Since there are 6 students out of which one needs to be selected, the principal chose two die on which there are six numbers each numbered from 1 , 2, 3, 4, 5, 6.
Since there are two dice, the total possible outcome is 36.
Hence, the probability of getting one number each is 1/36
Hence, the principal used a fair method because each result is an equally likely possible outcome.
Option B is correct.
You need to solve for one variable in equation 1 and substitute it in equation 2 to solve.
Equation 1: x+y=24
x= number of 3 pt questions
y= number of 5 pt questions
24= Total number of questions
Equation 2: 3x+5y=100
100= Total point value possible on test
3x= point value of 3 pt questions
5y= point value of 5 pt questions
x+y=24
Subtract y from both sides
x=24-y
Substitute in equation 2:
3x+5y=100
3(24-y) +5y=100
72-3y+5y=100
72+2y=100
Subtract 72 from both sides
2y=28
Divide both sides by 2
y=14
Substitute y=14 back in to solve for x:
3x+5y=100
3x+5(14)=100
3x+70=100
Subtract 70 from both sides
3x=30
Divide both sides by 3
x=10
So there are 10 three point questions
There are 14 five point questions.
Hope this helped! :)
One way to solve the system is to <u>substitute</u> a variable.
<u>Explanation:</u>
One approach to solve an equation is by substitution of one variable. Right now, a condition for one factor, at that point substitute that arrangement in the other condition, and explain. All value(s) of the variable(s) that fulfills a condition, disparity, arrangement of conditions, or arrangement of imbalances.
The technique for tackling "by substitution" works by settling one of the conditions (you pick which one) for one of the factors (you pick which one), and afterward stopping this go into the other condition, "subbing" for the picked variable and fathoming for the other. At that point you back-explain for the principal variable.
