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Galina-37 [17]
3 years ago
8

Which represents the polynomial below written in standard form? – 3x + 4x3 + 6

Mathematics
2 answers:
KengaRu [80]3 years ago
5 0

The answer is B , the second one

My name is Ann [436]3 years ago
5 0

Keywords:

<em>Polynomial, standard form, coefficients, exponents, degree </em>

For this case we must write the given polynomial in a standard way. We know that by definition, a polynomial, in its standard form, is given in the form: P (x) = ax ^ n + bx ^ {n-1} + ... + cx ^ 3 + dx ^ 2 + ex + f

Where:

a, b, c, d, e, f: They are the coefficients of the polynomial

n, n-1,3,2,1,0: They are the exponents. This polynomial is of degree "n", because "n" is the largest exponent.

x: It is the variable

If we have:

Q (x) = - 3x + 4x ^ 3 + 6

Resscribing the polynomial in a standard way we have:

Q (x) = 4x ^ 3-3x + 6

This is a polynomial of degree 3.

Answer:

Q (x) = 4x ^ 3-3x + 6

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3 years ago
A large airplane (plane A) flying at 26,000 feet sights a smaller plane (plane B) travelling at an altitude of 24,000 feet. The
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Answer:

The line of sight distance is 1285.58 feet.

Step-by-step explanation:

The situation is illustrated in the figure attached.

From the figure we see that the altitude difference of the planes and the distance between them form a right triangle with one angle of 40° .

The line of sight between the two planes is the hypotenuse of the triangle.

The altitude difference of the planes is

26,000ft -24,000ft = 2000 ft.

Therefore, if we call x the line of sight distance, from trigonometry we have

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\boxed{ x = 1285.58ft}

Therefore, the line of sight distance (x) is 1285.58 feet.

5 0
3 years ago
Find the probability that in five tosses of a fair die, a 3 will appear (a) twice, (b) at most once, (c) at least two times.
Jobisdone [24]

Answer:

A) 0.1612

B) 0.8031

C) 0.1969

Step-by-step explanation:

For each toss of the die, there are only two possible outcomes. Either it is a 3, or it is not. The probability of getting a 3 on each toss is independent from other tosses. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

Five tosses, so n = 5

The die has 6 values, from 1 to 6. The die is fair, so each outcome is equally as likely. The probability of a 3 appearing in a single throw is p = \frac{1}{6} = 0.167

(a) twice

This is P(X = 2)

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

P(X = 2) = C_{5,2}.(0.167)^{2}.(0.833)^{3} = 0.1612

(b) at most once

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

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

P(X = 0) = C_{5,0}.(0.167)^{0}.(0.833)^{5} = 0.4011

P(X = 1) = C_{5,1}.(0.167)^{1}.(0.833)^{1} = 0.4020

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.4011 + 0.4020 = 0.8031

(c) at least two times.

Either a 3 appears at most once, or it does at least two times. The sum of the probabilities of these events is decimal 1. So

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

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.8031 = 0.1969

6 0
3 years ago
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