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2 years ago

10

15 patients were discharged yesterday. 5 were hospitalized 4 days 6 were hospitalized 5 days and the rest were in for 2, 4, 8, 1

0. what was the average number of days for this group?

1 answer:

S_A_V [24]2 years ago

7 0

2+4+8+10=24

24 divided by 4 is 6

6

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$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

Step-by-step explanation:

Given

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Required

Solve:

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Simplify 6/10

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{9})^2$

Express 9 as 3^2

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{3^2})^2$

Apply law of indices

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{(3^2)^2}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{3^4}$

Express as a sing;e fraction

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3*5^2}{5^3*3^4}$

Apply law of indices:

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^{3-2}*3^{4-3}}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^1*3^1}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5*3}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

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2 years ago