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VikaD [51]
3 years ago
11

The price of one gallon of milk has increased from $3.24 to $4.05. What is the percentage increase?

Mathematics
2 answers:
Alisiya [41]3 years ago
5 0
Let initial price be = x = 3.24
final price = y = 4.05
Difference = 4.05 - 3.24
= 0.81
Percentage increase = 0.81 / 3.25 × 100
= 25 %
Vedmedyk [2.9K]3 years ago
3 0
(4.05 - 3.24) / 3.24......x 100
0.81 / 3.24...x 100
0.25.....x 100
25% increase <==
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if Dakota earned $15.75 in interest in account A and $28.00 in interest in account B after 21 months. if the simple interest rat
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keeping in mind that 21 months is more than a year, since there are 12 months in a year,  then 21 months is really 21/12 years.


\bf ~~~~~~ \stackrel{\textit{account A}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &15.75\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 15.75=P(0.03)\left( \frac{7}{4} \right)\implies \cfrac{15.75}{(0.03)\left( \frac{7}{4} \right)}=P\implies \boxed{300=P}


\bf ~~~~~~ \stackrel{\textit{account B}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &28\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 28=P(0.049)\left( \frac{7}{4} \right)\implies \cfrac{28}{(0.049)\left( \frac{7}{4} \right)}=P\implies \boxed{326.53\approx P}


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3 years ago
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SCORPION-xisa [38]

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Maria ran 17 miles at an average speed of 5 miles per hour.
mel-nik [20]
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