Last year your professor wanted to study the pattern in class grades. The following table contains a sample of grades collected
from the final exam last semester:
64 80 75 98 75
Using Table and looking at the relative positions of the Mean, Median and Mode for the sample, what can you say about the distribution?
64 80 75 98 75
Using Table and looking at the relative positions of the Mean, Median and Mode for the sample, what can you say about the distribution?
1 answer:
Answer:
mean = 78.4
median = 77.5
mode = 75
This is Right - skewed (positive skewness) distribution
Step-by-step explanation:
<u>Mean:-</u>
The mean (average) is found by adding all of the numbers together and dividing by the number of items and it is denoted by x⁻
mean =
mean (x⁻ ) = 78.4
The mean of the given data = 78.4
<u>Median:</u>
The median is found by ordering the set from lowest to highest and finding the exact middle.
64 ,75, 80, 98
The middle term of the given data set =
<u>Mode :</u>
The mode is the most common repeated number in a data set.
64 ,75, 75, 80, 98
in data the most common number = 75
<u>Conclusion</u>:-
mean = 78.4
median = 77.5
mode = 75
This is Right - skewed (positive skewness) distribution
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For a hyperbola 
where
the directrix is the line
and the focus is at (0, c).
Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
a = √9 = 3
b = √16 = 4
Your hyperbola's constants are ...
a = 3
b = 4
______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
where
the directrix is the line
and the focus is at (0, c).
Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
a = √9 = 3
b = √16 = 4
Your hyperbola's constants are ...
a = 3
b = 4
______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
Answer:
1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.
2) See attached picture.
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
Step-by-step explanation:
1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:
and we set the first derivative equal to zero so we get:
95-18t=0
and solve for t
-18t=-95
t=5.28s
so now we calculate the position of the car after 5.28 seconds, so we get:
so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.
2) For answer 2 I take the second derivative of the function so I get:
j"(t)=-18
and then we graph them. (See attached picture)
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
