7) What is the solution to the system of equations?
1 answer:
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You hit the right target!
Just to be sure, substitute with x = 2 in the equation.
Answer is 54^2.
since there are triangles in this one, it is easier to do this. all you have to do is make boxes in and since there are angles, outside the lines. like this picture.
the orange is 1
red and dotted is 2
blue is 3
purple is 4(a regular box which should be easy to count.
all you have to do is add up all the boxes within the boundaries.
boundary 1: 12
boundary 2: 18
boundary 3: 6
boundary 4: 36
now when the boundary has a triangle in it (1, 2, & 3) divide the number you got in half or 2.
boundary 1: 6
boundary 2: 9
boundary 3: 3
the ractangle box doesn't not get divided.
boundary 1: 6
boundary 2: 9
boundary 3: 3
boundary 4: 36
add all the numbers you got now for each boundary and that would be your area squared.
6+9+3+36=54^2
so your final answer is 54^2.
i hope this helps you.
since there are triangles in this one, it is easier to do this. all you have to do is make boxes in and since there are angles, outside the lines. like this picture.
the orange is 1
red and dotted is 2
blue is 3
purple is 4(a regular box which should be easy to count.
all you have to do is add up all the boxes within the boundaries.
boundary 1: 12
boundary 2: 18
boundary 3: 6
boundary 4: 36
now when the boundary has a triangle in it (1, 2, & 3) divide the number you got in half or 2.
boundary 1: 6
boundary 2: 9
boundary 3: 3
the ractangle box doesn't not get divided.
boundary 1: 6
boundary 2: 9
boundary 3: 3
boundary 4: 36
add all the numbers you got now for each boundary and that would be your area squared.
6+9+3+36=54^2
so your final answer is 54^2.
i hope this helps you.
Answer:
(a)$67
(b)You are expected to win 56 Times
(c)You are expected to lose 44 Times
Step-by-step explanation:
The sample space for the event of rolling two dice is presented below
Total number of outcomes =36
The event of rolling a 5 or a 6 are:
Number of outcomes =20
Therefore:
P(rolling a 5 or a 6)
The probability distribution of this event is given as follows.
First, we determine the expected Value of this event.
Expected Value
Therefore, if the game is played 100 times,
Expected Profit =$0.67 X 100 =$67
If you play the game 100 times, you can expect to win $67.
(b)
Probability of Winning
If the game is played 100 times
Number of times expected to win
Therefore, number of times expected to loose
= 100-56
=44 times