All categories

3 years ago

What is the difference? (2x+5)/(x^2-3x)-(3x+5)/x^3-9x)- (x+1)/x^2-9)

Mathematics

2 answers:

Gre4nikov [31]3 years ago

7 0

Answer: Option A is correct $\frac{(x+2)(x+5)}{(x^{3}-9)}$

Explanation:

Given equation $\frac{2x+5}{x^{2} -3x}$ $- \frac{3x+5}{x^{3} -9x}$ $- \frac{x+1}{x^{2} -9}$

will become as below after taking out the lcm x(x-3)(x+3)

$\frac{2x+5}{x(x-3)} -\frac{3x+5}{x(x-3)(x+3)} -\frac{x+1}{(x-3)(x+3)}$

after simplifying

we will get $\frac{(2x+5)(x+3)-(3x+5)-x(x+1)}{x(x-3)(x+3)}$

After further simplification we will get

$\frac{2x^{2} +11x+15-3x-5-x^{2}-x}{x( x^{2}-9)}$

On more simplification we will get

$\frac{x^{2} +7x+10}{x(x^{2}-9)}$

finally after simplification we will get

$\frac{(x+2)(x+5)}{x(x^{2}-9)}$

which will lead to the final result

$\frac{(x+2)(x+5)}{x^{3}-9x}$

professor190 [17]3 years ago

3 0

A)(x+5)(x+2)/(x^3-9x)

You might be interested in

The phone lines to an airline reservation system are occupied 40% of the time. Assume that the events that the lines are occupie

nikitadnepr [17]

Answer:

a) $P(X=3)$

And we can use the probability mass function and we got:

$P(X=3) = (10C3) (0.4)^3 (1-0.4)^{10-3} = 0.215$

b) For this case we want this probability:

$P(X \geq 1)$

and the probability that the one call would be not occupied is 1-0.4=0.6

And if we use the complement rule we got:

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

$P(X=0) = (10C0) (0.6)^0 (1-0.6)^{10-0} = 0.000105$

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

c) $E(X) = np = 10*0.4 = 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

a. What is the probability that for exactly three calls, the lines are occupied?

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.4)$