There are 600 students including the seventh and eighth graders at the party.
This problem uses the concept of percentages to define the conditions that are laid in front of us.
Let the original number of students be S , and the number of seventh graders be = 0.60S
We know that percent is used to convey the mathematical term of a fraction multiplied by 100.
Total students after 20 eighth graders arrive = S + 20
And we have that
Number of seventh graders / total number of students = 58%
.60S / [ S + 20 ] = .58 we multiply both sides by S + 20
0.60S =0 .58 [ S + 20]
.60S = .58S + 11.6 we subtract 0.58S from both the sides
0.02S = 11.6 we divide both the sides by .02
S = 11/6 / 0.02 = 580
So the total number of students = 580 + 20 = 600 .
Hence there are 600 students at the party at that time.
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You could just subtract it or you could make it into a plus negative
Answer:
Php2040.38
Step-by-step explanation:
Given
--- Principal
--- Rate
--- Time
--- monthly
Required
Determine the amount at the end of two years
This is calculated as:
So, we have:
<em>Hence, the final amount is: Php2040.38</em>
Answer: $6.50
If the number is below 5 (so 1-4) you round down. If its above 5 (5-9), you round up.
E.G. 78.89
Round this to the nearest 10th, you'll get 78.90
Answer: D) No; If x is 5, the expression on the left simplifies to 8, making the inequality false.
In other words, if x = 5, then x+3 becomes 5+3 which ultimately becomes 8, but this is not greater than 8 on the right side.
Your steps could look like this
x+3 > 8
5+3 > 8 ... replace x with 5
8 > 8 ... this is a false inequality
