The second side a triangle is 4 inches longer than the first side. The third side of the triangle is 3 inches longer than the se

cond side. If the perimeter of the triangle is 101 inches, find the length of the third side of the triangle.

Mathematics

1 answer:

ozzi3 years ago

8 0

$\begin{cases} \stackrel{\textit{1st side}}{a}\\[1em] \stackrel{\textit{2nd side}}{a + 4}\\[1em] \stackrel{\textit{3rd side}}{(a+4)+3} \end{cases}\qquad \stackrel{\textit{the perimeter or nam}\textit{ely their sum}}{(a)+(a+4)+[(a+4)+3] ~~ =~~101} \\\\\\ (a)+(a+4)+a+7=101\implies 3a+11=101\implies 3a=90 \\\\\\ a=\cfrac{90}{3}\implies a = 30~\hfill \stackrel{\textit{the 3rd side is}}{(30+4)+3\implies 37}$

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see the attachment

Step-by-step explanation:

We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).

Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.

That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".

The simulation of 70 different choices of a random class is shown in the attachment.

_____

Comment on the question

IMO, the use of 70 trials is coincidentally the same number as the first 70 days of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.