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3 years ago

[(x/x+3)+(3/x+3)+(2/x+3)] sum

2 answers:

8 0

You get the answer x+5/x+3.
This is because all of the fractions have the same denominator, so just add the numerators and put them over x+3.

AysviL [449]3 years ago

3 0

Answer:

$\frac{(x+5)}{(x+3)}$

Step-by-step explanation:

We have been given an expression

$\frac{x}{x+3}+\frac{3}{x+3}+\frac{2}{x+3}$

We need to find the sum of the given expression we will first take the LCM of the denominator which is $(x+3)$ then we get

$\frac{x+3+2}{x+3}$

$\Rightarrow \frac{(x+5)}{(x+3)}$

Therefore the required sum that is sum of the expression is

$\frac{(x+5)}{(x+3)}$

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2 years ago

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Step-by-step explanation:

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The National Safety Council publishes information about automobile accidents in Accident Facts. According to that document, the

vladimir1956 [14]

Answer:

$P(X = x) = C_{8,x}.(0.4)^{x}.(0.6)^{8-x}$

In which x is the number of which we want to find the probability.

Step-by-step explanation:

For each traffic fatality, there are only two possible outcomes. EIther it involved an intoxicated or alcohol-impaired driver or nonoccupant, or it didn't. Traffic fatalities are independent of other traffic fatalities, which means that 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.

The probability is .40 that a traffic fatality involves an intoxication or alcohol-impaired driver or nonoccupant.

This means that $p = 0.4$

Eight traffic fatalities

This means that $n = 8$

Find the probability that the number which involve an intoxicated or alcohol-impaired driver or nonoccupant is

This is P(X = x), in which x is the number of which we want to find the probability. So

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

$P(X = x) = C_{8,x}.(0.4)^{x}.(0.6)^{8-x}$

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3 years ago