What is 7 over 16 as a decimal
2 answers:
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The given function is
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)
Answer: 0.82
Step-by-step explanation:
The probability of the computer not containing neither a virus nor a worm is expressed as P(∩
) , where P(
) is the probability that the event V doesn't happen and P(
) is the probability that the event W doesn't happen.
P()= 1-P(V) = 1-0.17 = 0.83
P()=1-P(W) = 1-0.05 = 0.95
Since and
aren't mutually exclusive events, then:
P(∪
) = P(
) + P(
) - P(
∩
)
Isolating the probability that interests us:
P(∩
)= P(
) + P(
) - P(
∪
)
Where P(∪
) = 1 - 0.04 = 0.96
Finally:
P(∩
) = 0.83 + 0.95 - 0.96 = 0.82