What is [(7!)/(3!)(1/2)^3(1-1/2)^{7-3}]*1/2 ?

The solution to [(7!)/(3!)(1/2)^3(1-1/2)^{7-3}]*1/2 is 105/32

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simplify [(7!)/(3!)(1/2)^3(1-1/2)^{7-3}]*1/2

simplify

[\frac{7 !}{3 !} (\frac{1}{2})^{3} (1 - \frac{1}{2})^{7 - 3}] \cdot \frac{1}{2}

\frac{105}{32}

Decimal Notation

3.28125

Solution steps

(\frac{7 !}{3 !} (\frac{1}{2})^{3} (1 - \frac{1}{2})^{7 - 3}) \frac{1}{2}

(\frac{1}{2})^{3} = \frac{1}{8}

= \frac{7 !}{3 !} \cdot \frac{1}{8} (1 - \frac{1}{2})^{7 - 3} \frac{1}{2}

(1 - \frac{1}{2})^{7 - 3} = \frac{1}{16}

= \frac{7 !}{3 !} \cdot \frac{1}{8} \cdot \frac{1}{16} \cdot \frac{1}{2}

Cancel

\frac{7 !}{3 !} : 840

= 840 \cdot \frac{1}{8} \cdot \frac{1}{16} \cdot \frac{1}{2}

840 \cdot \frac{1}{8} \cdot \frac{1}{16} = \frac{105}{16}

= \frac{105}{16} \cdot \frac{1}{2}

\frac{105}{16} \cdot \frac{1}{2} = \frac{105}{32}

= \frac{105}{32}

- 2 u + 2 = - 7 u + 17

\frac{x ^{- 1} - y ^{- 1}}{x ^{- 2} - y ^{- 2}}

3 (x^{2} + 12) = 84

2 x^{2} = 17 x + 9

(e^{\frac{2 x}{3}})^{2}

What is [(7!)/(3!)(1/2)^3(1-1/2)^{7-3}]*1/2 ?
The solution to [(7!)/(3!)(1/2)^3(1-1/2)^{7-3}]*1/2 is 105/32