Natalie added 20 pieces of candy to a bowl. Two days later, half of the candy in the bowl is gone. If there are 14 pieces left i
1 answer:
Answer:
8 pieces of candy
Step-by-step explanation:
Take the original amount of candy in the bowl to be .
It says that 20 pieces of candy were added, therefore there are pieces of candy in the bowl.
Half of the candy is gone, therefore pieces of candy are left in the bowl.
After half of the candy is gone, there are 14 pieces left, so we can say that
Solve for :
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Answer:
1.02% probability of spending 0 dollars on coffee over the course of a five day week
7.68% probability of spending 3 dollars on coffee over the course of a five day week
23.04% probability of spending 6 dollars on coffee over the course of a five day week
34.56% probability of spending 9 dollars on coffee over the course of a five day week
25.92% probability of spending 12 dollars on coffee over the course of a five day week
7.78% probability of spending 12 dollars on coffee over the course of a five day week
Step-by-step explanation:
For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
In this problem we have that:
The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so
The week has 5 days, so
She spends 3 dollars on coffee each day the train is on time.
Probabability that she spends 0 dollars on coffee:
This is the probability of the train being late all 5 days, so it is P(X = 0).
1.02% probability of spending 0 dollars on coffee over the course of a five day week
Probabability that she spends 3 dollars on coffee:
This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).
7.68% probability of spending 3 dollars on coffee over the course of a five day week
Probabability that she spends 6 dollars on coffee:
This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).
23.04% probability of spending 6 dollars on coffee over the course of a five day week
Probabability that she spends 9 dollars on coffee:
This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).
34.56% probability of spending 9 dollars on coffee over the course of a five day week
Probabability that she spends 12 dollars on coffee:
This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).
25.92% probability of spending 12 dollars on coffee over the course of a five day week
Probabability that she spends 15 dollars on coffee:
Probability that the subway is on time all days of the week, so .
7.78% probability of spending 12 dollars on coffee over the course of a five day week