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Grade 7/8 POTW #2 I will start off with the second POTW since that is the quicker and easier one. We have 3 different die: 8 faces 12 faces 20 faces The question asks us the probability of at least 2 5s on the 3 die. To start things off, we have to find the number of possibilities or outcomes or the denominator. 8 x 12 x 20 = 1920 We now how to find all possibilities of at least 2 5s as shown. 1/8 x 1/12 = 1/96 = 20/1920 1/8 x 1/20 = 1/160 = 12/1920 1/12 x 1/20 = 1/240 = 8/1920 1/8 x 1/12 x 1/20 = 1/1920 However, there are 3 repeats of a 5 on each die. -(3/1920) 20/1920 + 12/1920 + 8/1920 + 1/1920 - 3/1920 = (20 + 12 + 8 + 1 - 3)/1920 = 38/1920 = 19/960 The probability of getting at least 2 5s on the die is 19/960. I will do the first POTW after this.
POTW 33: I didn't really know what to work with so this is what I put together A = 280 m squared I made a line where line GA is and extended it horizontally to the other side to make a rectangle I did some guess and check where I tried GA as 8 m and BC as 12 m to get the long rectangle area as 160 m squared Then since BC is 12 cm, I found the area of the shorter rectangle and got 120 m squared. I then added the two to find the area to confirm dimensions which is 280 m squared, so my guess was correct. Now that I know the dimensions of the room, I can find out where D is located on CE. CE is 18 m as AB is 10 m and FG is 8 m. I don't know what the question means by located, but I assume it's like "D is located __ m up CE. So I'm going to treat DC as a line segment for where D is located on CE. The rooms are split equally, so each trapezoidal area should be 140 m squared. The formula for trapezoid is A = (b1 + b2) /2 * h = A = (10 + _) /2 * 12. We can then work the formula backwards to find the value of DC. b2 = A/h * 2 - 10. b2 = 140 / 12 * 2 - 10. b2 = 11.6666... * 2 - 10 b2 = 23.3333... - 10 b2 = 13.3333... Using the formula: A = (10 + 13.3333...) /2 * 12 A = 140 m squared. Therefore DC = 13.3333... m long, where D is located
POTW 34: Combinations of numbers = 8 * 12 * 20 = 1920 combinations 1/8 chance for 5 on first die, 1/12 on second, 1/20 on third To get 5 on 2 or more dice = 1/8 * 1/12 or 1/8 * 1/20 or 1/20 * 1/12 Hoping for best combo would be 1/8 * 1/12 = 0.01 = 1/100 = 1% Therefore the best chance of getting 2 fives on two or more dice is 1% chance, 1/100, or 0.01.
Grade 7/8 POTW #1 I did my work on paper, but I will translate it onto here. Firstly, I took the figure, and changed it into 2 rectangles. The rectangle with length 10m would have an area less than 140m^2 because it got smaller than half the original figure. From there, I figured out that the rectangle with less area was 120m^2 with dimensions 10 by 12 while the one with a greater area was 160m^2 with dimensions 20 by 8. From there, I went back to the original shape and made the smaller rectangle again leaving a triangle, rectangle, and the other smaller figure. The triangle has an area of 20 m^2 because the figure + the rectangle (140 + 120) = 260 with 20 m remaining for the triangle. The triangle has a base of 12 m from the small rectangle meaning (20 x 2)/12 = h or height = 3.333... EC is 18 m as 8 + 10 (2 sides) = 18. CD is 10 + 3.333... = 13.333... while ED is 8 - 3.333... = 4.666... Therefore, D would be 4.333... m from E or 13.333...m from C.
POTW 34 There are 1920 different possibilities. I know this because the first cube has 8 different possibilities so there is 1/8 chance that I roll a certain number. The second cube has 12 sides so there is a 1/12 chance that I roll a certain number. The last cube has 20 sides so there is 1/20 chance that I pick a certain number. This means that there are (1/8 * 1/12 * 1/20) 1/1920 chances that I get a certain combination of numbers. I made a tree diagram and figured out that there are 38 possible ways that I can roll two or more fives. That means that there is a 38/1920 or 19/960 chance that a 5 is showing on the top of at least 2 dies.
POTW 33: I'm not sure exactly how to do this, but I'm guessing I would have to make a traingle with AD, AH (H is if GA was extended to intersect EC) and DH.
POTW 34: Info: - One 8 face - One 12 face - One 20 face - Probability that a 5 is showing for two of more dice?
Probabilty is almost always shown in a fraction (if not shown in a percentage), so to do this, we need to find the total number of possibilities for all dice and then find the "5 on two or more" possibilities. To find the total number of possibilities, we multiply 8, 12 and 20 which is 1920. Then, we need to calculate the total number of possibilities for "5 on two or more". To do this, we need to find out the possibilities of getting: - All 5 facing up: 1 possibility - 5 facing up for 8 and 12: 19 possibilities (there are 19 possibilities that the 20 isn't a 5 on that cube) - 5 facing up for 8 and 20: 11 possibilities - 5 facing up for 12 and 20: 7 possibilities
POTW 34: To do this, first you would have to figure out the denominator. Since you have an 8 sided die, a 12 sided die, and a 20 sided die, the total number of possibilities would be 8*12*20 = 1920. To get 5s on at least two dies, you have some choices. 5s on the 8 and 12 sided die, 5s on the 8 and 20 sided die, 5s on the 12 and 20 sided die, or 5s on all 3 sides.
If you were to get 5 on the 8 and 12 sided die, the third die can be any number, so that would give you 20 choices. If you were to get 5 on the 8 and 20 sided die, you would have 12 choices. If you were to get 5 on the 12 and 20 sided die, you would have 8 choices, and if you were to get 5 on all 3 dice you would get only 1 choice. Adding these up you would have 20+12+8+1 = 41 different ways, and 1920 in total. So the fraction would be 41/1920, or roughly 0.0021%.
POTW 33: Not sure if my answer is correct, as the answer is a repeating decimal. (All numbers are in meters) Basically, to do this question you must figure out AG, or FG, since they are the same length. To do this, you have to write an algebraic equation, using AG as the unknown variable, or as people like to say, x or y. What I did was I split the shape into two rectangles from a line, which creates a perpendicular line from point A to line CE.
Equation: 20*FG + 10(20-AG) = 280 Since FG is equal to AG, we will substitute FG for AG. 20*AG + 10(20-AG) = 280 20AG + 200-10AG = 280 10AG + 200 = 280 10 AG = 80 AG = 8 Thus, AG would equal to 8, and so would FG. Once we do that, BC would be 12. Now, if we still kept that line from A to line CE, then the top area would be 160 and the bottom would be 120. However, 280/2 = 140, so our goal is to make both areas be 140. So, what we do is make an extra triangle from A to D for the bottom area to make both sides equal. To get both areas to 140, there is a 20 meter squared difference. This means that the triangle has to be an area of 20 meters squared. Since our base is 12, then our height would be 20*2/12 = 3 1/3, following the triangle formula.
This means that D would be 4 2/3 m away from E, and 13 1/3 m away from C.
Grade 7/8 POTW #2
ReplyDeleteI will start off with the second POTW since that is the quicker and easier one. We have 3 different die:
8 faces
12 faces
20 faces
The question asks us the probability of at least 2 5s on the 3 die. To start things off, we have to find the number of possibilities or outcomes or the denominator.
8 x 12 x 20 = 1920
We now how to find all possibilities of at least 2 5s as shown.
1/8 x 1/12 = 1/96 = 20/1920
1/8 x 1/20 = 1/160 = 12/1920
1/12 x 1/20 = 1/240 = 8/1920
1/8 x 1/12 x 1/20 = 1/1920
However, there are 3 repeats of a 5 on each die.
-(3/1920)
20/1920 + 12/1920 + 8/1920 + 1/1920 - 3/1920 = (20 + 12 + 8 + 1 - 3)/1920 = 38/1920 = 19/960
The probability of getting at least 2 5s on the die is 19/960.
I will do the first POTW after this.
POTW 33:
ReplyDeleteI didn't really know what to work with so this is what I put together
A = 280 m squared
I made a line where line GA is and extended it horizontally to the other side to make a rectangle
I did some guess and check where I tried GA as 8 m and BC as 12 m to get the long rectangle area as 160 m squared
Then since BC is 12 cm, I found the area of the shorter rectangle and got 120 m squared.
I then added the two to find the area to confirm dimensions which is 280 m squared, so my guess was correct.
Now that I know the dimensions of the room, I can find out where D is located on CE. CE is 18 m as AB is 10 m and FG is 8 m.
I don't know what the question means by located, but I assume it's like "D is located __ m up CE.
So I'm going to treat DC as a line segment for where D is located on CE. The rooms are split equally, so each trapezoidal area should be 140 m squared.
The formula for trapezoid is A = (b1 + b2) /2 * h = A = (10 + _) /2 * 12.
We can then work the formula backwards to find the value of DC. b2 = A/h * 2 - 10.
b2 = 140 / 12 * 2 - 10.
b2 = 11.6666... * 2 - 10
b2 = 23.3333... - 10
b2 = 13.3333...
Using the formula:
A = (10 + 13.3333...) /2 * 12
A = 140 m squared.
Therefore DC = 13.3333... m long, where D is located
POTW 34:
Combinations of numbers = 8 * 12 * 20 = 1920 combinations
1/8 chance for 5 on first die, 1/12 on second, 1/20 on third
To get 5 on 2 or more dice = 1/8 * 1/12 or 1/8 * 1/20 or 1/20 * 1/12
Hoping for best combo would be 1/8 * 1/12 = 0.01 = 1/100 = 1%
Therefore the best chance of getting 2 fives on two or more dice is 1% chance, 1/100, or 0.01.
Grade 7/8 POTW #1
ReplyDeleteI did my work on paper, but I will translate it onto here.
Firstly, I took the figure, and changed it into 2 rectangles. The rectangle with length 10m would have an area less than 140m^2 because it got smaller than half the original figure. From there, I figured out that the rectangle with less area was 120m^2 with dimensions 10 by 12 while the one with a greater area was 160m^2 with dimensions 20 by 8.
From there, I went back to the original shape and made the smaller rectangle again leaving a triangle, rectangle, and the other smaller figure. The triangle has an area of 20 m^2 because the figure + the rectangle (140 + 120) = 260 with 20 m remaining for the triangle. The triangle has a base of 12 m from the small rectangle meaning (20 x 2)/12 = h or height = 3.333...
EC is 18 m as 8 + 10 (2 sides) = 18. CD is 10 + 3.333... = 13.333... while ED is 8 - 3.333... = 4.666...
Therefore, D would be 4.333... m from E or 13.333...m from C.
POTW 34
ReplyDeleteThere are 1920 different possibilities. I know this because the first cube has 8 different possibilities so there is 1/8 chance that I roll a certain number. The second cube has 12 sides so there is a 1/12 chance that I roll a certain number. The last cube has 20 sides so there is 1/20 chance that I pick a certain number. This means that there are (1/8 * 1/12 * 1/20) 1/1920 chances that I get a certain combination of numbers. I made a tree diagram and figured out that there are 38 possible ways that I can roll two or more fives. That means that there is a 38/1920 or 19/960 chance that a 5 is showing on the top of at least 2 dies.
19/960 = 0.019 = 1.9%.
POTW 33:
ReplyDeleteI'm not sure exactly how to do this, but I'm guessing I would have to make a traingle with AD, AH (H is if GA was extended to intersect EC) and DH.
POTW 34:
Info:
- One 8 face
- One 12 face
- One 20 face
- Probability that a 5 is showing for two of more dice?
Probabilty is almost always shown in a fraction (if not shown in a percentage), so to do this, we need to find the total number of possibilities for all dice and then find the "5 on two or more" possibilities. To find the total number of possibilities, we multiply 8, 12 and 20 which is 1920. Then, we need to calculate the total number of possibilities for "5 on two or more". To do this, we need to find out the possibilities of getting:
- All 5 facing up: 1 possibility
- 5 facing up for 8 and 12: 19 possibilities (there are 19 possibilities that the 20 isn't a 5 on that cube)
- 5 facing up for 8 and 20: 11 possibilities
- 5 facing up for 12 and 20: 7 possibilities
1+19+11+7= 38
38/1920
reduce
19/960
POTW 34:
ReplyDeleteTo do this, first you would have to figure out the denominator. Since you have an 8 sided die, a 12 sided die, and a 20 sided die, the total number of possibilities would be 8*12*20 = 1920. To get 5s on at least two dies, you have some choices.
5s on the 8 and 12 sided die, 5s on the 8 and 20 sided die, 5s on the 12 and 20 sided die, or 5s on all 3 sides.
If you were to get 5 on the 8 and 12 sided die, the third die can be any number, so that would give you 20 choices. If you were to get 5 on the 8 and 20 sided die, you would have 12 choices. If you were to get 5 on the 12 and 20 sided die, you would have 8 choices, and if you were to get 5 on all 3 dice you would get only 1 choice.
Adding these up you would have 20+12+8+1 = 41 different ways, and 1920 in total.
So the fraction would be 41/1920, or roughly 0.0021%.
POTW 33: Not sure if my answer is correct, as the answer is a repeating decimal.
ReplyDelete(All numbers are in meters)
Basically, to do this question you must figure out AG, or FG, since they are the same length. To do this, you have to write an algebraic equation, using AG as the unknown variable, or as people like to say, x or y.
What I did was I split the shape into two rectangles from a line, which creates a perpendicular line from point A to line CE.
Equation: 20*FG + 10(20-AG) = 280
Since FG is equal to AG, we will substitute FG for AG.
20*AG + 10(20-AG) = 280
20AG + 200-10AG = 280
10AG + 200 = 280
10 AG = 80
AG = 8
Thus, AG would equal to 8, and so would FG.
Once we do that, BC would be 12.
Now, if we still kept that line from A to line CE, then the top area would be 160 and the bottom would be 120. However, 280/2 = 140, so our goal is to make both areas be 140. So, what we do is make an extra triangle from A to D for the bottom area to make both sides equal. To get both areas to 140, there is a 20 meter squared difference. This means that the triangle has to be an area of 20 meters squared. Since our base is 12, then our height would be 20*2/12 = 3 1/3, following the triangle formula.
This means that D would be 4 2/3 m away from E, and 13 1/3 m away from C.
POTW #33
ReplyDeleteD is located 4.6(recurring)cm away from E and 13.3(recurring)cm away from C to accomplish the equal area split.
My work was done on paper.