Travis has decided to budget his spending money. He can spend no more than $143.15 every month. He has also decided to spend 5.2
times as much money on video games as he spends on movies. What linear inequality describes this?
1 answer:
Answer:
Let the amount of money Travis spends on movies be $.
Then the amount of money spent on video games will be .
We were told that Travis can spend no more than $.
The key word no more than means less than or equal to.
This means that, when we add the total money spent on movies and video games we should get something less or equal to $.
Hence, the inequality that describes the problem is given by;
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Answer:
With 99 % confidence, it can be said that the population mean driving distance to work (in miles) is between the interval's endpoints [19.91 miles, 31.49 miles] .
Step-by-step explanation:
<u>The complete question is</u>: In a random sample of six people, the mean driving distance to work was 25.7 miles and the standard deviation was 6.7 miles. Assuming the population is normally distributed and using the t-distribution, a 99% confidence interval for the population mean mu is left parenthesis 14.7 comma 36.7 right parenthesis (and the margin of error is 11.0).
Through research, it has been found that the population standard deviation of driving distances to work is 5.5 . Using the standard normal distribution with the appropriate calculations for a standard deviation that is known, find the margin of error and construct a 99 % confidence interval for the population mean mu .
Interpret the results. Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a decimal. Do not round.)
A. nothing % of all random samples of six people from the population will have a mean driving distance to work (in miles) that is between the interval's endpoints.
B. With nothing % confidence, it can be said that most driving distances to work (in miles) in the population are between the interval's endpoints.
C. It can be said that nothing % of the population has a driving distance to work (in miles) that is between the interval's endpoints.
D. With nothing % confidence, it can be said that the population mean driving distance to work (in miles) is between the interval's endpoints.
We are given that in a random sample of six people, the mean driving distance to work was 25.7 miles and the standard deviation was 6.7 miles.
Through research, it has been found that the population standard deviation of driving distances to work is 5.5 .
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = ~ N(0,1)
where, = sample mean driving distance to work = 25.7 miles
= population standard deviation = 5.5 miles
n = sample of people = 6
= population mean driving distance to work
<em>Here for constructing a 99% confidence interval we have used a One-sample z-test statistics because we know about the population standard deviation. </em>
<u>So, 99% confidence interval for the population mean, </u><u> is ;
</u>
P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level
of significance are -2.58 & 2.58}
P(-2.58 < < 2.58) = 0.99
P( <
<
) = 0.99
P( <
<
) = 0.99
<u>99% confidence interval for</u> = [
,
]
= [ ,
]
= [19.91, 31.49]
Therefore, a 99% confidence for the population mean is [19.91, 31.49] .
The margin of error here is =
= = 5.793
With 99 % confidence, it can be said that the population mean driving distance to work (in miles) is between the interval's endpoints [19.91, 31.49] .