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garri49 [273]
2 years ago
15

Kal decides to open a home-based cupcake business. His cost is an initial value of $10 in operating expenses plus $0.25 to make

each cupcake. He will sell each cupcake at $1.50 to try and make a profit. How many cupcakes will he need to sell before he makes a profit?
Also need an equation
Mathematics
1 answer:
pshichka [43]2 years ago
4 0

Answer:

4

Step-by-step explanation:

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weeeeeb [17]
You’re answer is G!
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3 years ago
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A man pays 20% of his salary as tax and has 24,000$ left. How much did he pay as tax
puteri [66]

Answer:

He paid $6,000 in taxes.

Step-by-step explanation:

His final amount after taxes is $24000.  In order to obtain this number we had to tax 20% of the total (Note that 100%-20%=80%).  In other words if he was taxed 20% he took home 80% of what he was paid.  Therefore mathematically we have:

0.8A_{pre-tax}=A_{taxed}

Since we know he was left with $24000 after taxes then:

0.8A_{pre-tax}=24000\\\\\frac{0.8A_{pre-tax}}{0.8}=\frac{24000}{0.8}\\\\A_{pre-tax}=\frac{24000}{0.8}\\\\A_{pre-tax}=30000

So originally he had $30,000.  So how much did he pay in taxes? We can calculate the difference of both pretax and after tax and we get:

$30,000 - $24,000= $6,000 taxed

4 0
4 years ago
Solve pls brainliest
Shtirlitz [24]

Answer:

1.8 or 1 4/5 should be the answer!

8 0
3 years ago
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Joe's annual income has been increasing in the same dollar amount. The first year his income was $15,200, and the 4th year his i
lidiya [134]

Answer:

Therefore 6th year his income was $19,700.

Step-by-step explanation:

Given, Joe's annual income has been increasing in some dollar amount . The first year his income was $15,200 and 4th  year his income was $17,900.

A=$17900, P= $15,200 and n = 3 year

A= P(1+\frac{r}{100} )^n

\Leftrightarrow 17900=15200(1+\frac{r}{100})^3

\Leftrightarrow (1+\frac{r}{100})^3=\frac{17900}{15200}

\Leftrightarrow (1+\frac{r}{100})=(\frac{17900}{15200})^{\frac{1}{3} }

\Leftrightarrow \frac{r}{100}=(\frac{17900}{15200})^{\frac{1}{3} }-1

\Leftrightarrow r=5.60

Let  t^{th} year Joe's income was $19,700.

\therefore 19700=15,200(1+\frac{5.60}{100} )^{t-1}

\Leftrightarrow 1.296= (1+0.056)^{t-1}

\Leftrightarrow 1.296= (1.056)^{t-1}

\Leftrightarrow  log(1.296)= (t-1)log(1.056)

\Leftrightarrow  \frac{log(1.296)}{log(1.056)}= (t-1)

\Leftrightarrow t-1= 4.75

\Leftrightarrow t= 4.75+1

\Leftrightarrow t = 5.75 ≈6

Therefore 6th year his income was $19,700.

8 0
3 years ago
6 divided by 7.6 times 44
Delvig [45]

Answer:55.73333333333

Step-by-step explanation:

7 0
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