What is the least possible sum of the digits displaying the time on a 12-hour digital clock?

1 answer:

4 0

So if you think about it, this problem isn't that hard.So say it's 12:59, the total sum of those digits would be 71. if it was 1:11, then the sum would be 3. if it would be 3:47, the sum would be 50. So now, you have to find the time where it would be the least possible sum. That time would be 1:00 because 1+0+0=1.

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Need answer ASAP thanks!

barxatty [35]

Answer:

assuming its an annual interest

Okay so 6 percent interest, the bank is paying you.
So with this it’s 6 percent of 1500 and add it to 1500.

You can always find 6 percent of 1500 and then add but here’s a short cut.
Your principle (beginning) balance is 1500.
That’s already 100 percent since thats yoru original value.
You then get added 6 percent interest.
We are jsut adding 6 percent to 100 percent so 106 percent.
Now we solve normally and you’d get the answer faster.

106 percent is 106/100 or 1 3/5 or 1.06

now we multiply
1500 * 1.06 = 1590

Your final balance would be 1590 after the 6 percent interest is added.

5 0

2 years ago

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp

Katena32 [7]

Let's move like a crab, backwards some.

after 2 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2$

after 3 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3$

is that enough to pay the $1100?

now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000 <---- standard form.

8 0

2 years ago

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12345 [234]

Answer:

mhm

Step-by-step explanation:

8 0

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Novosadov [1.4K]

Answer: