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tia_tia [17]
3 years ago
5

Ln (x-2) +ln (2x-3)=2ln x

Mathematics
1 answer:
goblinko [34]3 years ago
4 0
D:x-2\ \textgreater \ 0 \wedge 2x-3\ \textgreater \ 0 \wedge x\ \textgreater \ 0\\
D:x\ \textgreater \ 2 \wedge x\ \textgreater \ \dfrac{3}{2}\wedge x\ \textgreater \ 0\\
D:x\ \textgreater \ 2\\\\
\ln (x-2) +\ln (2x-3)=2\ln x \\
\ln (x-2)(2x-3)=\ln x^2\\
(x-2)(2x-3)=x^2\\
2x^2-3x-4x+6=x^2\\
x^2-7x+6=0\\
x^2-x-6x+6=0\\
x(x-1)-6(x-1)=0\\
(x-6)(x-1)=0\\
x=6 \vee x=1\\
1\not \in D\\
\boxed{x=6}
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Answer:

The standard deviation for the mean weigth of Salmon is 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount order stores.

Step-by-step explanation:

The mean sample of the sum of n random variables is

\overline{X} = \frac{X_1+X_2+...+X_n}{n}

If X_1, ..., X_n are indentically distributed and independent, like in the situation of the problem, then the variance of X_1 + .... + X_n will be the sum of the variances, in other words, it will be n times the variance of X_1 .

However if we multiply this mean by 1/n (in other words, divide by n), then we have to divide the variance by 1/n², thus \overkine{X} = \frac{V(X_1)}{n} and as a result, the standard deviation of \overline{X} is the standard deviation of X_1 divided by \sqrt{n} .

Since the standard deviation of the weigth of a Salmon is 2 lbs, then the standard deviations for the mean weigth will be:

  • Restaurants: We have boxes with 9 salmon each, so it will be \frac{2}{\sqrt{9}} = \frac{2}{3}
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We conclude that de standard deivation of the mean weigth of salmon of the types of shipment given is: 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount outlet stores.

7 0
2 years ago
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