A plumber charges 25$ for a service call plus 50$ per hour of service.write an equation for the cost,C,after h hours of service.
1 answer:
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<h3>Answer:</h3>
- g(x) has an axis of symmetry at x = 3
- g(x) is shifted right 3 units from the graph of f(x)
- g(x) is shifted up 4 units from the graph of f(x)
The vertex form of g(x) is ...
... g(x) = -(x -3)² +4
This is offset to the right by 3 and up by 4 from the parent function. (It is also first reflected across the x-axis.)
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<em>Vertex form</em>
You know the leading coefficient is -1 because that's what it is for x² in the given form. When you factor -1 from the first two terms, of the given form, you have ...
... g(x) = -1(x² -6x) -5
Half the x coefficient inside parentheses will be the constant in the squared binomial term, so that term is (x -3)². The constant in that square is +9, so adding that value inside and outside parentheses in g(x) gives ...
... g(x) = -1(x² -6x +9) -5 +9
... g(x) = -(x -3)² +4 . . . . . vertex form
_____
<em>About transformations</em>
g(x) = f(x -a) causes the graph of f(x) to be shifted "a" units to the right. For a function f(x) with an axis of symmetry at x=0, it moves the axis of symmetry to x=a.
g(x) = f(x) +a causes the graph of f(x) to be shifted "a" units up.
g(x) = -f(x) causes the graph of f(x) to be reflected across the x-axis.
Here, we have all three of these transformations. First is the reflection:
... f₁(x) = -f(x) = -x²
Then we have shifting to the right 3 units. (also moves the axis of symmetry)
... f₂(x) = f₁(x-3) = -(x -3)²
Finally, we have shifting up 4 units.
... g(x) = f₂(x) +4 = -(x -3)² +4
Answer:
a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).
b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).
c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).
d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.
Step-by-step explanation:
a) Develop a 90% confidence interval estimate of the population mean monthly rent.
Our sample size is 120.
The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So
Then, we need to subtract one by the confidence level and divide by 2. So:
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have .
Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So
Now, we multiply T and s
The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.
The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.
The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).
b) Develop a 95% confidence interval estimate of the population mean monthly rent.
Now we have that
So
With 119 and 0.025 in the t-distribution table, we have .
The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.
The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.
The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).
c) Develop a 99% confidence interval estimate of the population mean monthly rent.
Now we have that
So
With 119 and 0.025 in the t-distribution table, we have .
The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.
The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.
The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).
d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.
The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.