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Darina [25.2K]
4 years ago
8

Registration records show that 5 out of 8 students at college are older than 25 years. If there are 1200 students at the college

, how many would you expect to be older than 25 years?
Mathematics
1 answer:
stellarik [79]4 years ago
6 0
Take the total (1200) and divide it by the base number (8) and then multiply by how many there are in the ratio (5)
1200/8= 150    150*5=750
750 students are 25+
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Solve 3/8 = 6w-7/2w+14
Studentka2010 [4]
(PLEASE GIVE THIS BRAINLIEST)Multiply the numbers
3
8
=
6

−
1
⋅
7
2

+
1
4
3
8
=
6
w
−
1
⋅
7
2
w
+
14
83​=6w−1⋅27​w+14
3
8
=
6

−
7
2

+
1
4
Combine multiplied terms into a single fraction
3
8
=
6

−
7
2

+
1
4
3
8
=
6
w
−
7
2
w
+
14
83​=6w−27​w+14
3
8
=
6

+
−
7

2
+
1
4
3
8
=
6
w
+
−
7
w
2
+
14
83​=6w+2−7w​+14
3
Find common denominator
3
8
=
6

+
−
7

2
+
1
4
3
8
=
6
w
+
−
7
w
2
+
14
83​=6w+2−7w​+14
3
8
=
2
⋅
6

2
+
−
7

2
+
2
⋅
1
4
2
3
8
=
2
⋅
6
w
2
+
−
7
w
2
+
2
⋅
14
2
83​=22⋅6w​+2−7w​+22⋅14​
4
Combine fractions with common denominator
3
8
=
2
⋅
6

2
+
−
7

2
+
2
⋅
1
4
2
3
8
=
2
⋅
6
w
2
+
−
7
w
2
+
2
⋅
14
2
83​=22⋅6w​+2−7w​+22⋅14​
3
8
=
2
⋅
6

−
7

+
2
⋅
1
4
2
3
8
=
2
⋅
6
w
−
7
w
+
2
⋅
14
2
83​=22⋅6w−7w+2⋅14​
5
Multiply the numbers
3
8
=
2
⋅
6

−
7

+
2
⋅
1
4
2
3
8
=
2
⋅
6
w
−
7
w
+
2
⋅
14
2
83​=22⋅6w−7w+2⋅14​
3
8
=
1
2

−
7

+
2
⋅
1
4
2
3
8
=
12
w
−
7
w
+
2
⋅
14
2
83​=212w−7w+2⋅14​
6
Multiply the numbers
3
8
=
1
2

−
7

+
2
⋅
1
4
2
3
8
=
12
w
−
7
w
+
2
⋅
14
2
83​=212w−7w+2⋅14​
3
8
=
1
2

−
7

+
2
8
2
3
8
=
12
w
−
7
w
+
28
2
83​=212w−7w+28​
7
Combine like terms
3
8
=
1
2

−
7

+
2
8
2
3
8
=
12
w
−
7
w
+
28
2
83​=212w−7w+28​
3
8
=
5

+
2
8
2
3
8
=
5
w
+
28
2
83​=25w+28​
8
Multiply all terms by the same value to eliminate fraction denominators
3
8
=
5

+
2
8
2
3
8
=
5
w
+
28
2
83​=25w+28​
2
⋅
3
8
=
2
(
5

+
2
8
2
)
2
⋅
3
8
=
2
(
5
w
+
28
2
)
2⋅83​=2(25w+28​)
9
Cancel multiplied terms that are in the denominator
2
⋅
3
8
=
2
(
5

+
2
8
2
)
2
⋅
3
8
=
2
(
5
w
+
28
2
)
2⋅83​=2(25w+28​)
2
⋅
3
8
=
5

+
2
8
2
⋅
3
8
=
5
w
+
28
2⋅83​=5w+28
10
Multiply the numbers
2
⋅
3
8
=
5

+
2
8
2
⋅
3
8
=
5
w
+
28
2⋅83​=5w+28
3
4
=
5

+
2
8
3
4
=
5
w
+
28
43​=5w+28
11
Subtract
2
8
28
28
from both sides of the equation
3
4
=
5

+
2
8
3
4
=
5
w
+
28
43​=5w+28
3
4
−
2
8
=
5

+
2
8
−
2
8
3
4
−
28
=
5
w
+
28
−
28
43​−28=5w+28−28
12
Simplify
Subtract the numbers
Subtract the numbers
−
1
0
9
4
=
5

−
109
4
=
5
w
−4109​=5w
13
Divide both sides of the equation by the same term
−
1
0
9
4
=
5

−
109
4
=
5
w
−4109​=5w
−
1
0
9
4
5
=
5

5
−
109
4
5
=
5
w
5
5−4109​​=55w​
14
Simplify
Divide the numbers
Cancel terms that are in both the numerator and denominator
Move the variable to the left

=
−
1
0
9
2
0
w
=
−
109
20
w=−20109​
7 0
3 years ago
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