September 28 Challenge
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| Click here for a link to this week's Math Enrichment Problem of the Week DUE Friday, September 28 |
September 21 Challenge
Samantha is experimenting with three spinners. Each spinner is divided into three equal sections with a number in each section.
Spinner A shows 8, 6, 1;
Spnner B shows 7, 5, 3;
Spnner C shows 9, 4, 2.
Samantha spins two spinners and recored as winner the one that stopped on the greater number. She has already found that A usually wins over B and the B usually wins over C. She guesses that A will usually win over C.
Is her guess correct?
Find the probabilities that A wins over B and that B wins over C.
If spinners A and C are spun, which one do you think will win most often?
Answer:
Her guess is wrong.
P(A wins over B) = .56
P(B wins over C) = .56
P(A wins over C) = .44 So, C will win over A 56% of the time.
Spinner A shows 8, 6, 1;
Spnner B shows 7, 5, 3;
Spnner C shows 9, 4, 2.
Samantha spins two spinners and recored as winner the one that stopped on the greater number. She has already found that A usually wins over B and the B usually wins over C. She guesses that A will usually win over C.
Is her guess correct?
Find the probabilities that A wins over B and that B wins over C.
If spinners A and C are spun, which one do you think will win most often?
Answer:
Her guess is wrong.
P(A wins over B) = .56
P(B wins over C) = .56
P(A wins over C) = .44 So, C will win over A 56% of the time.
September 14 Challenge
How many domino pieces (bones) are in a double-fifteen set?
Remember there will be a double-zero, a double-one, and so on to double-fifteen, and that one-zero is the same as zero-one.
http://en.wikipedia.org/wiki/Dominoes
Remember there will be a double-zero, a double-one, and so on to double-fifteen, and that one-zero is the same as zero-one.
Answer: 136
For more information, go to:http://en.wikipedia.org/wiki/Dominoes
Due December 4
Answer to Last Week's Problem:
3:00 A.M., 3:00 P.M., 9:00 A.M., 9:00 P.M.
Today's Problem:
In a far-off land, three fish can be traded for two loaves of bread, and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?
3:00 A.M., 3:00 P.M., 9:00 A.M., 9:00 P.M.
Today's Problem:
In a far-off land, three fish can be traded for two loaves of bread, and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?
Due November 27
Answer to Yesterday's Problem:
25 triangles. The perimeter will be a + a + b = 99, where a represents the lengths of the equal sides and b represents the base, or the third side. It follows that a + a must be even, since the sum of two odd numbers or the sum of two even numbers will always be even. Therefore, the base b must be an odd number in length to allow an even number plus b to equal 99. Therefore, to test the numbers, the possible lengths of the base must include 1 to 97, inclusive. However, the sum of the two equal lengths must be greater than the length of the base for the three segments to form a triangle. When the base length reaches 51, this situation is no longer possible (24 + 24 = 48). Therefore, the greatest length of a base will be 49 inches. The base includes the odd numbers 1 to 49, inclusive, which will provide 25 triangles. The possibilities are listed below.
25, 25, 49
26, 26, 47
27, 27, 45
28, 28, 43
29, 29, 41
...
49, 49, 1
Today's Problem (5 points):
Find all the times, on the hour, when the minute and hour hands of a clock form a 90 degree angle.
25 triangles. The perimeter will be a + a + b = 99, where a represents the lengths of the equal sides and b represents the base, or the third side. It follows that a + a must be even, since the sum of two odd numbers or the sum of two even numbers will always be even. Therefore, the base b must be an odd number in length to allow an even number plus b to equal 99. Therefore, to test the numbers, the possible lengths of the base must include 1 to 97, inclusive. However, the sum of the two equal lengths must be greater than the length of the base for the three segments to form a triangle. When the base length reaches 51, this situation is no longer possible (24 + 24 = 48). Therefore, the greatest length of a base will be 49 inches. The base includes the odd numbers 1 to 49, inclusive, which will provide 25 triangles. The possibilities are listed below.
25, 25, 49
26, 26, 47
27, 27, 45
28, 28, 43
29, 29, 41
...
49, 49, 1
Today's Problem (5 points):
Find all the times, on the hour, when the minute and hour hands of a clock form a 90 degree angle.
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