The quadratic equation y = –6x2 + 100x – 180 models the store’s daily profit, y, for selling soccer balls at x dollars. The quad
ratic equation y = –4x2 + 80x – 150 models the store’s daily profit, y, for selling footballs at x dollars. Use a graphing calculator to find the intersection point(s) of the graphs, and explain what they mean in the context of the problem
2 answers:
Answer:
Step-by-step explanation:
The quadratic equation models the store’s daily profit, y, for selling soccer balls at x dollars.
The quadratic equation models the store’s daily profit, y, for selling footballs at x dollars.
Now we are supposed to find the intersection point(s) of the graphs using graphing calculator
Refer the attached graph.
-- Green line
-- Purple line
So, from the graph we can see there are two intersection points :
(8.333,236.667)
(2.053,0)
So, from solution (8.333,236.667) we depict that at x = 8.333 dollars the store's daily profit y = 236.667 for selling football or soccer ball.
From solution (2.053,0)we depict that at x = 0 dollars the store's daily profit y = 2.053 for selling football or soccer ball.
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Answer: 1.8%
Step-by-step explanation:
Answer:
(C) The slope of the regression equation is not significantly different from zero
Step-by-step explanation:
Let's suppose that we have the following linear model:
Where Y is the dependent variable and X the independent variable. represent the intercept and
the slope.
In order to estimate the coefficients we can use least squares estimation.
If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:
Null Hypothesis:
Alternative hypothesis:
Or in other wouds we want to check is our slope is significant.
In order to conduct this test we are assuming the following conditions:
a) We have linear relationship between Y and X
b) We have the same probability distribution for the variable Y with the same deviation for each value of the independent variable
c) We assume that the Y values are independent and the distribution of Y is normal
The significance level is provided and on this case is
The standard error for the slope is given by this formula:
Th degrees of freedom for a linear regression is given by since we need to estimate the value for the slope and the intercept.
In order to test the hypothesis the statistic is given by:
The confidence interval for the slope would be given by this formula:
And using the last formula we got that the confidence interval for the slope coefficient is given by:
IF we analyze the confidence interval contains the value 0. So we can conclude that we don't have a significant effect of the slope on this case at 5% of significance. And the best option would be:
(C) The slope of the regression equation is not significantly different from zero
Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>Step-by-step explanation:
Given the function j(x) = 11.6 and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6 and k(x) =
are inverse functions.