Use the number line determine the absolute value. enter the value, as mixed numbers in the simplest form, in the box.
1 answer:
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Answer:
the absolute maximum value is 89.96 and
the absolute minimum value is 23.173
Step-by-step explanation:
Here we have cotangent given by the following relation;
so that the expression becomes
f(t) = 9t +9/tan(t/2)
Therefore, to look for the point of local extremum, we differentiate, the expression as follows;
f'(t) = =
Equating to 0 and solving gives
Where n is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively
Or we have by chain rule
f'(t) = 9 -(9/2)csc²(t/2)
Equating to zero gives
9 -(9/2)csc²(t/2) = 0
csc²(t/2) = 2
csc(t/2) = ±√2
The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or
in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2
We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;
f(x) for x = π/4, π/2, 3π/2, 7π/2
f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965
f(π/2) = 9·π/2 +9/tan(π/4) = 23.137
f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412
f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96
Therefore the absolute maximum value = 89.96 and
the absolute minimum value = 23.173.
Answer:
Step-by-step explanation:
We are given the following in the question:
The needle size should not be too big and too small.
The diameter of the needle should be 1.65 mm.
We design the null and the alternate hypothesis
Sample size, n = 35
Sample mean, = 1.64 mm
Sample standard deviation, s = 0.07 mm
Type I error:
- It is the error of rejecting the null hypothesis when it is true.
- It is also known as false positive error.
- It is the rejecting of a true null hypothesis.
Thus, type I error in this study would mean we reject the null hypothesis that the average diameter is 1.65 mm but actually the average diameters of the needle is 1.65 mm.
Thus, average diameter is 1.65 mm and we decide that it is not 1.65 mm.