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3 years ago

(Will l give 25 points!) Solve the system of linear equations by substitution. -8x + y = 5 and 5x - y = 3

Mathematics

1 answer:

Serggg [28]3 years ago

6 0

Answer:

x = -8/3 and y = -49/3

Step-by-step explanation:

(Step 1) To solve using substitution, you first need to isolate a variable in one of these equations. I chose to isolate "y" in the following equation.

-8x + y = 5 <--- Original equation

y = 5 + 8x <--- Add 8x to both sides

(Step 2) Now, substitute (5 + 8x) for "y" in the second equation.

5x - y = 3 <--- Original equation

5x - (5 + 8x) = 3 <--- Substitute 5 + 8x for "y"

5x - 5 - 8x = 3 <--- Distribute (-) in the parentheses

-3x - 5 = 3 <--- Combine terms with "x"

-3x = 8 <--- Add 5 to both sides

x = -8/3 <--- Divide both sides by -3

(Step 3) Since you know the value of "x", plug it into one of the equations to solve for "y".

y = 5 + 8x <--- Original (rearranged) equation

y = 5 + 8(-8/3) <--- Plug (-8/3) for "x"

y = 5 - 64/3 <--- Multiply

y = 15/3 - 64/3 <--- Convert 5 into a fraction

y = -49/3 <--- Subtract

(Step 3.5) You can check your work by plugging "y" into the other equation to see if the answer for "x" is the same.

5x - y = 3 <--- Original equation

5(-8/3) - y = 3 <--- Plug (-8/3) for "x"

-40/3 - y = 3 <--- Multiply

-40/3 - y = 9/3 <--- Convert 3 into a fraction

-49/3 - y = 0 <--- Subtract both sides by 9/3

-49/3 = y <--- Add both sides by "y"

The value of the system of equations is x = -8/3 and y = -49/3.

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So on this case the 95% confidence interval would be given by (-1.152;5.152).

$-1.152 < \mu_{safe1}-\mu_{safe2}$

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for ?D contains 0.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution

Let put some notation

x=value for Safe 1 , y = value for Safe 2

x: 103, 90, 64, 120, 104, 92, 145, 106, 76

y: 101, 94, 58, 112, 103, 90, 140, 110, 74

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: 2, -4, 6, 8, 1, 2, 5, -4, 2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{18}{9}=2$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =4.093$

The next step is calculate the degrees of freedom given by:

$df=n-1=9-1=8$

Now we need to calculate the critical value on the t distribution with 8 degrees of freedom. The value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , so we need a quantile that accumulates on each tail of the t distribution 0.025 of the area.