Answer:
C
Explanation:
On the scatter plot, there is a dot where the lines for 6 and 3 meet. when looking for the ordered pair you would go along first then up (> then ^) so it would be (6,3). Another example, for the third dot on the plot, if you go along first you get 12 then when you go up you get 9 so the ordered pair for that would be (12,9). I really hope you understand what i mean and this helps as i struggled with explaining haha!
Answer:
True
Step-by-step explanation:
angle 3 and 6 are alternate angles.
alternate angles have the following characteristics:
1. they are inside the parallel lines
2. they are on opposite sides of the transversal
3. they are EQUAL
another example of alternate angles is angle 4 and angle 5
Answer:
a) 1 game
b) 41 goals
c) median = 2
Step-by-step explanation:
a)
As we can see in the line graph, where we have the 0 for the number of goals scored, the graph indicates only 1 in the number of games, so we have only 1 game where no goals were scored.
b)
To find the total number of goals scored, we multiply the goals scored by the number of games for that score, and then sum them all:
total goals = 1*0 + 4*1 + 5*2 + 6*3 + 1*4 + 1*5 = 41 goals
c)
To find the median, we put all the goals in crescent order, and then find the value in the middle. As we have 18 games, the middle value will be an average of the 9th and 10th terms.
We have 1 number 0, 4 numbers 1 and 5 numbers 2 in the beginning, so for these 10 numbers, the 9th and the 10th are the score 2, so the median is 2.
Earlier today I set you ten questions from this year’s International Singapore Maths Competition. Here are the questions and the answers. On the whole you did very well - smarter than a 10-year-old Singaporean! (With the caveat that they didn’t have multiple choice answers to choose from, and they are only ten). The only questions where your most popular answer the wrong one were 6 and 8. (C in Q6, and B in Q8). Thanks for taking part - now look through your workings...
For Year 5 pupils:
1. Mary cut off 2/5 of a piece of string. Later, she cut off another 14 m. The ratio of the length of string remaining to the total length cut off is 1 : 3. What is the length of the remaining string?
A. 5 m
B. 7 m
C. 10 m
D. 14 m
Solution is C. [73 per cent of readers got it right]
Oh Mary! This is how I would have solved it, using equations. Let L be the original length of the string, and R be what is remaining once you have cut the string twice. We know that R = (L x 3/5) – 14m, and that ((L x 2/5) +14) /R = 3, or 2L/5 + 14 = 3R. By substituting the first equation in the second we have 2L/5 + 14 = 9L/5 –42. Which rearranges to:7L/5 = 56, or L = 40. So R = 10m.
Interestingly, the Singapore method of solution is different. It requires us to think more visually about the string: We cut 2/5 of it. Then 14m, and are left with a piece that is a third of the size of what was cut. In other words, we are left with 1/4 of the original length. In order to compare the fractions 2/5 and then 1/4, lets change them to the lowest common denominator, which is 20. So, we cut off 8/20, subtract 14m and are left with 5/20. Let’s now draw the string divided into twentieths:
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The 14m must be 7/20 of the string, which mean each twentieth is 2m. The remaining piece of string is 5/20, i.e 10m
2. The areas of the faces of a rectangular box are 84 cm2, 70 cm2 and 30 cm2. What is the volume of the box?
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Photograph: ISMC
A. 300 cm3
B. 420 cm3
C. 490 cm3
D. 504 cm3
Solution is B. [85 per cent of readers got it right]
First we need to work out the possible side lengths, by seeing which two numbers multiply to be the area of each face. The 84 face could be 1 x 84, 2 x 42, 3 x 28, 4 x 21, 6 x 14 or 7 x 12. The 70 face could be 1 x 70, 2 x 35, 5 x 14 or 7 x 10. The 30 face could be 1 x 30, 2 x 15, 3 x 10 or 5 x 6.
The common factors between 84 and 70 are 1, 2, 7 and 14.
The common factors between 84 and 30 are 1, 2, 3 and 6.
The only way to make 84 with one each of these common factors are 14 from the top line and 6 from the bottom. So the edge bordering the 84 and 70 faces has length 14, and the edge bordering the 84 and 30 edges has length 6. Which means the height must be 30/6, or 70/14 = 5. Thus the volume is 14 x 6 x 5 = 420cm.
3. There are four numbers. If we leave out any one number, the average of the remaining three numbers will be 45, 60, 65 or 70. What is the average of all four numbers?
A. 50
B. 55
C. 60
D. 65
Solution is C. [82 per cent of readers got it right]
If the four numbers are A, B, C and D, then we know that
A + B + C = 45 × 3
A + B + D = 60 × 3
A + C + D = 65 × 3
B + C + D = 70 × 3
Now add them up to get 3A + 3B + 3C + 3D = (45 + 60 + 65 + 70) × 3
Which is A + B + C + D = (45 + 60 + 65 + 70) = 240. So their average is 240/4 = 60
Answer:
Answer explained below
Step-by-step explanation:
Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds
1.START T=0.....
1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.
T=1.....
EACH OF THE M/C SENDS 10 NEW MAILS ....
.................................
LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.
SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.
HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=
M[N+1]=10*M[N].......................1
THIS IS THE RECURRENCE RELATION.....
2.SOLUTION .....
M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........
M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]
M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2
THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....
M[N]=[10^(N+3)].............................................3
..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....
3.AFTER N=20 SECONDS , THE ANSWER IS ....
M[20]=10^(20+3)=10^23
