Discount Rental Cars charges a daily fee plus a mileage fee for renting its cars. Barney was charged $145.00 for 3 days and 200
miles, while Mary was charged $275.50 for 5 days and 430 miles. What does Discount Rental Cars charge per day and per mile?
1 answer:
Answer:
The charge per day is $25 and the charge per mile is $0.35
Step-by-step explanation:
Let
x -----> the charge per day
y ----> the charge per mile
we know that
3x+200y=145 -----> equation A
5x+430y=275.50 -----> equation B
Solve the system of equations by graphing
Remember that the solution of the system is the intersection point both graphs
using a graphing tool
The intersection point is (25,0.35)
see the attached figure
therefore
The charge per day is $25 and the charge per mile is $0.35
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Answer:
Therefore, the sampling distribution of is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.
The 95% interval estimate of the population mean is
LCL = 7.431 hours to UCL = 10.569 hours
Step-by-step explanation:
Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.
The population standard deviation is
A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.
The sample size is
The sample mean is
Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.
The population mean would be the same as the sample mean that is
The sample standard deviation would be
Where is the population standard deviation and n is the sample size.
Therefore, the sampling distribution of is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.
The population mean confidence interval is given by
Where the margin of error is given by
Where n is the sampling size, s is the sample standard deviation and is the t-score corresponding to a 95% confidence level.
The t-score corresponding to a 95% confidence level is
Significance level = α = 1 - 0.95 = 0.05/2 = 0.025
Degree of freedom = n - 1 = 250 - 1 = 249
From the t-table at α = 0.025 and DoF = 249
t-score = 1.9695
So the required 95% confidence interval is
The 95% interval estimate of the population mean is
LCL = 7.431 hours to UCL = 10.569 hours