Does the line whose equation is y=-2x+13 pass through the point (4,5)? Explain how you arrived at
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Answer:
<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>
<em>The null hypothesis is accepted . </em>
<em>Assume the population variances are approximately the same</em>
<u><em>Step-by-step explanation:</em></u>
<u>Explanation</u>:-
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
<em>The first sample size 'n₁'= 20</em>
<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>
<em>standard deviation of first sample S₁ = 3.2 pounds</em>
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
<em>The second sample size n₂ = 24</em>
<em>mean of the second sample "x₂⁻"= 14.89 pounds</em>
<em>standard deviation of second sample S₂ = 2.7 poun</em>ds
<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>
<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>
<em>Level of significance ∝ =0.05</em>
<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>
<em>Test statistic :-</em>
<em> </em>
<em> where </em>
<em> substitute values and we get S² = 40.988</em>
<em> </em><em></em>
<em> </em> 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
<u><em>Conclusion</em></u>:-
<em>The null hypothesis is accepted </em>
<em>Assume the population variances are approximately the same.</em>
<em> </em>
<em> </em>
<em> </em>
The acceleration is given by the slope of the line. This line passes through the points (0, 2) and (3, 5), so it has slope
which means the acceleration over this 3 second interval is 1 m/s^2.
The answer is D. 4x + 3y = -6
I first found the slope using the slope formula. Then I plugged in the values of the given coordinate and the slope I found into the point slope form equation, then I simplified. Then I manipulated the equation to get x and y on the right side, and the constant on the left side. Finally, I multiplied both sides by 3 to get rid of the fraction.
I first found the slope using the slope formula. Then I plugged in the values of the given coordinate and the slope I found into the point slope form equation, then I simplified. Then I manipulated the equation to get x and y on the right side, and the constant on the left side. Finally, I multiplied both sides by 3 to get rid of the fraction.