Answer:
+
Step-by-step explanation:
That would be called the quotient.
Answer:
Jayce should not use either option. Option 1 is likely to be based so that paperbacks are overrepresented, while Option 2 is likely to be biased so that e-books are overrepresented.
Step-by-step explanation:
<em>Given:</em>
<em>Jayce volunteers for the local library. The librarian wants to find out whether the patrons prefer paperback books or e-books. Jayce cannot decide which method to use for polling the patrons.</em>
<em>Option 1: Poll every fifth patron who enters the library on Paperback Lovers Day.</em>
<em>Option 2: Poll every third patron who enters the library on Technology Appreciation Day.</em>
<em />
<em>Since, Option 1: Poll every fifth patron who enters the library on Paperback Lovers Day. It biased because since it paperback lovers day thus, it would be likely that most people like the paperback which make it biased.</em>
<em>Since, Option 2: Poll every third patron who enters the library on Technology Appreciation Day. It biased because since it technology appreciation day thus, it would be likely that most people don't care about these stuff since people coming in are for technology appreciation.</em>
<em />
<em />
<em>Therefore, the answer is:</em>
Jayce should not use either option. Option 1 is likely to be based so that paperbacks are overrepresented, while Option 2 is likely to be biased so that e-books are overrepresented.
<u><em>Kavinsky</em></u>
Answer:
B) Angle 2 and Angle 3
Step-by-step explanation:
Angle 2 and Angle 3 are equal since they're vertical angles and aren't supplementary
Angle 7 and 8 are supplementary
Angle 1 and 7 are supplementary
Honestly, i have no clue with angle 4 and angle 4. I just know that vertical angles aren't supplementary because they're equal to each other so I'll assume this is the case as well.
These ordered pairs<span> are in the </span>solution set<span> of the equation </span>x<span> > </span>y. ... (2<span>, </span>0<span>). </span>3(2<span>) + </span>2(0<span>) ≤ 6. 6 + </span>0<span> ≤ 6. 6 ≤ 6. (</span>4, −1<span>). </span>3(4<span>) + </span>2(−1) ≤ 6. 12 + (−2<span>) ≤ 6 ... </span>3<span>). </span>
