Answer:
6.2y - 3.7
Step-by-step explanation:
−y+5.3+7.2y−9
Subtract 9 from 5.3 to get −3.7.
−y−3.7+7.2y
Combine −y and 7.2y to get 6.2y.
6.2y−3.7
There are 2 variables in this problem. One variable is the class number and other variable is the participation in extracurricular activities. Each variable has further two categories. There are two classes: Class 10 and 11. And students either participate or do not participate in extracurricular activities, which makes 2 categories.
The best approach to solve this question is to build a table and start entering the given information in it. When the given data has been entered fill the rest on basis of the data you have.
18 students from grade 11 participate in at least one Extracurricular activities. This means the rest students i.e. 22 students from grade 11 do not participate in Extracurricular activities.
32 students from grade 10 participate in at least one Extracurricular activities. This means total students who participate in at least one Extracurricular activities are 18 + 32 = 50 students.
The rest 50 students do not participate in at least one Extracurricular activities. From these 22 are from class 11. So the rest i.e. 28 are from class 10.
Answer:
To find the cost per yard, divide the cost by the amount:
p: 6.25 / 6.5 = 0.96 --> The cost per yard is $0.96
r: 3 /4 = 0.75 --> The cost per yard is $0.75
b: 8.1 /8.5 = 0.95 --> The cost per yard is $0.95
s: 7.2 / 6 = 1.2 --> The cost per yard is $1.20
In order from cheapest to most expensive:
Red
Brown
Purple
Silver
90 days.
For this question you just need to find the least common multiple of 18 and 30. That can be found by first finding the prime factorization of each number. 18 = 2 * 3 * 3 and 30 = 2 * 3 * 5. Then you multiply the 2 * 3 * 3 * 5, you ignore the 3 from 30 because 18 has more 3s. The answer is 90.
