The left-handed students12% are left handed, percent means per hundred. Change percent into per 100.
12% =

Then simplify the fraction by dividing the numerator and denominator by 4

=

=

This is the fraction showing the proportion of the left-handed students
The right-handed studentsDetermine the fraction showing the proportion of the right-handed students
1 - the left-handed
= 1 -

Equalize the denominator to 25

Simplify

This is the proportion of the right-handed students
<span>
Are there more right-handed or left-handed students?There are more right-handed students than the left-handed students, because 21/25 is more than 3/25</span>
Answer:
12
Step-by-step explanation:
This thing to know about this question is that angle 1 and angle 3 are actually the same.
When two lines intersect, the opposite angles such as that shown in 1 and 3 are equal to each other.
So what you can do is set angle 1 and 3 equal to each other like so:
5x+10 = 70
Then solve for x
5x+10 = 70
5x = 60
x = 12
I'm not going to do the entire question, but I'll do the first part.
For the first part, you have to set up an inequality. Use context clues, such as
no more than (the one used in this sentence) to figure out what the symbol should be for the inequality.
In this case, because it is NO MORE THAN 6400, it would be:
6400

250r + 475p
where r represents the # refrigerators, and p represents the # pianos
Now that you have the equation, make the graph, then for the 3rd part, plug in 12 and 8 to r and p to see if it keeps the inequality true.
Hope this helps! :)
Let the third angle of the triangle be y.
y + 64° + 45° = 180° (angle sum of triangle is 180°)
x + y = 180° (angle of a straight line is 180°)
Notice that they both equal 180°. This means that they are the same thing, hence:
y + 64° + 45° = x + y
Since y = y, it is not necessary to write them down. Remove y from both sides of the equation:
64° + 45° = x
Surprise! exterior angle = sum of interior opposite angles.
I hope this clears up any confusion you might have. If not, feel free to leave a comment with your question.