Answer:
A. 0.009899
B. 0.005624
Step-by-step explanation:
Data:
Let the probability that an item is defective = 
The probability that the item is not defective = 
The probability that the fifth item is defective = 
= 0.009899
Probability that one in 5 items is defective = 0.005624
Answer:
about 11.5 cm.
Step-by-step explanation:
I know the measurement of the other leg is about 11.5 cm. I know because I used the Pythagorean theorem.
a^2+b^2=c^2.
"a" and "b" are the values of the legs of the triangle, while "c" is the measure of the hypotenuse. We know that 8cm is the measure of one of the legs, and 14 cm is a measure of the hypotenuse.
8^2+b^2=14^2 simplified: 64+b^2=196
Then, I subtracted 64 on both sides, so I would have "b" by itself.
b^2=132
Next, I found the square root of both b^2 and 132, so I would find the true value of "b."
b=11.4891252931
So, the measure of the other leg rounded to the nearest tenth is about 11.5.
So.
.
Let's get started with formulas.
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I KNOW you hate formulas!! So do I, but we have to use them. No "buts".
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So, how many minutes are in 2 weeks?
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Search it up. For your convenience, it is 20,160 minutes.
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Well, I think you have your answer, since each quiz is 5 minutes, 5 times 50 (5 x 50) is 2,500 minutes.
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Yes, you can do 50 quizzes and MUCH MUCH more in 2 weeks.
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Hope I helped!!
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
