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3 years ago

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Which system of equations is represented by the matrix below? 1 5 11 4 -6-3 A. y + 5x = 11 4y - 6x = -3 B. X+ 5y = 11 4x - 6y =

-3 C. 1 + 5y = 11 4x - 6y = -3

1 answer:

Vitek1552 [10]3 years ago

3 0

Answer: Choice B

$x+5y = 11\\4x-6y = -3$

=====================================

Explanation:

The given matrix is

$\left[\begin{array}{cc|c}1&5&11\\4&-6&-3\end{array}\right]$

The numbers to the right of the vertical bar represent the values on the right hand side of each equation in the final answer.

The numbers to the left of the bar represent the x and y coefficients of the equations. The first number of any given row is the x coefficient. The second number is the y coefficient.

For instance, the first row has $\begin{array}{cc}1&5\end{array}$ to indicate the x coefficient is 1, and the y coefficient is 5. We end up with 1x+5y or simply x+5y. Putting everything together, the first equation would be x+5y = 11

Through similar steps, the second equation is 4x-6y = -3

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A major credit card company has determined that customers charge between $100 and $1100 per month. If the monthly amount charged

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Answer:

a) $E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

b) First we need to calculate the variance given by this formula:

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And the deviation would be:

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And using the cdf we got:

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Step-by-step explanation:

For this case we define the random variable X who represent the customers charge, and the distribution for X on this case is:

$X \sim Unif (a= 100, b=1100)$

Part a

For this case the average is given by the expected value and we can use the following formula:

$E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

Part b

First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

Part c

For this case we want to find the percent between 600 and 889, so we can use the cumulative distribution function given by:

$F(x) = \frac{X -100}{1100-100}= \frac{x-100}{1000}, 100 \leq X \leq 1100$

And we can find this probability:

$P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

Part d

$P(311 < X< 889) = P(X$

And using the cdf we got:

$P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

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3 years ago

Read 2 more answers

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Answer:

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Step-by-step explanation:

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$\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {8^2 - 4\cdot-4.9\cdot1} }}{{2\cdot-4.9}}} \end{array}$

$\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {64 + 19.6} }}{{-9.8}}} \end{array}$

$\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {83.6} }}{{-9.8}}} \end{array}$

$\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {9.14} }}{{-9.8}}} \end{array}$

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Hope this helped!

3 0

3 years ago