These three individual items take 1/4, 1/6, and 1/8 of the budget. The part of the budget that they take all together is the sum of those individual parts. You need to add up 1/4, 1/6, and 1/8. To do that, you'll need to put all three fractions over a common denominator first, the
Answer:
C: x^4+6x^3-12x-72
Step-by-step explanation:
For the function : f(x)=√x^2+12x+36
It can be factorized to
f(x)=√(x+6)²
=x+6
f(x)*g(x)
=(x+6)(x^3-12)
Expand the brackets
=x^4-12x+6x^3-72
rearrange the numbers
x^4+6x^3-12x-72
Answer:
10(5x - 6).
8(2m + 5).
9(3 - 5c)
Step-by-step explanation:
Given that
50x-60
16m+40
27-45c
The greatest common factor is a measure of the highest number present inside, or common to all the numbers that were listed. To take this one at a time, we have....
50x - 60. The highest number common to both of them is 10, and thus, we have
50x - 60 = 10(5x - 6), and this is the solved part.
16m + 40. The greatest number that's present inside both numbers is 8, and thus we have
16m + 40 = 8(2m + 5) and this is the solved part.
27 - 45c. The greatest common number is 9, and as a result, we have
27 - 45c = 9(3 - 5c), and this is the solution we were searching for
Answer:
Sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.
Step-by-step explanation:
To yield a more accurate estimate of the population mean, margin of error should be minimized.
margin of error (ME) of the mean can be calculated using the formula
ME= where
- z is the corresponding statistic in the given confidence level(z-score or t-score)
- s is the standard deviation of the sample (or of the population if it is known)
for a given confidence level, and the same standard deviation, as the sample size increases, margin of error decreases.
Thus, random sample of 50 people from population A, has smaller margin of error than the sample of 20 people from population B.
Therefore, sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.