In the data set, what is the variance? 6 8 1 9 4

Mathematics

1 answer:

Juli2301 [7.4K]4 years ago

3 0

The answer is s² ≈ 10.3

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Please this is due tonight!!! Express given logarithm in terms of a, b and c.

KATRIN_1 [288]

$\textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_5(18)\implies \cfrac{\log(18)}{\log(5)}\implies \cfrac{\log(2\cdot 3\cdot 3)}{\log(5)}\implies \cfrac{\log(2)+\log(3)+\log(3)}{\log(5)}$

$\begin{cases} \log(3)=a\\ \log(2)=b\\ \log(5)=c \end{cases}\implies \cfrac{b+a+a}{c}\implies \cfrac{b+2a}{c} \\\\[-0.35em] ~\dotfill\\\\ \log_5^2(18)\implies [\log_5(18)]^2\implies \left( \cfrac{b+2a}{c} \right)^2\implies \cfrac{(b+2a)^2}{c^2} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{b^2+4ab+4a^2}{c^2}~\hfill$

8 0

3 years ago

Determine whether the given system has a unique solution, no solution, or infinitely many solutions.

victus00 [196]

Answer:

The system has no solution.

Step-by-step explanation:

To find the solution to this system of linear equations

$\left\begin{array}{ccccc}-3x_1&+x_2&-2x_3&=&8&\\x_1&+5x_2&-x_3&=&4&\\-x_1&+11x_2&-4x_3&=&1&\end{array}\right$

First, state the problem in matrix form, this means, extracting only the numbers, and putting them in a box.

$\left[ \begin{array}{ccc|c} -3 & 1 & -2 & 8 \\\\ 1 & 5 & -3 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belong

Next, transform the augmented matrix to the reduced row echelon form with the help of Row Operations.

Row Operation 1: multiply the 1st row by -1/3

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 1 & 5 & -1 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 2: add -1 times the 1st row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 3: add 1 times the 1st row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ 0 & \frac{32}{3} & - \frac{10}{3} & - \frac{5}{3} \end{array} \right]$

Row Operation 4: multiply the 2nd row by 3/16

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 5: add -32/3 times the 2nd row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 6: multiply the 3rd row by -1/15

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & 1 \end{array} \right]$