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>
Looking at this in terms of sets, let's call O the set of all owls, and F the set of all things that can fly. What this original statement is saying every animal that's a member of the set of all owls is also a member of the set of all things that can fly, or in other words, O⊂F (O is a subset of F). Negating this tells us that, while there's <em>at least one</em> element of O that also belongs to F, O is not contained entirely in F (O⊆F, in notation), so a good negation or our original statement might be:
<em>Not all owls can fly.</em>
Answer:
y = 9
Step-by-step explanation:
1) Multiply both sides by 3.
3 × 3 = y
2) Simplfiy 3 × 3 to 9.
9 = y
3) Switch sides.
9
Simplify means solve and to solve we first need to do 2 to the 2nd power which is two * 2 that equal 4 than do 2 to the 10th power that equal 1024 and than we can just do
4 divided by 1024 = 0.0390625
