Question 1467: Heteroscedasticity means that the variability of y values is larger for some x values than for others. True or False?
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Final answer:
Heteroscedasticity does mean that the variability of y-values is larger for some x values than for others, which is a condition that can impact the efficiency of your estimator and lead to incorrect conclusions in regression analysis.
Explanation:
The statement 'Heteroscedasticity means that the variability of y values is larger for some x values than for others' is True. In the context of regression analysis, heteroscedasticity refers to the variability of the random disturbance (the y-values) being different across elements of an independent variable (the x-values). For instance, the variance of errors might increase or decrease with the level of the dependent variable. This violates the assumption of homoscedasticity in ordinary least squares (OLS) regression, which presumes that the variation around the regression line is the same for all values of the independent variable. The presence of heteroscedasticity could impact the efficiency of your estimator and could lead to incorrect conclusions about the relationship between the dependent and independent variables.
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Answer:
AB-DE
Step-by-step explanation:
hi I m from India nice to meet you bro
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Answer:
The maximum is P=112.4 at (23.4,7)
Step-by-step explanation:
From the graph, the coordinates of the vertices of the feasible region are:
(0,25)
(9,7)
(23.4, 7)
(15,17.5)
Substituting these values in the objective function, P.
At (0,25), P = 6x − 4y=6(0)-4(25)=-100
At (9,7), P = 6x − 4y=6(9)-4(7)=26
At (23.4,7), P = 6x − 4y=6(23.4)-4(7)=112.4
At (15,17.5), P = 6x − 4y=6(15)-4(17.5)=20
Since the objective is to maximize,
The maximum is P=112.4 at (23.4,7)
Final answer:
To solve the linear programming problem, graph the inequalities to find the feasible region, then compute the function P = 6x − 4y at each corner point of the feasible region to find the maximum value. The values of x and y must also uphold all the inequalities.
Explanation:
The subject of the problem is a linear programming problem, and to solve it, we first identify the feasible region by graphing inequalities. This involves graphing x + 2y ≤ 50, 5x + 4y ≤ 145, 2x + y ≥ 25, y ≥ 7, and x ≥ 0. The feasible region would be formed by the area enclosed within those lines.
Next, we find the corner points of the feasible region because, in a linear programming problem, the maximum and minimum always occur at the vertices or corner points. Let's calculate these corner points.
Finally, we evaluate the function P = 6x − 4y at each corner point and find the value of P that would be maximized. It's crucial to remember that the values of x and y must satisfy all the given inequalities.
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Answer:
Step-by-step explanation:
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Not only is there something wrong with the way you've written the equation,
there's no way it could be true even after you fix the little mistake.
First let's divide each side by 2 :
h(t) = 10 - 8t + 16t
This is the equation of the height of a ball that starts from 10-ft above the ground, not 20, begins with 8 ft/sec of downward speed as soon as it's kicked, and what to do with that mysterious ' 16t ' at the end ? Well, if it were ' 16t-squared ', it very well could reflect the acceleration of gravity.
BUT ... since the '10' is positive, we know that the upward direction is the
positive direction for this problem, and then the sign of the acceleration term
can't be positive. That would mean that the ball is accelerating upward at the
rate of 32 ft/sec every second, and if we just wait a few minutes, that ball is on
its way to the moon !
-- If we accept the equation exactly as it's written in the question, then the ball
is kicked from an initial height of 10-ft, it has an upward speed of +8 ft/sec forever,
it never sinks lower than the initial 10-ft, and it has no maximum height.
-- If we make the last term ' 16t² ', then the ball is kicked from an initial height
of 10-ft, the kicker aims down and gives it an initial speed of 8 ft/sec DOWNward,
but it has an upward acceleration of 32 ft/sec every second, The lowest it ever gets
is 1-ft below the balcony, at exactly 0.25 second after the kick, then begins rising,
faster and faster. In this case also, the ball is headed for the moon, and has no
maximum height.
The question needs some serious work.
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he earn: 0.25*1440*12=4320
[because he sells 0.25/each and buy 1440 dozen so 1440*12]
4320-3600=720