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3 years ago

Given the following roots build a polynomial function in standard form: -3 and +2

Mathematics

1 answer:

swat323 years ago

3 0

Answer:

f(x) = x² +x -6

Step-by-step explanation:

The standard form will look like ...

f(x) = x² +bx +c

where b is the opposite of the sum of the roots, and c is their product.

f(x) = x² -(-3+2)x +(-3)(2)

f(x) = x² +x -6

_____

<em>Additional comment</em>

In general, "standard form" is ax²+bx+c. In this case, the coefficient 'a' can be 1 since neither of the roots is expressed as a fraction. The sum of roots is (-b/a) and the product of roots is (c/a).

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Salim had a flock of 300 chickens 15%

Lera25 [3.4K]

Answer:

what is the question

Step-by-step explanation:

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3 years ago

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A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st

Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that $n = 15, p = \frac{1}{5} = 0.2$

What is the probability that he answers at least 10 questions correctly?

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

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

$P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001$

$P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011$

$P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0$

$P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0$

$P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0$

$P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0$

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111$

0.0111% probability that he answers at least 10 questions correctly

3 0

3 years ago

Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

nikitadnepr [17]

You can start with the form
∆y(x -x1) -∆x(y -y1) = 0
Here, we have
∆y = 11-(-3) = 14
∆x = -3-1 = -4
and we can choose (x1, y1) = (1, -3). This gives
14(x -1) -(-4)(y -(-3)) = 0
14x +4y -2 = 0
All these terms have a common factor of 2 that we can remove. Adding 1 to the result puts it in standard form:

7x +2y = 1