(11/x^2-25)+(4/x+5)= 3/x-5

1 answer:

8 0

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\
\cfrac{11}{x^2-25}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\implies \cfrac{11}{x^2-5^2}+\cfrac{4}{x+5}=\cfrac{3}{x-5}$

$\bf \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\impliedby 
\begin{array}{llll}
\textit{notice, LCD is }(x-5)(x+5)\\
\textit{so let's multiply all by it}\\
\textit{to toss the denominators}
\end{array}$

$\bf (x-5)(x+5)\left( \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5} \right)=(x-5)(x+5)\left( \cfrac{3}{x-5} \right)
\\\\\\
11+4(x-5)=3(x+5)\implies 11+4x-20=3x+15
\\\\\\
4x-9=3x+15\implies x=24$

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A student has a 30% chance of receiving chocolate ice cream for dessert at lunch. A statistics class designs a simulation to app

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<h3>How to calculate the probability of an event?</h3>

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

For the considered case, it is given that:

Probability of an student getting Chocolate Ice Cream for dessert at lunch = 30% = 0.3

The count of the sections the class should label as "Chocolate Ice Cream" should be such that spinner getting "Chocolate Ice Cream" as option should have the probability as 0.3

Let the count of the sections the class should label as "Chocolate Ice Cream" be $x$