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babymother [125]
4 years ago
9

1. Determine whether -3 is a root of x3 + 3x2 + x + 1 = 0.

Mathematics
1 answer:
umka21 [38]4 years ago
4 0

Answer:

1. The given equation does not have root at x = - 3.

2. (x - 3) (x + 1) (x² + 4) = 0

This curve will intersect twice the x-axis.

Step-by-step explanation:

1. If the right hand side of the equation x^{3} +3x^{2} +x+1 =0  ......... (1) becomes same as the left hand side by putting x = - 3, then only we can conclude that x = - 3 is a root of the equation.

But in this case (-3)^{3} + 3(-3)^{2}+(-3) +1 \neq  0.

Therefore, x = - 3 is not a root of the above equation (1).

2. The equation of the polynomial with roots 3, -1, 2i, and -2i is

(x - 3) (x + 1) (x - 2i) (x + 2i) = 0

⇒ (x - 3) (x + 1) (x² + 4) = 0 (Answer)

Therefore, the graph of the above curve will intersect twice the x-axis in a real coordinate plane, at x = 3 and at x = -1. (Answer)

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a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

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

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

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

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

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(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

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P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

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P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

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