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>
20x + 15y = 2000
x + y = 115
55+60=115
20 (55) + 15 (60)=2000
1100+900=2000
2000=2000
Answer:
i think it's x=1
Step-by-step explanation:
To solve 4(x-2)^2 = 16
divide each side by 4 to have (x-2)^2 = 4
take the square root of both sides
x-2 = -2 or x-2 = 2
add 2 to each side
x = 0 or x = 4
4 ( 0-2)^2 = 16
4 (-2)^2 = 16
4 (4) = 16 true
4 ( 4-2)^2 = 16
4 (2)^2 = 16
4 (4) = 16 true.
the values for x are 0 and 4.
