A football team scored fewer than 16 points in Saturday’s game. Than wants to write an inequality for the number of points. He u
2 answers:
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Answer:
<em>The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance</em>
<em>Null hypothesis is rejected </em>
<em>Alternative hypothesis is accepted</em>
<em>The extra carbonation of cola results in a higher average compression strength</em>
<u><em /></u>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>Given data</em>
<em>First sample size (n₁) = 10 </em>
<em>mean of the first sample(x₁⁻) = 537</em>
<em>standard deviation of the first sample (S₁) = 22</em>
<em>second sample size (n₂) = 10 </em>
<em>mean of the second sample (x₂⁻) = 559</em>
<em>standard deviation of the second sample (S₂) = 17</em>
<u><em>Step(ii)</em></u><em>:-</em>
<u><em>Null hypothesis : H₀:</em></u><em>- The extra carbonation of cola results in a lower average compression strength</em>
<u>Alternative Hypothesis :H₁</u>
<em>The extra carbonation of cola results in a higher average compression strength</em>
<u><em>Step(iii)</em></u><em>:-</em>
<em>By using student's t -test for difference of means</em>
<u><em>Test statistic</em></u>
<em> </em><em />
<em> where </em>
<em> </em><em />
<em> </em><em />
<em> </em><em />
<em> t = -2.375</em>
<em>|t| = |-2.375| = 2.375</em>
<em>Degrees of freedom</em>
<em>γ = n₁+n₂ -2 = 10+10-2 =18</em>
<em /><em />
<em>The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance</em>
<em>Null hypothesis is rejected </em>
<em>Alternative hypothesis is accepted</em>
<u><em>Final answer:-</em></u>
<em>The extra carbonation of cola results in a higher average compression strength</em>
<u><em /></u>
<em />
<em> </em>
<em />
Answer:
The equation determine a relation between x and y
x = ±
y = ±
The domain is 1 ≤ y ≤ 3
The domain is -1 ≤ x ≤ 1
The graphs of these two function are half circle with center (0 , 2)
All of the points on the circle that have distance 1 from point (0 , 2)
Step-by-step explanation:
* Lets explain how to solve the problem
- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where
R is read as "has distance 1 of"
- This relation can also be read as “the point (x, y) is on the circle
of radius 1 with center (0, 2)”
- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”
* <em>Lets solve the problem</em>
- The equation of a circle of center (h , k) and radius r is
(x - h)² + (y - k)² = r²
∵ The center of the circle is (0 , 2)
∴ h = 0 and k = 2
∵ The radius is 1
∴ r = 1
∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²
∴ The equation is ⇒ x² + (y - 2)² = 1
∵ A circle represents the graph of a relation
∴ The equation determine a relation between x and y
* Lets prove that x=g(y)
- To do that find x in terms of y by separate x in side and all other
in the other side
∵ x² + (y - 2)² = 1
- Subtract (y - 2)² from both sides
∴ x² = 1 - (y - 2)²
- Take square root for both sides
∴ x = ±
∴ x = g(y)
* Lets prove that y=h(x)
- To do that find y in terms of x by separate y in side and all other
in the other side
∵ x² + (y - 2)² = 1
- Subtract x² from both sides
∴ (y - 2)² = 1 - x²
- Take square root for both sides
∴ y - 2 = ±
- Add 2 for both sides
∴ y = ±
∴ y = h(x)
- In the function x = ±
∵ ≥ 0
∴ 1 - (y - 2)² ≥ 0
- Add (y - 2)² to both sides
∴ 1 ≥ (y - 2)²
- Take √ for both sides
∴ 1 ≥ y - 2 ≥ -1
- Add 2 for both sides
∴ 3 ≥ y ≥ 1
∴ The domain is 1 ≤ y ≤ 3
- In the function y = ±
∵ ≥ 0
∴ 1 - x² ≥ 0
- Add x² for both sides
∴ 1 ≥ x²
- Take √ for both sides
∴ 1 ≥ x ≥ -1
∴ The domain is -1 ≤ x ≤ 1
* The graphs of these two function are half circle with center (0 , 2)
* All of the points on the circle that have distance 1 from point (0 , 2)