Neach row, colour the one with more number of pictures

Mathematics

1 answer:

saul85 [17]3 years ago

5 0

Answer

neach row, colour the one with more number of pictures

Step-by-step explanation:

neach row, colour the one with more number of pictures

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Answer: Lost 44%, Games won: 28/50

Step-by-step explanation:

Percentage = (part / whole) * 100

The total number of games the team played is 28 + 22, which equals 50.

Percentage of losses = (22/50) * 100 = 44%

Fraction of wins = 28/50

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I hope this helps! ツ

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3 years ago

Suppose that you have $6000 to invest. Which investment yields the greater return over four years: 8.25% compounded quarterly or

WARRIOR [948]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.25\%\to \frac{8.25}{100}\to &0.0825\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.0825}{4}\right)^{4\cdot 4}\implies A=6000(1.020625)^{16}\\\\
-------------------------------\\\\$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.3\%\to \frac{8.3}{100}\to &0.083\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus two}
\end{array}\to &2\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.083}{2}\right)^{2\cdot 4}\implies A=6000(1.0415)^8$

compare them away.

4 0

3 years ago

Neach row, colour the one with more number of pictures​

Neach row, colour the one with more number of pictures