Answer:
theres no question so we cant answer nothing
Answer:
Step-by-step explanation:
Answer:
![P( eleventh\: \: |\: \: FT)= 0.2453 \\\\P( eleventh\: \: |\: \: FT)= 24.53\% \\\\](https://tex.z-dn.net/?f=P%28%20eleventh%5C%3A%20%5C%3A%20%7C%5C%3A%20%5C%3A%20FT%29%3D%20%200.2453%20%5C%5C%5C%5CP%28%20eleventh%5C%3A%20%5C%3A%20%7C%5C%3A%20%5C%3A%20FT%29%3D%20%2024.53%5C%25%20%5C%5C%5C%5C)
Step-by-step explanation:
We are given a joint probability table.
There are four different graders in a school
1. Grade Ninth
2. Grade Tenth
3. Grade Eleventh
4. Grade Twelfth
Field trip refers to the students who will attending the amusement park field trip.
No field trip refers to the students who will not be attending the amusement park field trip.
We want to find out the probability that the selected student is an eleventh grader given that the student is going on a field trip.
![P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)}](https://tex.z-dn.net/?f=P%28%20eleventh%5C%3A%20%5C%3A%20%7C%5C%3A%20%5C%3A%20FT%29%3D%20%5Cfrac%7BP%28%20eleventh%5C%3A%20%5C%3A%20and%20%5C%3A%20%5C%3A%20FT%29%7D%7BP%28FT%29%7D)
Where P(eleventh and FT) is the probability of students who are in eleventh grade and will be going to field trip
![P(eleventh\: and \: FT) = \frac{13}{92} \\\\P(eleventh \: and \: FT) = 0.1413](https://tex.z-dn.net/?f=P%28eleventh%5C%3A%20%20and%20%5C%3A%20FT%29%20%3D%20%5Cfrac%7B13%7D%7B92%7D%20%20%5C%5C%5C%5CP%28eleventh%20%5C%3A%20%20and%20%5C%3A%20FT%29%20%3D%200.1413)
Where P(FT) is the probability of students who will be going to field trip
![P(FT) = \frac{12}{92} + \frac{9}{92} + \frac{13}{92} + \frac{19}{92}\\\\P(FT) = 0.1304 + 0.0978 + 0.1413 + 0.2065\\\\P(FT) = 0.576 \\\\](https://tex.z-dn.net/?f=P%28FT%29%20%3D%20%5Cfrac%7B12%7D%7B92%7D%20%2B%20%5Cfrac%7B9%7D%7B92%7D%20%2B%20%5Cfrac%7B13%7D%7B92%7D%20%2B%20%5Cfrac%7B19%7D%7B92%7D%5C%5C%5C%5CP%28FT%29%20%3D%200.1304%20%2B%200.0978%20%2B%200.1413%20%2B%200.2065%5C%5C%5C%5CP%28FT%29%20%3D%200.576%20%5C%5C%5C%5C)
So the required probability is
![P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)} \\\\P( eleventh\: \: |\: \: FT)= \frac{0.1413}{0.576} \\\\P( eleventh\: \: |\: \: FT)= 0.2453 \\\\P( eleventh\: \: |\: \: FT)= 24.53\% \\\\](https://tex.z-dn.net/?f=P%28%20eleventh%5C%3A%20%5C%3A%20%7C%5C%3A%20%5C%3A%20FT%29%3D%20%5Cfrac%7BP%28%20eleventh%5C%3A%20%5C%3A%20and%20%5C%3A%20%5C%3A%20FT%29%7D%7BP%28FT%29%7D%20%5C%5C%5C%5CP%28%20eleventh%5C%3A%20%5C%3A%20%7C%5C%3A%20%5C%3A%20FT%29%3D%20%5Cfrac%7B0.1413%7D%7B0.576%7D%20%5C%5C%5C%5CP%28%20eleventh%5C%3A%20%5C%3A%20%7C%5C%3A%20%5C%3A%20FT%29%3D%20%200.2453%20%5C%5C%5C%5CP%28%20eleventh%5C%3A%20%5C%3A%20%7C%5C%3A%20%5C%3A%20FT%29%3D%20%2024.53%5C%25%20%5C%5C%5C%5C)
Let
x---------> the length of the sides that have the same length in meters
we know that
the perimeter of the figure is equal to
![45.56=(24.2+2x)](https://tex.z-dn.net/?f=45.56%3D%2824.2%2B2x%29)
<u>solve for x</u>
![2x=45.56-24.2](https://tex.z-dn.net/?f=2x%3D45.56-24.2)
![x=21.36/2](https://tex.z-dn.net/?f=x%3D21.36%2F2)
![x=10.68\ m](https://tex.z-dn.net/?f=x%3D10.68%5C%20m)
therefore
<u>the answer is</u>
Each of the same length side is
long