Answer:
ok this is a good question
Step-by-step explanation:
i don't even know it
Given the domain {-4, 0, 5}, what is the range for the relation 12x 6y = 24? a. {2, 4, 9} b. {-4, 4, 14} c. {12, 4, -6} d. {-12,
xz_007 [3.2K]
The domain of the function 12x + 6y = 24 exists {-4, 0, 5}, then the range of the function exists {12, 4, -6}.
<h3>How to determine the range of a function?</h3>
Given: 12x + 6y = 24
Here x stands for the input and y stands for the output
Replacing y with f(x)
12x + 6f(x) = 24
6f(x) = 24 - 12x
f(x) = (24 - 12x)/6
Domain = {-4, 0, 5}
Put the elements of the domain, one by one, to estimate the range
f(-4) = (24 - 12((-4))/6
= (72)/6 = 12
f(0) = (24 - 12(0)/6
= (24)/6 = 4
f(5) = (24 - 12(5)/6
= (-36)/6 = -6
The range exists {12, 4, -6}
Therefore, the correct answer is option c. {12, 4, -6}.
To learn more about Range, Domain and functions refer to:
brainly.com/question/1942755
#SPJ4
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:

And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Yes its right
hope this helps
Answer:
idk
Step-by-step explanation:
idk