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3 years ago

8. Solve the system of equations using substitution.

Mathematics

1 answer:

Elodia [21]3 years ago

8 0

Answer:

Step-by-step explanation:

In order to solve this we need to plug in the answer choices and make sure they work for BOTH answers!

Let us start with option A. (6,3) 3 = 4(6) + 6 = 3 = 30 This is automaticly wrong because 3 doesnt equal 30. There is no need to try it on the second equation since the first one was wrong.

Next answer choice, B (3,6) 6 = 2(3) = 6=6. The second equation works perfect for this. Now we have to make sure that it works for the first one as well. 6 = 4(3) + 6 = 6 = 18 which is wrong.

Next, C. (1,2) 2 = 2(1) = 2=2 Right now lets check the other equation.

2 = 4(1) + 6 = 2=10. Wrong

Last, D. (-3, -6) -6 = 2(-3) = -6 = -6 Right now lets check the other equation.

-6 = 4(-3) + 6 = -6 = -6. CORRECT. Answer choice D works for both equations.

Sorry for the wait of response.

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Make b the subject of the formula a = square root b+6

dexar [7]

Answer:

b = (a - 6)^2

Step-by-step explanation:

Solve for b:

a = sqrt(b) + 6

a = sqrt(b) + 6 is equivalent to sqrt(b) + 6 = a:

sqrt(b) + 6 = a

Subtract 6 from both sides:

sqrt(b) = a - 6

Raise both sides to the power of two:

Answer: b = (a - 6)^2

6 0

3 years ago

WHATS X AND Y?? Y = 3x+ 4 and 2x – 4y = -6

Georgia [21]

Answer:

X=-1, y=1

Step-by-step explanation:

Substitute value of y in first equation (3x+4) with y of second equation.

2x-4(3x+4)=-6

2x-12x-16=-6

-10x=10x=-1

solve for y but putting -1 for x value in either equation

y=3(-1)+4

y=-3+4

y=1

4 0

3 years ago

The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for

Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, $s^{2}$ = sample variance = 1.88

$\sigma^{2}$ = population variance

n = sample of windshields = 83

So, 95% confidence interval for population variance, $\sigma^{2}$ is;

P(58.85 < $\chi^{2} __8_2$ < 108.9) = 0.95 {As the table of $\chi^{2}$ at 82 degree of freedom

gives critical values of 58.85 & 108.9}

P(58.85 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 108.9) = 0.95

P( $\frac{ 58.85}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{108.9}{(n-1)s^{2} }$ ) = 0.95

P( $\frac{ (n-1)s^{2}}{108.9 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{58.85 }$ ) = 0.95

95% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{108.9 }$ , $\frac{ (n-1)s^{2}}{58.85 }$ )

= ( $\frac{ (83-1)\times 1.88}{108.9 }$ , $\frac{ (83-1)\times 1.88}{58.85 }$ )

= (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

8 0

3 years ago

Help please:):):):):):)

soldi70 [24.7K]

do eat popcorn in a tub

poopy face

4 0

3 years ago

Read 2 more answers

Y=-3×+1 is it an independent dependent or a inconsistent

scoundrel [369]

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4 0

3 years ago