Segment Addition Postulate Using the following image, solve for x.
1 answer:
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Since g(6)=6, and both functions are continuous, we have:
![\lim_{x \to 6} [3f(x)+f(x)g(x)] = 45\\\\\lim_{x \to 6} [3f(x)+6f(x)] = 45\\\\lim_{x \to 6} [9f(x)] = 45\\\\9\cdot lim_{x \to 6} f(x) = 45\\\\lim_{x \to 6} f(x)=5](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2Bf%28x%29g%28x%29%5D%20%3D%2045%5C%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2B6f%28x%29%5D%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B9f%28x%29%5D%20%3D%2045%5C%5C%5C%5C9%5Ccdot%20lim_%7Bx%20%5Cto%206%7D%20f%28x%29%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20f%28x%29%3D5)
if a function is continuous at a point c, then
,
that is, in a c ∈ a continuous interval, f(c) and the limit of f as x approaches c are the same.
Thus, since
, f(6) = 5
Answer: 5
if a function is continuous at a point c, then
that is, in a c ∈ a continuous interval, f(c) and the limit of f as x approaches c are the same.
Thus, since
Answer: 5
Using conditional probability, it is found that there is a 0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.
Conditional Probability
In which
- P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
- Event A: Person has the flu.
- Event B: Person got the flu shot.
The percentages associated with getting the flu are:
- 20% of 30%(got the shot).
- 65% of 70%(did not get the shot).
Hence:
The probability of both having the flu and getting the shot is:
Hence, the conditional probability is:
0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.
To learn more about conditional probability, you can take a look at brainly.com/question/14398287
Answer:
Step-by-step explanation:
We have been given two expressions and
Now we need to find out least common multiple of these two expressions.
First we need to find out what is common factor of both expressions. and
Least common multiple means find the expression which can be divided by both expressions.
15 and 6 both goes into 30
so 30 is part of LCM (least common multiple )
Now pickup the highest exponent of each variable.
So we get
Hence required least common multiple is