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Yuliya22 [10]
3 years ago
15

Working together, Sam and Charlie can change all the light bulbs in the office building in 3 hours. It would have taken Charlie

8 hours to do the job alone.
A table showing Rate in part per hour, Time in hours, and Part of Light Bulbs Changed. The first row shows,Sam, and has r, 3, and r times 3. The second row shows Charlie, and has StartFraction 1 Over 8 EndFraction, 3 and StartFraction 1 Over 8 EndFraction times 3.
Which equation can be used to solve for the rate, r, for Sam to change all the light bulbs in the office building by himself?

3r = 3 r equals StartFraction 3 Over 8 EndFraction.
3r + 3 r plus StartFraction 3 Over 8 EndFraction equals 1. = 1
r + StartFraction 1 Over 3 EndFraction r plus StartFraction 3 Over 8 EndFraction r equals 1.r = 1
r + StartFraction 1 Over 3 EndFraction r plus StartFraction 1 Over 8 EndFraction r equals 1.r = 1
Mathematics
2 answers:
Virty [35]3 years ago
6 0

Answer:

B

3r + 3/8=1

Step-by-step explanation:

EDGE

vitfil [10]3 years ago
4 0

Answer:

b

Step-by-step explanation:

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vampirchik [111]
\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$5000\\
P=\textit{original amount deposited}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, four times}
\end{array}\to &4\\
t=years\to &7
\end{cases}
\\\\\\
5000=P\left(1+\frac{0.05}{4}\right)^{4\cdot 7}\implies 5000=P(1.0125)^{28}\\\\\\ \cfrac{5000}{(1.0125)^{28}}=P
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