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3 years ago

Nikki saved 15% on a $100 purchase. How much did she pay

Mathematics

1 answer:

Gwar [14]3 years ago

3 0

NIKKI PAID A WHOPPING 85

SAVING 15

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$220 marked up 60% ????????????????????????

maria [59]

Answer:

352$

Step-by-step explanation:

6 0

3 years ago

Ben has a part time job at the fun station. Suppose he worked 13.5 hours last week and $81. How much does Ben earn per hour

scoundrel [369]

To get the answer, you will just use this operation: Division.
Let's divide it. 81 ÷ 13.5 = 6.

So, the answer is
Ben earns $6 per hour at the fun station.

8 0

3 years ago

Read 2 more answers

Help me solve this and please explain to me. :(:

aalyn [17]

Answer:

E = 50

D = 150

Step-by-step explanation:

We know that the exterior angle is equal to the sum of the opposite interior angles

D + E = 200

D = 3E

Substituting that into the equation

3E + E = 200

Combing terms

4E = 200

Divide by 4

4E /4 = 200/4

E = 50

D = 3E

D = 3*50 = 150

8 0

3 years ago

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I need help I don’t understand this at all

Alina [70]

Answer:

it is 60

Step-by-step explanation:

1= what, well its 60

7 0

2 years ago

Read 2 more answers

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

3 years ago