Answer:
d, 1/6
Step-by-step explanation:
so i did a diagram of some sorts to do this and got around six possible outcomes.
h is heads, t is tails.
h-h-h
t-t-t
h-t-h
t-h-t
t-t-h
h-h-t
feel free to correct me if any of this is wrong.
Answer:
Adult ticket: $7
Child ticket: $2
Step-by-step explanation:
Set up a system of equations where a represents the cost of one adult ticket and c is the cost of one child ticket:
2a + 3c = 20
a + 4c = 15
Solve by elimination by multiplying the bottom equation by -2:
2a + 3c = 20
-2a -8c = -30
Add them together:
-5c = -10
c = 2
Now, we can plug in 2 as c to find the value of a:
2a + 3c = 20
2a + 3(2) = 20
2a + 6 = 20
2a = 14
a = 7
Sometimes the best way to tell whether two variables are associated is to ask yourself whether they are not associated. Think backward. In a two-way frequency table, if the relative frequencies for one variable are the same (or close) for all categories of another variable, there is no (or little) association.
Answer:
the square of the length of the hypotenuse equals the sum of the squares of the lengths of the other two sides.
yeah i guess you re right sorry for making confusion
