Let w = number of weeks.
In week w,
Laura has: 720 + 30w
Taylor has: 1200 - 30w
Set the two amounts equal and solve for w, the number of weeks.
720 + 30w = 1200 - 30w
60w = 480
w = 8
They will have the same amount of money in 8 weeks.
Laura will have in 8 weeks:
720 + 30w = 720 + 30 * 8 = 720 + 240 = 960
Taylor will have in 8 weeks:
1200 - 30w = 1200 - 30 * 8 = 1200 - 240 = 960
They will both have $960 in 8 weeks.
Answer: 14
<u>Step-by-step explanation:</u>
Consecutive even numbers would be {2, 4, 6, 8, 10, ...}
Let x represent the smallest number, then x + 2 is the middle number and x + 4 is the largest number.
1st number: x
2nd number: x + 2
3rd number: x + 4
1st number + 2nd number + 3rd number = Sum
(x) + (x + 2) + (x + 4) = 48
3x + 6 = 48
3x = 42
x = 14
1st number: x ⇒ x = 14
2nd number: x + 2 ⇒ 14 + 2 = 16
3rd number: x + 4 ⇒ 14 + 4 = 18
<u>Check:</u>
14 + 16 + 18 = 48 
Answer:
4. 11/6, 121/36
5. 5/8, 25/64
6. 4/7, 16/49
7. 14/9, 196/81
Step-by-step explanation:
The ratio of perimeters is the same as the ratio of corresponding legs:
perimeter ratio = red leg/blue leg
The area ratio is the square of that:
area ratio = (perimeter ratio)²
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<h3>4.</h3>
perimeter ratio = 11/6
area ratio = (11/6)² = 121/36
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<h3>5.</h3>
perimeter ratio = 5/8
area ratio = (5/8)² = 25/64
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<h3>6.</h3>
perimeter ratio = 4/7
area ratio = (4/7)² = 16/49
__
<h3>7.</h3>
perimeter ratio = 14/9
area ratio = (14/9)² = 196/81
Answer:
0.00939495805 about 0.01
Step-by-step explanation:
basically solving for arccos(8.5/9.9) in degrees, had to run through calculator
Answer: -1
The negative value indicates a loss
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Explanation:
Define the three events
A = rolling a 7
B = rolling an 11
C = roll any other total (don't roll 7, don't roll 11)
There are 6 ways to roll a 7. They are
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
Use this to compute the probability of rolling a 7
P(A) = (number of ways to roll 7)/(number total rolls) = 6/36 = 1/6
Note: the 36 comes from 6*6 = 36 since there are 6 sides per die
There are only 2 ways to roll an 11. Those 2 ways are:
5+6 = 11
6+5 = 11
The probability for event B is P(B) = 2/36 = 1/18
Since there are 6 ways to roll a "7" and 2 ways to roll "11", there are 6+2 = 8 ways to roll either event.
This leaves 36-8 = 28 ways to roll anything else
P(C) = 28/36 = 7/9
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In summary so far,
P(A) = 1/6
P(B) = 1/18
P(C) = 7/9
The winnings for each event, let's call it W(X), represents the prize amounts.
Any losses are negative values
W(A) = amount of winnings if event A happens
W(B) = amount of winnings if event B happens
W(C) = amount of winnings if event C happens
W(A) = 18
W(B) = 54
W(C) = -9
Multiply the probability P(X) values with the corresponding W(X) values
P(A)*W(A) = (1/6)*(18) = 3
P(B)*W(B) = (1/18)*(54) = 3
P(C)*W(C) = (7/9)*(-9) = -7
Add up those results
3+3+(-7) = -1
The expected value for this game is -1.
The player is expected to lose on average 1 dollar per game played.
Note: because the expected value is not 0, this is not a fair game.
