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>
C
That’s the answer to your question .
Answer:
(-2, -4)
Step-by-step explanation:
You can complete the square of the equation to get
y+(4/2)^2 = x^2+4x+(4/2)^2
y+4 = x^2 + 4x + 4
y+4 = (x+2)^2
y = (x+2)^2 - 4
This gives the form y = a(x-h)^2 + k where (h, k) is the vertex of the equation. You can also arrive at the same conclusion by making some observations of the equation. (x+2)^2 minimum value is going to be 0 since and negative values resulting from x+2 is going to become positive because of the square. So the minimum value is when x+2 is 0 or when x is equal to -2 and when it's at that minimum value of 0 it's going to have 4 subtracted from it which gives the vertex of (-2, -4)
Answer: x=1
-3x=x-6+2x
-6x=-6
x=1
