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>
3√1024<span>= 10.079368399159
That's your answer.
</span>
Answer:
The playground is 160ft by 360ft.
The model is 4in by 9in
First, 1ft = 12 inches.
Then the measures of the playground, in inches, is:
160*12in = 1920 in
360*12in = 4320in
The playground is 1920in by 4320in.
Then, the ratios between the measures of the playground and the model are:
1920in/4in = 480
4320in/9in = 480
This means that each inch in the model, represents 480 inches in the actual playground.
1 Cancel <span>33</span>
<span>x+\frac{6}{x}+4+x-1<span>x+<span><span>x</span><span>6</span><span></span></span>+4+x−1</span></span>
2 Collect like terms
<span>(x+x)+\frac{6}{x}+(4-1)<span>(x+x)+<span><span>x</span><span>6</span><span></span></span>+(4−1)</span></span>
3 Simplify
<span><span>2x+\frac{6}{x}+3<span>2x+<span><span>x</span><span>6</span><span></span></span>+3</span></span><span>
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