Prove that cos^4∆ - sin^4∆ = 2cos^2∆ - 1
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Answer:
We conclude that her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company states.
Step-by-step explanation:
We are given;
Population mean; μ = 30 minutes
Population standard deviation; σ = 3 minutes
Sample size; n = 10
Sample mean; x¯ = 25
Significance level = 5% = 0.05
Let's define the hypotheses;
Null hypothesis; H0: μ = 30
Alternative hypothesis: Ha: μ ≠ 30
Let's find the test statistic;
z = (x¯ - μ)/(σ/√n)
z = (25 - 30)/(3/√10)
z = -5.27
From online p-value from z-score calculator attached using, z = -5.27, significance level = 0.05, two tailed hypothesis, we have;
p < 0.00001
This is less than the significance level, and so we will reject the null hypothesis and conclude that Her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated.
This question is missing some parts. Here is the complete question.
Consider the oriented path which is a straight line segment L running from (0,0) to (16,16).
(a) Calculate the line inetrgal of the vector field F = (3x-y)i + xj along line L using the parameterization B(t) = (2t,2t), 0 ≤ t ≤ 8.
Enter an exact answer.
(b) Consider the line integral of the vector field F = (3x-y)i + xj along L using the parameterization C(t) = , 16 ≤ t ≤ 32.
The line integral calculated in (a) is ____________ the line integral of the parameterization given in (b).
Answer: (a) 384
(b) the same as
Step-by-step explanation: <u>Line</u> <u>Integral</u> is the integral of a function along a curve. It has many applications in Engineering and Physics.
It is calculated as the following:
in which (.) is the dot product and r(t) is the given line.
In this question:
(a) F = (3x-y)i + xj
r(t) = B(t) = (2t,2t)
interval [0,8] are the limits of the integral
To calculate the line integral, first substitute the values of x and y for 2t and 2t, respectively or
F(B(t)) = 3(2t)-2ti + 2tj
F(B(t)) = 4ti + 2tj
Second, first derivative of B(t):
B'(t) = (2,2)
Then, dot product between F(B(t)) and B'(t):
F(B(t))·B'(t) = 4t(2) + 8t(2)
F(B(t))·B'(t) = 12t
Now, line integral will be:
<u>Line integral for the conditions in (a) is 384</u>
<u />
(b) same function but parameterization is C(t) = :
F(C(t)) =
F(C(t)) =
C'(t) =
Limits are 16 and 32, so line integral will be:
<u>With the same function but different parameterization, line integral is the same.</u>