Two events are occurring:
1) Rolling a die
Sample Space = {1,2,3,4,5,6}
Total number of outcomes in sample space = 6
Favorable outcomes = Odd number
Number of Favorable outcomes = 3
Probability of getting an odd number = 3/6
2) Tossing a coin
Sample Space = {H, T}
Probability of getting a head= 1/2
The probability of getting odd number and head will be the product of two probabilities, which will be = 3/6 x 1/2 = 3/12
Thus there is 3/12 = 1/4 (0.25 or 25%) probability of getting an odd number and a head in given scenario.
So correct answer is option C
Answer:
Option C is correct.
She is not correct as she use the wrong y intercepts when graphing the equations.
Step-by-step explanation:
Given the system of equations:
.....[1]
. .....[2]
equate these two equations we get;
Add 5 to both sides we get;
Add 2x to both sides we have;
Multiply both sides by 3 we get;
8x = 24
Divide both sides by 8 we get;
x = 3
Substitute the value of x in [2] we have;
y = -2(3)+3 = -6+3
Simplify:
y = -3
Therefore, the correct solution for this given system of equation is (3, -3)
You can see the graph of these equations below.
So i think in the school there may be 4800 in the school. Hope this helps
Remember that if something is raised to a fraction exponent, the denominator of the fraction is the radical and the numerator stays a power. So it becomes:
fourth root of (48c)^3
Hope this helps
