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Aleks04 [339]
3 years ago
13

Solve the system of linear equations. separate the x- and y- values with a coma.

Mathematics
1 answer:
Katarina [22]3 years ago
7 0
The Answer is x=7, y=6

Solve the following system:
{14 x - 4 y = 74 | (equation 1)
{7 x - 7 y = 7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{14 x - 4 y = 74 | (equation 1)
{0 x - 5 y = -30 | (equation 2)
Divide equation 1 by 2:
{7 x - 2 y = 37 | (equation 1)
{0 x - 5 y = -30 | (equation 2)
Divide equation 2 by -5:
{7 x - 2 y = 37 | (equation 1)
{0 x+y = 6 | (equation 2)
Add 2 × (equation 2) to equation 1:
{7 x+0 y = 49 | (equation 1)
{0 x+y = 6 | (equation 2)
Divide equation 1 by 7:
{x+0 y = 7 | (equation 1)
{0 x+y = 6 | (equation 2)
Collect results:
Answer:  {x = 7
                {y = 6

P{lease note the brackets are supposed to go over both equations but I haven'y been able to find a way to make it work, see attachment.

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A sample of salary offers (in thousands of dollars) given to management majors is: 48, 51, 46, 52, 47, 48, 47, 50, 51, and 59. U
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Answer:

Step-by-step explanation:

number of samples, n = 10

Mean = (48 + 51 + 46 + 52 + 47 + 48 + 47 + 50 + 51 + 59)/10 = 49.9

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (48 - 49.9)^2 + (51 - 49.9)^2 + (46 - 49.9)^2+ (52 - 49.9)^2 + (47 - 49.9)^2 + (48 - 49.9)^2 + (47 - 49.9)^2 + (50 - 49.9)^2 + (51 - 49.9)^2 + (59- 49.9)^2 = 128.9

Standard deviation = √128.9/10 = 3.59

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 10 - 1 = 9

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.262

Margin of error = 2.262 × 3.59/√10

= 2.57

the lower limit of this confidence interval is

49.9 - 2.57 = 47.33

the lower limit of this confidence interval is

49.9 + 2.57 = 52.47

So it is false

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