7/9 would be irrational, I think.
<h3>Answer:</h3>
1.9
Step-by-step explanation:
<h3>The above is direct and inverse variation.</h3>
A = kB/C -----------(1)
A=12
B = 3
C= 2
substitute A, B and C into equation (1).
12 = K × 3/2
12 = 3k/2
12×2 = 3k
3K = 24
dividing bothsides by 3
3K/3 = 24/3
K = 8
substitute K = 8 into equation (1)
A = 8B /C --------------(2)
Equation (2) is the equation connecting
A,B and C.
Finding B when A = 10 and C = 1.5
10 = 8B / 1.5
10× 1.5 = 8B
15 = 8B
Dividing bothsides by 8 :
B = 15/8
B = 1.875
B = 1. 9 ( approximately)
A biased example: Asking students who are in line to buy lunch
An unbiased example: Asking students who are leaving/going to lunch(<em>NOT buying </em><em>lunch</em><em />).
But in this case, the answer choices can be... confusing.
Don't panic! You're given numbers and, of course, your use of logic.
Answer choice A: 100 students grades 6-8
Answer choice B: 20-30 students any <em>one</em> grade<em></em><em>
</em>Answer choice C: 5 students
<em></em>Answer choice D: 50 students grade 8
An unbiased example would be to choose students from <em>any grade.</em> So we can eliminate choices B and D.
Now, the question wants to <em>estimate how many people at your middle school buy lunch.</em> This includes the whole entire school, and if you are going to be asking people, you aren't just going to assume that if 5 people out of 5 people you asked bought lunch, the whole school buys lunch.
So, to eliminate all bias and/or error by prediction, answer choice A, the most number of students, is your answer.
Answer: Associative property of addition
I hope this helps :D
Hello!
Let 'a' stand for adult tickets and 'c' for children tickets.
a + c = 15
(Both adult and children tickets make up a total of 15 tickets).
30a + 20c = 330
a = 3
c = 12
A N S W E R:
William bought 3 adults tickets and 12 children's tickets.
