To win at LOTTO in one state, one must correctly select 6 numbers from a collection of 62 numbers (1 through 62). The order in w
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Answer:
a) f(x) = 4 - x²
The linear approximation of the function at a=1 is
L(x) = 5 - 2x at a = 1
b) The graph of the function and the linear approximation at that point is attached to this solution.
The curve represent the real function,
f(x) = 4 - x²
The straight line represents the linear approximation of the function at a=1.
L(x) = 5 - 2x
The curve and the function evidently cross paths at x=1 and understandably so.
c) Using the linear approximation obtained at a = 1.
f(1.1) = 2.8
Using the actual function, the actual value of f(1.1) = 2.79
d) Percent error = 0.358%
Step-by-step explanation:
f(x) = 4 - x²
a) The linear approximation of the function at the given point is given as
L(x) = f(a) + f'(a) [x - a]
f(x) = 4 - x²
a = 1
f(a) = 4 - 1² = 3
f'(x) = -2x
f'(a) = -2(1) = -2
L(x) = f(a) + f'(a) [x - a]
L(x) = 3 + (-2)(x - 1)
L(x) = 3 -2x + 2
L(x) = 5 - 2x
L(x) = -2x + 5
f(x) = 4 - x²
L(x) = 5 - 2x at a = 1
b) The graph of the function and the linear approximation at that point is attached to this solution.
The curve represent the real function,
f(x) = 4 - x²
The straight line represents the linear approximation of the function at a=1.
L(x) = 5 - 2x
The curve and the function evidently cross paths at x=1 and understandably so.
c) Use the linear approx. to estimate the given fxn value.
f(1.1)
L(x) = 5 - 2x
L(1.1) = 5 - 2(1.1) = 2.8
Using the function, the actual value of f(1.1) = 4 - 1.1² = 2.79
d) Compute the percent error in your approximation, 100*Iapprox-exactI/IexactI, where the exact value is given by a calculator
Percent error
= 100% × (|approx - exact|)/exact
Approximated value = 2.8
Exact value = 2.79
Percent error = 100% × (2.8-2.79)/2.79
Percent error = 0.358%
Hope this Helps!!!
Answer:
The expected sum of the numbers on the upward faces of the two dice is 7.
Step-by-step explanation:
Consider the provided information.
If two pair of dice tossed the possible out comes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,6)
Now we need to find the expected sum of the numbers on the upward faces of the two dice.
The expected sums can be:
Sum: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Prob: 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36
As we know that the expectation of experiment can be calculated as:
Here S represents the numerical outcomes and P(S) is the respective probability.
Substitute the respective values in the above formula.
Hence, the expected sum of the numbers on the upward faces of the two dice is 7.