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kompoz [17]
3 years ago
6

How to solve and respond.

Mathematics
1 answer:
son4ous [18]3 years ago
3 0

Step-by-step explanation:

the error:

it is stated that

\frac{5x}{x+7} +\frac{7}{x} = \frac{5x}{x+7} +\frac{7(7)}{x+7}

subtract (5x)/(x+7) from both sides

\frac{7}{x}  = \frac{7(7)}{x+7}

multiply both sides by x + 7

7(x+7) = 49x

7x + 49 = 49x

subtract 7x from both sides to isolate x and its coefficient

49 = 42x

thus, this is only true when 49 = 42x. in order for these two equations to be equal, they must <em>always </em>be true, so this is wrong

the solution:

we want to express 7/x as (something) / (x+7). to do this, we can multiply 7/x  by 1.

anything divided by itself = 1. thus, if we multiply both the numerator and the denominator by something that turns x into (x+7), we can do what we want to do.

(x+7)/x * x turns x into (x+7), so we multiply both the numerator and denominator by (x+7)/x to get

\frac{7}{x} = \frac{7(x+7)/x}{x+7}

substitute this for 7/x in our original problem

\frac{5x}{x+7} +\frac{7(x+7)/x}{x+7} =  \frac{5x}{x+7} +\frac{(7x+49)/x}{x+7} = \frac{5x}{x+7} +\frac{7+49/x}{x+7} = \frac{5x+7+49/x}{x+7}

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\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

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In point b)

\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

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using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

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