Jamal works in a city and sometimes takes a taxi to work. The taxicabs charge $1.50 for the first 1/5 mile and $0.25 for each ad
2 answers:
Answer:
9 miles
Step-by-step explanation:
Jamal works in a city and sometimes takes a taxi to work.
The taxicabs charge $1.50 for the first mile and $0.25 for each additional
mile.
Let Jamal can travel by taxi = x miles
the charges of x miles = 0.25x
He has $3.75 in his pocket.
So the equation to represent the distance he can travel by taxi will be
1.50 + 0.25x = 3.75
0.25x = 3.75 - 1.50
0.25x = 2.25
x =
x = 9 miles
He can travel 9 miles by taxi.
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Answer:
We accept the null hypothesis and the population mean is $120.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 100
Sample mean, = $120
Alpha, α = 0.01
Sample standard deviation, s = $25
First, we design the null and the alternate hypothesis
We use two-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have
p-value one tail= 0.024
p-value two tail= 0.048
Conclusion:
Since the p-value for two tailed test is greater than the significance level, we fail to reject the null hypothesis and accept it.
Thus, the population mean is $120.
Check the picture below.
Answer:
Step-by-step explanation:
Let us record the station number 1, 2 or 3 for each family member A, B or C.
I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.
From table it is clear that the total outcomes possible are 27.
We know that, probability can be defined as,
a) All Members Assigned to the Same Station.
Cases for all members being assigned to same station are as follows:
[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).
Therefore,
b) At Most Two Members Assigned to the Same Station.
It means that maximum of 2 members can have the same station. Cases for this situation are as follows:
[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],
[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]
(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).
Therefore,
c) All Members Assigned to a Different Station.
For this scenario, we have the following results:
[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).
Therefore,
