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3 years ago

Which of the following is the correct factorization of the polynomial below? x^3-15

Mathematics

2 answers:

stealth61 [152]3 years ago

5 0

Answer with explanation:

Use the identity to solve the problem

$a^3 -b^3=(a-b)(a^2+ab+b^2)\\\\x^3-15=x^3-(15^{\frac{1}{3})^3}\\\\=[x-(15)^{\frac{1}{3}}][x^2+(15)^{\frac{1}{3}}x+((15)^{\frac{1}{3}})^{2}]\\\\=[x-(15)^{\frac{1}{3}}][x^2+(15)^{\frac{1}{3}}x+(15)^{\frac{2}{3}}]$

Anuta_ua [19.1K]3 years ago

4 0

The polynomial is irreducible.

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iren2701 [21]

Answer:

125/36

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2 years ago

What are three ratios equal to 5/40

kiruha [24]

1/8,2/16,3/24 that should be the answer

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3 years ago

There are 4 semi- trailer truck park a line at the stop. After the first truck, each in the line weights 2 tons more than the tr

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Let's say the first truck weighs x tons

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3 years ago

medical tests. Task Compute the requested probabilities using the contingency table. A group of 7500 individuals take part in a

uysha [10]

Probabilities are used to determine the chances of an event

The probability that a person is sick is: 0.008
The probability that a test is positive, given that the person is sick is 0.9833
The probability that a test is negative, given that the person is not sick is: 0.9899
The probability that a person is sick, given that the test is positive is: 0.4403
The probability that a person is not sick, given that the test is negative is: 0.9998
A 99% accurate test is a correct test

(a) Probability that a person is sick

From the table, we have:

$\mathbf{Sick = 59+1 = 60}$

So, the probability that a person is sick is:

$\mathbf{Pr = \frac{Sick}{Total}}$

This gives

$\mathbf{Pr = \frac{60}{7500}}$

$\mathbf{Pr = 0.008}$

The probability that a person is sick is: 0.008

(b) Probability that a test is positive, given that the person is sick

From the table, we have:

$\mathbf{Positive\ and\ Sick=59}$

So, the probability that a test is positive, given that the person is sick is:

$\mathbf{Pr = \frac{Positive\ and\ Sick}{Sick}}$

This gives

$\mathbf{Pr = \frac{59}{60}}$

$\mathbf{Pr = 0.9833}$

The probability that a test is positive, given that the person is sick is 0.9833

(c) Probability that a test is negative, given that the person is not sick

From the table, we have:

$\mathbf{Negative\ and\ Not\ Sick=7365}$

$\mathbf{Not\ Sick = 75 + 7365 = 7440}$

So, the probability that a test is negative, given that the person is not sick is:

$\mathbf{Pr = \frac{Negative\ and\ Not\ Sick}{Not\ Sick}}$

This gives

$\mathbf{Pr = \frac{7365}{7440}}$

$\mathbf{Pr = 0.9899}$

The probability that a test is negative, given that the person is not sick is: 0.9899

(d) Probability that a person is sick, given that the test is positive

From the table, we have:

$\mathbf{Positive\ and\ Sick=59}$

$\mathbf{Positive=59 + 75 = 134}$

So, the probability that a person is sick, given that the test is positive is:

$\mathbf{Pr = \frac{Positive\ and\ Sick}{Positive}}$

This gives

$\mathbf{Pr = \frac{59}{134}}$

$\mathbf{Pr = 0.4403}$

The probability that a person is sick, given that the test is positive is: 0.4403

(e) Probability that a person is not sick, given that the test is negative

From the table, we have:

$\mathbf{Negative\ and\ Not\ Sick=7365}$

$\mathbf{Negative = 1+ 7365 = 7366}$

So, the probability that a person is not sick, given that the test is negative is:

$\mathbf{Pr = \frac{Negative\ and\ Not\ Sick}{Negative}}$

This gives

$\mathbf{Pr = \frac{7365}{7366}}$

$\mathbf{Pr = 0.9998}$

The probability that a person is not sick, given that the test is negative is: 0.9998

(f) When a test is 99% accurate

The accuracy of test is the measure of its sensitivity, prevalence and specificity.

So, when a test is said to be 99% accurate, it means that the test is correct, and the result is usable; irrespective of whether the result is positive or negative.