18 divided by 2 equals 9 because
coldgirl [10]

Answer:

anwer a

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