The greatest number can be as large as 81.
<u>Step-by-step explanation:</u>
Given that,
- A set of five different positive integers has a mean of 33 and a median of 40.
- We need to find the set of five different positive integers.
We already know that,
- The term "median" is the middle term which is 40.
- Therefore, if you do not include 0 in positive integers, then the first two positive integers below the median value of 40 to be as low as possible are 1 and 2.
- The median 40 will be the third positive integer of the set.
- Therefore, the fourth positive integer should be the next lowest possible value of 40 which is 41.
With simple algebra you can figure out the last greater number.
-
The set of five different positive integers is given as {1,2,40,41,x}.
- Let, x be the last greater number in the set.
The term "mean" is defined as the sum of all the integers in the set divided by the number of integers in the set.
⇒ Mean = (1+2+40+41+x) / 5
⇒ 33 = (84+x) / 5
⇒ 33×5 = 84 + x
⇒ 165 - 84 = x
⇒ 81 = x
∴ The greatest number can be as large as 81.
The system of equation
y=8x and y=2x
To solve this using graphical method
The solution is the point of intersection of the two-lines
As shown from the graph above, the solution is (0,0)
Certin, because they can be any two cards.
Sort the data from the smallest one.
42, 55, 69, 74, 78, 83, 83, 97, 99
There are 9 data, n = 9
Find the location of Q₂
Q₂ location = (n + 1)/2
Q₂ location = (9 + 1)/2
Q₂ location = 10/2
Q₂ location = 5
The location of Q₂ is in the fifth order
42 first order
55 second order
69 third order
74 fourth order
78 fifth order
etc
So the Q₂ or the median is 78