What is the upper quartile, Q3, of the following data set? 54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41
scZoUnD [109]
The original data set is
{<span>54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41}
Sort the data values from smallest to largest to get
</span><span>{38, 41, 43, 46, 48, 52, 53, 54, 55, 56, 60, 62, 65, 67, 70}
</span>
Now find the middle most value. This is the value in the 8th slot. The first 7 values are below the median. The 8th value is the median itself. The next 7 values are above the median.
The value in the 8th slot is 54, so this is the median
Divide the sorted data set into two lists. I'll call them L and U
L = {<span>38, 41, 43, 46, 48, 52, 53}
U = {</span><span>55, 56, 60, 62, 65, 67, 70}
they each have 7 items. The list L is the lower half of the sorted data and U is the upper half. The split happens at the original median (54).
Q3 will be equal to the median of the list U
The median of U = </span>{<span>55, 56, 60, 62, 65, 67, 70} is 62 since it's the middle most value.
Therefore, Q3 = 62
Answer: 62</span>
Answer:
2.63
Step-by-step explanation:
29/11=2.63
Answer:
20 masks and 100 ventilators
Step-by-step explanation:
I assume the problem ask to maximize the profit of the company.
Let's define the following variables
v: ventilator
m: mask
Restictions:
m + v ≤ 120
10 ≤ m ≤ 50
40 ≤ v ≤ 100
Profit function:
P = 10*m + 65*v
The system of restrictions can be seen in the figure attached. The five points marked are the vertices of the feasible region (the solution is one of these points). Replacing them in the profit function:
point Profit function:
(10, 100) 10*10 + 65*100 = 6600
(20, 100) 10*20 + 65*100 = 6700
(50, 70) 10*50 + 65*70 = 5050
(50, 40) 10*50 + 65*40 = 3100
(10, 40) 10*10 + 65*40 = 2700
Then, the profit maximization is obtained when 20 masks and 100 ventilators are produced.
