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Thepotemich [5.8K]
3 years ago
11

Solve the system of linear equations by substitution. 2x-y=3 x+5y=14

Mathematics
1 answer:
Arturiano [62]3 years ago
6 0

Answer:

x= 29/11 or y= 25/11

Step-by-step explanation:

2x - y= 3 ...........equation 1

x + 5y= 14 ............equation 2

Make x the subject of the formula in equation 2

x= 14 - 5y ..............equation 3

Substitute x=14 - 5y in equation 1

2(14- 5y) - y=3

28 - 10y - y=3

Collect like terms

-10y - y=3 - 28

-11y= -25

divide both sides by coefficient of y

-11y/-11 = -25/-11

y= 25/11

Substitute y= 25/11 in equation 3

x=14 - 5(25/11)

x= 14 - 125/11

x= 29/11

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Answer:

56

Step-by-step explanation:

You add 91 and 33 and you get 124. Subtract 180 and 124 to get 56.

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The three angle bisectors of ΔABC intersect at point P. Point P is the
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<h3>Answer: C) incenter</h3>

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

Explanation:

If you were to intersect the angle bisectors (at least two of them), then you would locate the incenter. The incenter is the center of the incircle which is a circle where it is as large as possible, but does not spill over and outside the triangle. Therefore this circle fits snugly inside the triangle.

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extra notes:

* The centroid is found by intersecting at least two median lines

* The circumcenter is found by intersecting at least two perpendicular bisector lines

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* The incenter is always inside the triangle; hence the "in" as part of the name. The centroid shares this property as well because the medians are completely contained within any triangle. The other two centers aren't always guaranteed to be inside the triangle.

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3 years ago
-2x +8 &gt;4? pls help me
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Answer:

x > 2

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A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for
tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of <em>t</em> are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29

s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

  =\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}

  =-4.123

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}

   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

   =26.55\\\approx 27

Compute the critical value for <em>α</em> = 0.05 as follows:

t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

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2 years ago
Solve the equation.<br> 5x4 - 2x2 - 3 = 0
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