$\bf 7275(2^k)^3~~+~~5723\cdot 2^k~~+~~4725k(2^k)^3~~+~~3717k2^k~\hfill \begin{cases} n=2^k\\ k=\frac{ln(n)}{ln(2)}\\[-0.5em] \hrulefill\\ n=2^{\frac{ln(n)}{ln(2)}} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf 7275n^3~~+~~5223n~~+~~4725\cdot \cfrac{ln(n)}{ln(2)}(n^3)~~+~~3717\cfrac{ln(n)}{ln(2)}(n) \\\\\\ 7275\left( 2^{\frac{ln(n)}{ln(2)}} \right)^3+5273\left( 2^{\frac{ln(n)}{ln(2)}} \right)+4275\cfrac{ln(n)}{ln(2)}\left( 2^{\frac{ln(n)}{ln(2)}} \right)^3+3717\cfrac{ln(n)}{ln(2)}\left( 2^{\frac{ln(n)}{ln(2)}} \right)$

8 0

3 years ago

It was recently estimated that homes without children outnumber homes with children by about seven to five. If there are 2556 ho

nekit [7.7K]

Answer:

929

Step-by-step explanation:

For every 11 children, 7 will have homes and 4 will not.

The fraction of children without homes is 4/11.

The number of children without homes is

4/11 × 2556 = 929

Check:

The number of children with homes is 2556 - 929 = 1627

With/Without = 1627/929 = 7.005/4 ≈ 7/4

OK.

4 0

3 years ago