Basic Math Examples

5

$5$ ,

0

$0$ ,

0

$0$ ,

0 (\frac{0.005}{1 - {(1 + 0.005)}^{- 60}})

$0 (\frac{0.005}{1 - {(1 + 0.005)}^{- 60}})$

Step 1

Simplify the denominator.

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Step 1.1

Add

1

$1$ and

0.005

$0.005$ .

5, 0, 0, 0 (\frac{0.005}{1 - {1.005}^{- 60}})

$5, 0, 0, 0 (\frac{0.005}{1 - {1.005}^{- 60}})$

Step 1.2

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

5, 0, 0, 0 ⎛ ⎝ \frac{0.005}{1 - \frac{1}{{1.005}^{60}}} ⎞ ⎠

$5, 0, 0, 0 (\frac{0.005}{1 - \frac{1}{{1.005}^{60}}})$

Step 1.3

Raise

1.005

$1.005$ to the power of

60

$60$ .

5, 0, 0, 0 (\frac{0.005}{1 - \frac{1}{1.34885015}})

$5, 0, 0, 0 (\frac{0.005}{1 - \frac{1}{1.34885015}})$

Step 1.4

Divide

1

$1$ by

1.34885015

$1.34885015$ .

5, 0, 0, 0 (\frac{0.005}{1 - 1 \cdot 0.74137219})

$5, 0, 0, 0 (\frac{0.005}{1 - 1 \cdot 0.74137219})$

Step 1.5

Multiply

- 1

$- 1$ by

0.74137219

$0.74137219$ .

5, 0, 0, 0 (\frac{0.005}{1 - 0.74137219})

$5, 0, 0, 0 (\frac{0.005}{1 - 0.74137219})$

Step 1.6

Subtract

0.74137219

$0.74137219$ from

1

$1$ .

5, 0, 0, 0 (\frac{0.005}{0.2586278})

$5, 0, 0, 0 (\frac{0.005}{0.2586278})$

5, 0, 0, 0 (\frac{0.005}{0.2586278})

$5, 0, 0, 0 (\frac{0.005}{0.2586278})$

Step 2

Simplify the expression.

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Step 2.1

Divide

0.005

$0.005$ by

0.2586278

$0.2586278$ .

5, 0, 0, 0 \cdot 0.0193328

$5, 0, 0, 0 \cdot 0.0193328$

Step 2.2

Multiply

0

$0$ by

0.0193328

$0.0193328$ .

5, 0, 0, 0

$5, 0, 0, 0$

5, 0, 0, 0

$5, 0, 0, 0$

Step 3

The mean of a set of numbers is the sum divided by the number of terms.

\frac{5 + 0 + 0 + 0}{4}

$\frac{5 + 0 + 0 + 0}{4}$

Step 4

Simplify the numerator.

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Step 4.1

Add

5 + 0 + 0

$5 + 0 + 0$ and

0

$0$ .

\frac{5 + 0 + 0}{4}

$\frac{5 + 0 + 0}{4}$

Step 4.2

Add

5 + 0

$5 + 0$ and

0

$0$ .

\frac{5 + 0}{4}

$\frac{5 + 0}{4}$

Step 4.3

Add

5

$5$ and

0

$0$ .

\frac{5}{4}

$\frac{5}{4}$

\frac{5}{4}

$\frac{5}{4}$

Step 5

Divide.

1.25

$1.25$

Step 6

The mean should be rounded to one more decimal place than the original data. If the original data were mixed, round to one decimal place more than the least precise.

1.3

$1.3$