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3 years ago

4) 4x + 3y = 56 2x + y = 24 Solve using elimination please show work

Mathematics

1 answer:

lesya [120]3 years ago

7 0

Answer:

x = 8 y = 8

Step-by-step explanation:

Multiply 2nd equation by -3 to get -6x - 3y = -72

Now 4x + 3y = 56

-6x - 3y = -72

-2x = -16

x = 8

Substitute x = 8 into the 2nd equation

2(8) + y = 24

16 + y = 24

y = 8

Check: Substitute the values into the 1st equation

4(8) + 3(8) = 32 + 24 = 56 So we have the correct values for x and y

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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

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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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