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Inga [223]
3 years ago
11

Solve for x, y, and z

Mathematics
1 answer:
zloy xaker [14]3 years ago
3 0

Answer:

The solution is x=1,y=2,z=3

Step-by-step explanation:

The given system of equations is ;\

2x−3y+4z=8...(1)

3x+4y−5z=−4...(2)

4x−5y+6z=12...(3)

Make x the subject in equation (1)

x=\frac{8+3y-4z}{2}...(4)

Put equation (4) into equation (2) and (3)

3(\frac{8+3y-4z}{2})+4y-5z=-4

Multiply through by;

3(8+3y-4z)+8y-10z=-8

Expand;

24+9y-12z+8y-10z=-8

Simplify;

17y-22z=-32...(5)

Equation (4) in (3)

4(\frac{8+3y-4z}{2})-5y+6z=12

2(8+3y-4z)-5y+6z=12

16+6y-8z-5y+6z=12

y-2z=-4

y=2z-4...(6)

Put equation (6) into equation (5)

17(2z-4)-22z=-32

34z-68-22z=-32

34z-22z=-32+68

12z=36

z=3

Put z=3 into equation (6)

y=2(3)-4=2

Put y=2 and z=3 into equation 4

x=\frac{8+3(2)-4(3)}{2}=1

The solution is x=1,y=2,z=3

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Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

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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.  

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Mean =\displaystyle\frac{340119}{10} = 34011.9

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

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GenaCL600 [577]
We know that

At sea level, the height is 0 and
the pressure is 98 kilopascals

At 1000 ft, the height is 1000 and
the pressure <span>decreases about 11.41%

</span>i(100%-11.41%)/100----> (0.8859)

therefore
1) at 1000 ft--------> the pressure is 98*(0.8859)----------> 86.82 kilopascals
2) at 2000 ft--------> the pressure is 86.82*(0.8859)-------> 76.91 kilopascals
3) at 3000 ft--------> the pressure is 76.91*(0.8859)-------> 68.14 kilopascals
4) at 4000 ft---------> the pressure is 68.14*(0.8859)-------> 60.36 kilopascals


the answer is
<span>The pressure at an altitude of 4000 m is about </span>60.36 kilopascals
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