If there is 2 field trips with 23 students and each student costs $5 dollars. So for the 1st trip it will be 23x$5
1st trip- $115
And on the second trip its the same because there is 23 students that cost 5 dollars so it will be 23x5
2nd trip- $115
To find out the total of both trips youd add 115+115/
In total, for both trips it will cost $230 dollars.
u need to finish 4 assignments in 1 hour
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)
Answer:
5:5 (first box, pencils to pens)
7:3 (second box, coloured pencils to crayons)
The probability of picking a pen (1st box): 5/10
The probability of picking a crayon (2nd box): 3/10
Probability of picking both: 5/10*3/10 = 15/100