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2 years ago

Describe the error made in subtracting the two rational expressions shown

Mathematics

2 answers:

Andru [333]2 years ago

8 0

Answer:

The minus sign between the two fractions was not distributed correctly.

Step-by-step explanation:

The error made in this solution of rational fractions' subtraction is that the minus sign between the two terms will change the signs of numerator of second fraction so the numerator in the answer will be 3.

The correct solution is:

$\frac{1}{x-2} - \frac{1}{x+1}\\=\frac{x+1}{(x-2)(x+1)} - \frac{x-2}{(x-2)(x+1)}\\=\frac{x+1-(x-2)}{(x+1)(x-2)}\\=\frac{x+1-x+2}{(x+1)(x-2)}\\=\frac{3}{(x+1)(x-2)}$

Hence,

The error was in the last step of subtraction.

aev [14]2 years ago

6 0

Answer:

You must subtract all terms of the numerator being subtracted out, not just the first term.

The negative 2 should have been subtracted out to get a numerator of x+1 - x+2

The simplified numerator of the difference should be 3, not –1.

Step-by-step explanation:

got it right on edge 2020

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What is the answer to page 156-157 on chapter 4 proportions relationship.

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3 years ago

Does (2,0) work as a solution to the systems of equations to lines 3x+y=6 and 3x-y=6

fenix001 [56]

Answer:

Yes

Step-by-step explanation:

To see if (2,0) works as a solution to the systems of equations, we plug in the values of x and y and simplify. If the results are equal, then (2,0) is a solution.

3x + y = 6:

$3(2) + 0 = 6$
$6 + 0 = 6$
$6 = 6$
(2,0) is a solution to this equation.

3x - y = 6:

$3(2) - 0 = 6$
$6 - 0 = 6$
$6 = 6$
(2,0) is a solution to this equation.

Therefore, the answer is yes, it does work as a solution.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

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2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. 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.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

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