There is an association because the value 0.15 is not similar to the value 0.55
For the nutritionist to determine whether there is an association between where food is prepared and the number of calories the food contains, there must be an association between two categorical variables.
The conditions that satisfy whether there exists an association between conditional relative frequencies are:
1. When there is a bigger difference in the conditional relative frequencies, the stronger the association between the variables.
2. When the conditional relative frequencies are nearly equal for all categories, there may be no association between the variables.
For the given conditional relative frequency, we can see that there exists a significant difference between the columns of the table in the picture because 0.15 is significantly different from 0.55 and 0.85 is significantly different from 0.45
We can conclude that there is an association because the value 0.15 is not similar to the value 0.55
The volume of such a can with base radius
and height
would be
We desire the base's circumference and the can's height to add to 120, i.e.
Substituting this into
allows us to reduce the volume to a function of a single variable
:
Taking the derivative with respect to
yields
Set this equal to 0 and find any critical points:
This suggests the can will have maximum volume when its radius is
cm, which would give a volume of about 20,371 sq. cm.
Answer: y-7 = 4(x+3), choice B
============================================
Point slope form is generally
y-y1 = m(x-x1)
we have m as the slope and (x1,y1) as the point this line goes through. In this case,
m = 4
(x1,y1) = (-3,7) so x1 = -3 and y1 = 7
So,
y-y1 = m(x-x1)
y - 7 = 4(x - (-3))
y - 7 = 4(x + 3)
which is why choice B is the answer
I'll walk you through the first one, and then you should be able to do the rest.
The first step is to right out your numbers from least to greatest
Problem #1: 2,4,5,6,8,10,13,17,19,20
To find the MEDIAN, you're simply crossing out a number from each end until you meet in the middle. In this case, you have an even number of data, which means once you get to the middle, you're have to find the average.
In this problem, your two middle numbers are 8 & 10. Since only one number can be the median, you add them together, and divide by 2:
8+10=18 18/2=9 < this is your middle, or MEDIAN
Next, your first and third quartiles-- they're the median of the lower half and data, and upper half.
For the lower quartile, find the mean of 2,4,5,6, and 8. again, cross out one from each side until you get to the middle. FIRST QUARTILE = 5
Do the same process for the upper half of data (10,13,17,19 & 20).
THIRD QUARTILE = 17
The MIN is the lowest number of data = 2
The MAX is the highest number of data = 20
Best of luck!
