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Fiesta28 [93]
3 years ago
11

expression to approximate log a of x for all positive numbers a, b, and x, where a is not equal to 1 and b is not equal to one

Mathematics
1 answer:
amid [387]3 years ago
7 0

Question:

Approximate log base b of x, log_b(x).

Of course x can't be negative, and b > 1.


Answer:

f(x) = (-1/x + 1) / (-1/b + 1)


Step-by-step explanation:

log(1) is zero for any base.

log is strictly increasing.

log_b(b) = 1

As x descends to zero, log(x) diverges to -infinity


Graph of f(x) = (-1/x + 1)/a is reminiscent of log(x), with f(1) = 0.


Find a such that f(b) = 1

1 = f(b) = (-1/b + 1)/a

a = (-1/b + 1)


Substitute for a:

f(x) = (-1/x + 1) / (-1/b + 1)

f(1) = 0

f(b) = (-1/b + 1) / (-1/b + 1) = 1




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Answer:

a) P(X∩Y) = 0.2

b) P_1 = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability P_1 that he must stop at the first signal but not at the second one can be calculated as:

P_1 = P(X) - P(X∩Y)

P_1 = 0.36 - 0.2 = 0.16

At the same way, the probability P_2 that he must stop at the second signal but not at the first one can be calculated as:

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P_2 = 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is:

P = P_1+P_2\\P=0.16+0.31\\P=0.47

7 0
3 years ago
I need help on this 6th grade math problem
vivado [14]

Answer: x<17.5

Step-by-step explanation:

Subtract  from both sides: x+2.5-2.5<20-2.5

Simplify the arithmetic: x<20-2.5

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Hope it helps!

4 0
1 year ago
Read 2 more answers
What is a length of 6 ft and a width of 4 ft in area in meters
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Vinvika [58]

Answer:

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Aneli [31]

Answer:

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