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3 years ago

Solve the system using elimination.

2 answers:

7 0

We can't eliminate as is so we have to change something up there in the equations to get either the x values the same number but opposite signs, or the y values the same number but opposite signs. I chose to change the y values to the same number but different signs. In the first equation y is -3y and in the second one, y is -8y. The LCM of both of those numbers is 24, so we will multiply the first equation by an 8 (8*3=24) and the second equation by 3 (3*8=24) but since they are both negative right now, one of those multiplications has to involve a negative because - * - = +. Set it up like this:
8(-10x - 3y = -18)
-3(-7x - 8y = 11)
Multiply both of those all the way through to get new equations:
-80x - 24y = -144
21x +24y = -33
Now the y's cancel each other out leaving only the x's:
-59x = -177 and x = 3. Now plug that 3 into either one of the original equations to find the y value. Either equation will work; you'll get the same answer using either one. Promise. -7(3) - 8y = 11 gives a y value of -4. so your solution is (3, -4) or B above.

fiasKO [112]3 years ago

4 0

Answer:(3,-4)

Step-by-step explanation:

Given equations are

-10x-3y=-18---------------1

-7x-8y=11-------------------2

From 1

18-3y=10x

x=1.8-0.3y

substitute the value of x in equation 2

$-7\left ( 1.8-0.3y\right )-8y=11$

-5.9y=23.6

y=-4

thus x=1.8+1.2

x=3

Thus (x,y)=(3,-4)

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Construct the confidence interval for the population mean mu. c = 0.90, x = 16.9, s = 9.0, and n = 45. A 90% confidence inte

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Answer:

The 90% confidence interval for population mean is $14.7 < \mu < 19.1$

Step-by-step explanation:

From the question we are told that

The sample mean is $\= x = 16.9$

The confidence level is $C = 0.90$

The sample size is $n = 45$

The standard deviation

Now given that the confidence level is 0.90 the level of significance is mathematically evaluated as

$\alpha = 1-0.90$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the standardized normal distribution table. The values is $Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining critical values for $\frac{\alpha }{2}$ instead of that of $\alpha$ is because $\alpha$ represents the area under the normal curve where the confidence level 1 - $\alpha$ (90%) did not cover which include both the left and right tail while $\frac{\alpha }{2}$ is just considering the area of one tail which is what we required calculate the margin of error