Sergio and Lizeth have a very tight vacation budget. They plan to rent a car from a company that charges $75 a week plus $0.25 a
1 answer:
Answer: 500 miles
Step-by-step explanation:
Given : Sergio and Lizeth have planned to rent a car from a company that charges $75 a week plus $0.25 a mile.
i.e. Fixed charge= $75
Rate per mile = $0.25
Let x denotes the number of miles.
Then, Total charges = Fixed charge+ Rate per mile x No. of miles traveled
= $75+ $0.25x
To keep budget within $200, we have following equation.
Hence, they can travel 500 miles and still keep within their $200 budget.
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Answer:
One can be 99% confident the true mean shell length lies within the above interval.
The population has a relative frequency distribution that is approximately normal.
Step-by-step explanation:
We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.
The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;
P.Q. = ~
where, = sample mean shell length = 51.3 cm
s = sample standard deviation = 6.6 cm
n = sample of turtles = 6
= true mean shell length
Now, the 99% confidence interval for =
Here, = 1% so
= 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.
<u>So, 99% confidence interval for</u> =
= [51.3 - 10.864 , 51.3 + 10.864]
= [40.44 cm, 62.16 cm]
The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].
The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;
C. The population has a relative frequency distribution that is approximately normal.
This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.