if we take 64 to be the 100%, how much is 6¼% off of it?
![\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 64&100\\ x&6\frac{1}{4} \end{array}\implies \cfrac{64}{x}=\cfrac{100}{6\frac{1}{4}}\implies \cfrac{64}{x}=\cfrac{\frac{100}{1}}{\frac{25}{4}}\implies \cfrac{64}{x}=\cfrac{100}{1}\cdot \cfrac{4}{25} \\\\\\ \cfrac{64}{x}=16\implies 64=16x\implies \cfrac{64}{16}=x\implies 4=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{it had}}{64}-\stackrel{\textit{leakage}}{4}\implies \stackrel{\textit{remaining}}{60}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20amount%26%5C%25%5C%5C%20%5Ccline%7B1-2%7D%2064%26100%5C%5C%20x%266%5Cfrac%7B1%7D%7B4%7D%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B6%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B%5Cfrac%7B100%7D%7B1%7D%7D%7B%5Cfrac%7B25%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B1%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B25%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B64%7D%7Bx%7D%3D16%5Cimplies%2064%3D16x%5Cimplies%20%5Ccfrac%7B64%7D%7B16%7D%3Dx%5Cimplies%204%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bit%20had%7D%7D%7B64%7D-%5Cstackrel%7B%5Ctextit%7Bleakage%7D%7D%7B4%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bremaining%7D%7D%7B60%7D)
Answer:
yes, 1 can never equal 2 unless it is manipulated (ex: 1(2) = 2)
because one is not being manipulated to equal 2 here, it is a false statement and will always be a false statement.
Answer:
median: 10 <h <13
mean: 12.12
Step-by-step explanation:

<em><u>Solution:</u></em>
Given that we have to find the sum of 
<em><u>Let us first convert the mixed fractions to improper fractions</u></em>
Multiply the whole number part by the fraction's denominator.
Add that to the numerator.
Then write the result on top of the denominator
Therefore,

Now we can add both the fractions

Make the denominators same

Now convert the fraction to mixed number
When we divide 53 by 10 , the remainder is 3
Therefore, write 5 as a whole number and and 3 as numerator and 10 as denominator

Thus the sum of given mixed fraction is 
Answer:
5
Step-by-step explanation:
Substitute 2 in for x
|2+3| = 5
This would take the absolute value of the answer, or in other words once you solve the equation it will remove any negatives to return a positive number. I took the test and I have links to prove it.