1) The graph
The corresponding graph shows a growing curve, its shape is kind of the right half of a parabola that opens upward and starts at the point (0, 2500).
The vertical axis corresponds to C(j), it contains divisions of 2500 units, and are marked 2500, 5000, 7500, 10000, 12500, 15000 and 17500.
The horizontal axis corresponds to j, and the marks are 75, 150, 225, 300, 375, and 450.
2) Domain
Domain is the set of possible values for the independent variable, which is placed on the horizontal axis. This is the possible values of j.
They are all the positive numbers and zero, the the domain is:
All real numbers, j, such that j ≥ 0
3) Range
Range is the set of possible images (dependent variable); this is the possible values of C(j).
As you can see on the graph C(j) ≥ 2500
Then, the range is [2500, ∞).
<span>Answer: 18/25 ; or, write as: 0.72 .
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Explanation:
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9/20 * 4/5 = (</span>9*4) / (10*5) = 36/50 ;
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36/50 = </span>(36÷2) / (50÷2) = 18/25 ; or, write as: 0.72 .
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Answer:
Average annual tuition is $16991 for the year 2020.
Step-by-step explanation:
By using excel for the equation of the line of best fit,
Slope of the line from given data = 380.0286
y - intercept of the line = 11290.1
Therefore, equation of the regression line will be,
y = 380.0286x + 11290.1
Where x = number of years after 2005
Now we have to calculate the average annual tuition in the year 2020,
By substituting the value of x = 2020 - 2005
= 15 years
y = 380.0286×(15) + 11290.1
y = 16990.529
≈ 16991
Therefore, average annual tuition is $16991 for the year 2020.
<h2><u>
Answer with explanation</u>
:</h2>
Let
be the distance traveled by deluxe tire .
As per given , we have
Null hypothesis : 
Alternative hypothesis : 
Since
is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.
Test statistic : 
where, n= sample size
= sample mean
= Population mean
=sample standard deviation
For
, we have

By using z-value table,
P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23) [∵P(Z≤-z)=1-P(Z≤z)]
=1-0.9871=0.0129
Decision : Since p value (0.0129) < significance level (0.05), so we reject the null hypothesis .
[We reject the null hypothesis when p-value is less than the significance level .]
Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.