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Colt1911 [192]
3 years ago
12

HEY CAN ANYONE HELP ME PLZ

Mathematics
2 answers:
elixir [45]3 years ago
6 0

\Large \underline{\tt Given} :

\\

  • Total number of students = 35

  • Students present on Monday = \tt \dfrac{5}{7}\: of \:35

\\

\Large \underline{\tt To \: Find} :

\\

  • Students not present on Monday = ?

\\

\Large \underline{\tt Solution} :

\\

As, we have :

Students present on Monday = \tt \dfrac{5}{7} of 35

\\

\tt : \implies Students \: present \: on \: Monday = \dfrac{5}{7} \times 35

\tt : \implies Students \: present \: on \: Monday = \dfrac{5}{\cancel{7}} \times \cancel{35}

\tt : \implies Students \: present \: on \: Monday = 5 \times 5

\tt : \implies Students \: present \: on \: Monday = 25

\\

Now, we have to find students not present on Monday :

As, we have :

  • Total number of students = 35
  • Students present on Monday = 25

\\

Therefore,

Students absent on Monday = 35 - 25 = 10

\\

Hence, 10 students were not present on Monday.

Nataly [62]3 years ago
3 0

Answer:

2/7 is correct answer

oky na

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\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
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he then turns around and grabs that money and sticks it for another 9 years,

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
~~
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$16010.32\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\
t=years\to &9
\end{cases}
\\\\\\
A=16010.32\left(1+\frac{0.05}{2}\right)^{2\cdot 9}\implies A=16010.32(1.025)^{18}
\\\\\\
A\approx 24970.64

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