To test the claim that the resistance of electric wire can be reduced by more than 0.050 ohm by alloying, thirty two (32) values
were obtained for standard wire, and the sample average calculated was equal to 0.136 ohm. In addition, thirty two (32) values were obtained for alloyed wire, and the sample average calculated was equal to 0.083 ohm. The standard deviation resistance for standard wire is known to be 0.004 ohm, whereas the standard deviation resistance for alloyed wire is known to be 0.005. At the level of significance=0.05, does this support the claim?
The z score for this test is z0 =
The p-value for this test is p =
Based on the p-value calculated above, the decision is to fail to reject H0 at a level of significance =0.05 and conclude that the resistance of electric wire can't be reduced by more than 0.05 ohm by alloying. Enter 0 if this statement is FALSE or 1 otherwise:
The z score for this test is z0 =
The p-value for this test is p =
Based on the p-value calculated above, the decision is to fail to reject H0 at a level of significance =0.05 and conclude that the resistance of electric wire can't be reduced by more than 0.05 ohm by alloying. Enter 0 if this statement is FALSE or 1 otherwise:
1 answer:
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Answer:
α(t) = (-a*Sin(t), b*Cos (t)) where t ∈ [0, 2π]
Step-by-step explanation:
Given an arbitrary ellipse (x²/a²) + (y²/b²) = 1 (a, b > 0)
The parametrization can be as follows
x = -a*Sin(t)
y = b*Cos (t)
then
α(t) = (-a*Sin(t), b*Cos (t)) where t ∈ [0, 2π]
If t = 0
α(0) = (-a*Sin(0), b*Cos (0)) = (0, b)
If t = π/2
α(π/2) = (-a*Sin(π/2), b*Cos (π/2)) = (-a, 0)
If t = π
α(π) = (-a*Sin(π), b*Cos (π)) = (0, -b)
If t = 3π/2
α(3π/2) = (-a*Sin(3π/2), b*Cos (3π/2)) = (a, 0)
If t = 2π
α(2π) = (-a*Sin(2π), b*Cos (2π)) = (0, b)
We can see the sketch in the pic.
