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3 years ago

Solve the following system of equations by substitution and select the correct answer below: 4x − 6y = 28 8x − 4y = 24

Mathematics

2 answers:

Zarrin [17]3 years ago

8 0

According to my calculation it should be C: x= 1 and y = - 4
NOTE: I had it right on my online class.

Damm [24]3 years ago

4 0

Answer:

the solution is the point $(1,-4)$

Step-by-step explanation:

we have

$4x-6y=28$

isolate the variable x

$4x=28+6y$

$x=7+1.5y$ -----> equation A

$8x-4y=24$ -----> equation B

substitute the equation A in equation B

$8(7+1.5y)-4y=24$

$56+12y-4y=24$

$12y-4y=24-56$

$8y=-32$

$y=-4$

Find the value of x

$x=7+1.5(-4)=1$

the solution is the point $(1,-4)$

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Show your work on the number line 3-1 3/4?

BigorU [14]

Answer:

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Step-by-step explanation:

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3 years ago

Please answer now correct answer fast

Vikki [24]

Answer:

$Area = 112.1 m^2$

Step-by-step Explanation:

Given:

∆WXY

m < X = 130°

WY = x = 31 mm

m < Y = 26°

Required:

Area of ∆WXY

Solution:

Find the length of XY using Law of Sines

$\frac{w}{sin(W)} = \frac{x}{sin(X)}$

X = 130°

x = WY = 31 mm

W = 180 - (130 + 26) = 24°

w = XY = ?

$\frac{w}{sin(24)} = \frac{31}{sin(130)}$

Multiply both sides by sin(24) to solve for x

$\frac{w}{sin(24)}*sin(24) = \frac{31}{sin(130)}*sin(24)$

$w = \frac{31*sin(24)}{sin(130)}$

$w = 16.5 mm$ (approximated)

$XY = w = 16.5 mm$

Find the area of ∆WXY

$area = \frac{1}{2}*w*x*sin(Y)$

$= \frac{1}{2}*16.5*31*sin(26)$

$= \frac{16.5*31*sin(26)}{2}$

$Area = 112.1 m^2$ (to nearest tenth).

8 0

3 years ago

Please help for brainliest

finlep [7]

Answer:

8a : No because only the boys have a say and girls are part of the school to

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Step-by-step explanation:

4 0

3 years ago

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size 'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size n₂ = 24

mean of the second sample "x₂⁻"= 14.89 pounds

standard deviation of second sample S₂ = 2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

$t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

where $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

substitute values and we get S² = 40.988

$t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$

t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

Conclusion:-

The null hypothesis is accepted

Assume the population variances are approximately the same.

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Naya [18.7K]

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3 years ago