Given:
The piecewise function is
To find:
The range of given piecewise function.
Solution:
Range is the set of output values.
Both functions and as linear functions.
Starting value of is at x=-4 and end value is at x=3.
Starting value:
End value:
Starting value of is at x=3 and end value is at x=6.
Starting value:
End value:
Least range value is 0 at x=-4 and 0 is included in the range because -4 is included in the domain.
Largest range value is 11 at x=6 and 11 is not included in the range because 6 is not included in the domain.
So, the range of the given piecewise function is [0,11).
Therefore, the correct option is A.
Answer:
(1) The correct option is (A).
(2) The probability that Aadi will get Tails is .
Step-by-step explanation:
It is provided that:
- Eric throws a biased coin 10 times. He gets 3 tails.
- Sue throw the same coin 50 times. She gets 20 tails.
The probability of tail in both cases is:
(1)
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
In this case we need to compute the proportion of tails.
Then according to the Central limit theorem, Sue's estimate is best because she throws it <em>n = </em>50 > 30 times.
Thus, the correct option is (A).
(2)
As explained in the first part that Sue's estimate is best for getting a tail, the probability that Aadi will get Tails when he tosses the coin once is:
Thus, the probability that Aadi will get Tails is .
Parallel lines have same slope but different y-intercept. So they gave the equation y=4x-5 and the (0,3) where 3 is the y-intercept.
So the equation would be y=4x+3. HOPE THIS HELPS!!!!!!!!
If is a number that is both divisible by 4 and 5, then
4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that is a solution to the system, where is any integer. Simply put, any multiple of 20 fits the bill.
Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have when , so the sum we want to compute is