4+3+2=9 then 189 divided by 9 = 21 divided by 3 = 7
1 answer:
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Answer:
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.
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<em>Comment on the question</em>
IMO, the use of <em>70 trials</em> is coincidentally the same number as the first <em>70 days</em> 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.
Total discount for 10 group tickets = 10 · 0.05t
Amount of discount per ticket = 0.05t
Number of tickets = 10
Cost of buying 10 single tickets = 10t
Solution:
Cost for single ticket = t dollars
Discount percent per ticket = 5%
Number of tickets bought = 10
Total discount for 10 group tickets = Number of tickets × Discount amount
= 10 × 0.05t
Total discount for 10 group tickets = 10 · 0.05t
Amount of discount per ticket = 5%t
= t
= 0.05t
Amount of discount per ticket = 0.05t
Number of tickets = 10
Cost of buying 10 single tickets = 10t