<span>3[2(6 + 4)-8]+8÷2
Inner most bracket (6 + 4):
= </span><span>3[2(10)-8]+8÷2
Multiplication within the bracket (2(10)):
</span>= 3[20-8]+8÷2
Bracket (20 - 8):
= 3[12]+8÷2
Multiplication (3(12)):
= 36+8÷2
Division (8÷2):
= 36+4
Add (36 + 4):
=40
Answer: 40
Answer:
FALSE
Step-by-step explanation:
The central limit theorem tell us that for a random sample of size n, the average of the sample will be approximately normally distributed with mean 
and variance 
, where and 
are the mean a variance respectively of the sampling distribution. This always that the mean and the variance are both finite and the sample size n is greater than 30. We could use the central limit theorem, but in this case does not help us because we are considering 25 randomly selected data values, besides, we do not know the distribution of the original population.
10000(1+0.01)^10 = <span>11046.22
The value would be $11046.22</span>
Answer:
The answer to your question is 25%
Step-by-step explanation:
Data
Number of games won last year= 16
Number of games won this year = 20
Process
1.- Use the number of games won last year as the 100% and use proportions to find the answer.
16 games ---------------- 100 %
20 games -------------- x
x = (20 x 100) / 16
x = 2000/16
x = 125
2.- Subtract the percent obtain to 100% to obtain the percent increase.
Percent increase = 125 - 100
= 25%
D = m
----
v
2.336 g/cm^3 = m
----
350
(350) 2.336 = m
817.6g = m
