For this case, the first thing we must do is define variables.
We have then:
n: number of cans that each student must bring
We know that:
The teacher will bring 5 cans
There are 20 students in the class
At least 105 cans must be brought, but no more than 205 cans
Therefore the inequation of the problem is given by:
Answer:
105 <u><</u> 20n + 5 <u><</u> 205
the possible numbers n of cans that each student should bring in is:
105 <u><</u> 20n + 5 <u><</u> 205
I have answered this question before.
Given:
Bob has three kids whose ages has the product of 72. I will only be able to give you a set of 3 numbers to represent the ages. Had the sum of the ages been given, the specific ages would be provided.
I just did a prime factorization of 72.
72 ÷ 2 = 36
36 ÷ 2 = 18
18 ÷ 2 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1
1 x 2 x 2 x 2 x 3 x 3 = 72 ⇒ 2³ x 3²
There are a lot of possible combinations, here are a few.
1 x 8 x 9 = 72 ⇒ 1 + 8 + 9 = 18
1 x 4 x 18 = 72 ⇒ 1 + 4 + 18 = 23
<span>2 x 4 x 9 = 72 ⇒ 2 + 4 + 9 = 15</span>
Answer:
a) S= (1324
1342
1423
1432
2314
2341
2413
2431
3124
3142
3214
3241
4123
4132
4213
4231)
b) A= (1324
1342
1423
1432)
Step-by-step explanation:
The number of total outcome is given by,
T= 4×2×2×1=16
Listing them we have;
S=
(1324
1342
1423
1432
2314
2341
2413
2431
3124
3142
3214
3241
4123
4132
4213
4231)
b) A= (1324
1342
1423
1432)
In my opinion I think you would be able to make atleast 2 outfits
