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3 years ago

6

Solve the following three variable system.

1 answer:

Ket [755]3 years ago

6 0

$1)\ \ \ 2x + 3y + 4z = 9\\2)\ \ \ -x + 2y - z = 0\\3)\ \ \ -2x + 4y + z = 3\\===================================\\3)\ \ \ -2x + 4y + z = 3\ \ \ \Rightarrow\ \ \ z=2x-4y+3\\\\1)\ \ \ 2x + 3y + 4z = 9\ \ \ \Rightarrow\ \ \ 2x + 3y + 4(2x-4y+3) = 9\\2x+3y+8x-16y+12=9\\10x-13y=-3\\\\2)\ \ \ -x + 2y - z = 0\ \ \ \Rightarrow\ \ \ -x+2y-(2x-4y+3)=0\\-x+2y-2x+4y-3=0\\-3x+6y=3\ /:(-3)\\x-2y=-1\ \ \ \Rightarrow\ \ \ x=2y-1\\\\$

$1)\ \ \ 10x-13y=-3 \ \ \ \Rightarrow\ \ \ 10(2y-1)-13y=-3\\20y-10-13y=-3\\7y=7\ /:7\ \ \ \Rightarrow\ \ \ y=1\\\\2)\ \ \ x=2y-1\ \ \ \Rightarrow\ \ \ x=2\cdot2-1=1\\\\3)\ \ \ z=2x-4y+3\ \ \ \Rightarrow\ \ \ z=2\cdot1-4\cdot1+3=2-4+3=1\\\\Ans.\ x=y=z=1$

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Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

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Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

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Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

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