Answer:
or one seven eighteenths
Step-by-step explanation:
we know that
12 one half times one nine is the same that multiply 12 one half by one nine
so

Convert mixed number to an improper fraction

substitute

Convert to mixed number

therefore
12 one half times one nine is one seven eighteenths
Answer:
84 is the highest possible course average
Step-by-step explanation:
Total number of examinations = 5
Average = sum of scores in each examination/total number of examinations
Let the score for the last examination be x.
Average = (66+78+94+83+x)/5 = y
5y = 321+x
x = 5y -321
If y = 6, x = 5×6 -321 =-291.the student cannot score -291
If y = 80, x = 5×80 -321 =79.he can still score higher
If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.
If y= 100
The average cannot be 100 as student cannot score 179(maximum score is 100)
While fire clay bricks have a density of 2400 kg/m3, for common red bricks it is 1900 kg/m3 . [
Answer:
10 hours
Step-by-step explanation:
5.75x 10= 57.50
Answer:
The range of the function is:
Range R = {14, 17, 20}
Step-by-step explanation:
Given the function

We also know that range is the set of values of the dependent variable for which a function is defined.
In other words,
- Range refers to all the possible sets of output values on the y-axis.
We are given that the domain of the function is:
Domain D = {4, 5, 6}
Now,
substituting x = 4 in the function
f(4) = 3(4) + 2
f(4) = 12 + 2
f(4) = 14
substituting x = 5 in the function
f(5) = 3(5) + 2
f(5) = 15 + 2
f(5) = 17
substituting x = 6 in the function
f(6) = 3(6) + 2
f(6) = 18 + 2
f(6) = 20
Thus, we conclude that:
at x = 4, y = 14
at x = 5, y = 17
at x = 6, y = 20
Thus, the range of the function is:
Range R = {14, 17, 20}