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When you've completed a problem, always go back and compare your answer against the known facts and reason whether it could be correct or not. Grace by herself can grand 15 papers in 25 minutes or 30 papers in 50 minutes. Why why it take 5 minutes longer when both of them work together? (maybe they're sisters who fight a lot!)
Emma can grade 10 papers in 20 minutes = 1 paper in 2 minutes = 5 papers in 10 minutes.
Grace can grade 15 papers in 25 minutes = 3 papers in 5 minutes = 6 papers in 10 minutes.
Together they can grade 11 papers every 10 minutes. That's less than 1 minute per test so you know your final answer will be less than 30 minutes.
30 tests / (11 tests / 10 minutes) = 30 tests * (10 minutes / 11 tests) = 300 / 11 minutes = 27.2727... minutes. Another helpful hint when doing these type of questions is to always include your units, which is "tests" and "minutes" in this case. If your final answer doesn't work out to have "minutes" as the unit, you know you messed up somewhere.
42. The equation is not a line so it doesn't have a slope. To figure out the intercepts, you plug in 0 for one of the variables and compute the other.
There is no value of x where y = 0. So there isn't an x intercept. The function is undefined for x=2 since you would be dividing by 0.
I'll take on the last one..
47. Are you sure the first set of numbers isn't supposed to be x=64, y=8? If so then the an equation which contains those points would be x = y^2.
a) The graph is not a line (y=mx+b) but rather a curve. Only lines have slopes.
b) x = y^2 (assuming you made a type on the first set of numbers)
c) No. It would be (4, -2) since (-2)^2 = 4
I've always been fascinated with primes. There's got to be some sort of pattern in there somewhere. Many, many, years ago I tried using different bases (binary, hex, etc.) to see if any patterns jumped out. Then the use of the internet began to growth quickly and I discovered that primes were fascinating to many. Brilliant minds have spent countless hours on this and I realized that finding a pattern is not going to be as simple as me converting the numbers to a different power-based system.
I still think that a completely different numbering system might be the answer. For example, instead of a power-based system, how about a prime-based system? So instead of expressing a number in ones, tens and hundreds..., express them in terms of 2's, 3's, 5's, 7's, etc.
Decimal PrimeBased
0 0
1 1
2 10
3 100
4 20
5 1000
6 110
7 10000
8 30
9 200
10 1010
11 100000
12 120
Basically this numbering system is expressing the prime factorization in a numeric form. From right to left, the columns are the prime numbers: 1, 2, 3, 5, 7, 11. So 6 in this system (3^1) * (2^1) * (1^0). 12 is (3^1) * (2^2) * (1^0). This system has not helped much and it has its flaws. Although the columns are prime-based, the actual digits are decimal. So 4 is represented as 20 but "2" in this system is 10. So 4 should really be 10\0 with a "" separating the digits. Make any sense?
I would like to write more but our New Year's guests have arrived. Happy New Year's to all you mathematicians!
With the stipulation that the answers alternate between True and False, there are only the 2 solutions you mentioned.
Even if you remove that restriction, there are only 252 (10 choose 5) ways to arrange the 5 True's and 5 False's. And even if you remove the restriction of 5 True's and 5 False's, there are only 1024 ways to complete the exam (2^10).
So it's definitely a typo.
You won't ever have a pair of 4's in a row. Let's say X is a prime number. We know it's odd as they all are except for 2. Every odd number is going to be either:
A) divisible by 3 (x mod 3 = 0);
B) have a remainder of 1 when divided by 3 (x mod 3 = 1);
C) have a remainder of 2 when divided by 3 (x mod 3 = 2).
Let's consider X, X+4 and X + 8 when X is prime:
A) X is divisible by 3 - can't happen when X is prime
B) X mod 3 = 1
Then (X+4) mod 3 = 2
and (X+8) mod 3 = 0.
Therefore X+8 is divisible by 3 and is not prime.
C) X mod 3 = 2
Then (X+4) mod 3 = 0
Therefore X+4 is divisible by 3 and is not prime.
So for any prime number X, either X+4 or X+8 will be divisible by 3.
You submitted this same question about a week ago.
http://www.mathsisfun.com/forum/viewtopic.php?id=5489
Yes and no to LQ's second reply. He showed that the distance is 10 and that 1/5 of 10 is 2. But now you have to move 2 units from 4 towards -6. Which, coincidentally, takes you to 2. So LQ's answer is correct, just needed to show you one more step.
I like combinations also. But it can be done your way also. You had 6! / (4! * 2!) but you didn't know how to exclude the numbers that began with a 0. Well the easiest way to do that is to not include them to begin with. The valid numbers have to begin with a 1 so that leaves three 1's and two 0's to arrange. You know how to do that. There's 5 digits so you start on with 5!. Divide by 3! because the 1 occurs 3 times. And then divide by 2! because the 0 occurs twice. You now have 5! / (3! * 2!) = 120 / (6*2) = 120/12 = 10.
Let's call the hourly wage z
46 hours at company Y: 40z + (46-40) * 1.5z = 49z
how long does it take someone at X to make 49z? Let's call Q the number of overtime hours, i.e. those in excess of 37 in this case.
37z + 1.5Qz = 49z
1.5Qz = 12z
1.5Q = 12
Q = 8
So he would have to work 45 hours.
I agree with the 2880. I used a different approach. There are 7! = 5040 different ways to arrange the 7 kids. Now subtract out the number of ways that there are either 2 boys or 2 girls on the ends.
2 Boys in the end seats: 4C2 * 2! * 5! = 6 * 2 * 120 = 1440
4C2 denotes (4 choose 2) and is the number of ways to pick the 2 boys for the end seats. Multiply that by 2! because that's the number of ways to arrange those 2 boys. The 5! is the number of ways to arrange the 5 kids (2 boys, 3 girls) in the inner seats.
2 Girls in the end seats: 3C2 * 2! * 5! = 3 * 2 * 120 = 720
3C2 denotes (3 choose 2) and is the number of ways to pick the 2 girls for the end seats. Multiply that by 2! because that's the number of ways to arrange those 2 girls. The 5! is the number of ways to arrange the 5 kids (4 boys, 1 girl) in the inner seats.
5040 - 1440 - 720 = 2880
Is there a standard way to denote (X choose Y) in text? I'll have to read through Latex notes to see if it's capable of doing the standard notation.
You're right Mathsyperson! Good catch.
Going with Mathsyperson's idea, there are 12 ways to distribution the additional 4 dollars. It can be divided $4, $0, $0 (3 different ways to do that), $3, $1, $0 (6 different ways to do that) or $2, $2, $0 (3 different ways).
D G L
5 1 1
1 5 1
1 1 5
4 2 1
4 1 2
2 4 1
2 1 4
1 4 2
1 2 4
3 3 1
3 1 3
1 3 3
And someone is always going to get at least $3 so that requirement is also met.
The area of a circle is always
. So for a circle of radius 10, the area is or 314.159268The area of a rectangle is height * width. But you don't know the height or the width, only the perimeter. There is no limit to the number of rectangles with a perimeter of 10. The rectangle could have height and width of 1 and 4 (1+1+4+4=10) or 1.5 and 3.5 (1.5+1.5+3.5+3.5=10) and so on.
Since your questions doesn't specify any more details about the height and width, I'm only going to so solutions where both of them are integers.
Length Width Perimeter Area
1 4 10 4
2 3 10 6
3 2 10 6
4 1 10 4
Assuming you meant to say each RED rose costs $8...
You have 8x + 3y = 80. x and y have to be integers since you're not going to buy your lovely wife/mom/girlfriend a half of a rose. So now your just try out some numbers. If you buy one $8 rose, that leaves $72 for 24 yellow roses. So that's one possibility.
If you buy 2 red roses for $16, that leaves $64 dollars for 21.333 yellow roses. No good.
If you buy 3 red roses for $18, that leaves $56 dollars for 18.666 yellow roses. No good.
If you buy 4 red roses for $24, that leaves $48 dollars for 16 yellow roses. Good.
Continue this and hopefully you'll pick up the pattern. You can buy the following bouquets for $80:
1 red and 24 yellow
4 red and 16 yellow
7 red and 8 yellow
10 red and 0 yellow
If I understand your notation currently, this is what you're saying:
That just means that each term is equal to the previous term plus 4. So the second term is 7 (3+4) and the 3rd term is 11 (7+ 4). So to get the 100 term you would have to add 4 ninety times to the first term of 3. 3 + (4*99) = 399.
A cubic meter is a cube with each side measuring 1 meter or 100 centimeters. Volume of a cube is length * width * height. So 1 meter * 1 meter * 1 meter = 1 cubic meter. Or 100 cm * 100 cm * 100 cm = 1000000 cubic centimeters.
A cube with sides of 10 cm has a colume of 1000 cubic centimeters (10 * 10 * 10). 1000 cubic centimeters is 1/1000 of 1000000 cubic centimeters (cubic meter).
The initial value was 10.08 (3.04 + 7.04) and the decrease was 3.04. The percentage decrease would be the amount of the drop divided by the initial amount. 3.04 / 10.08 ~= 30%. That's not one of your choices though.
A six letter word with no repeated letters (ABCDEF) can be arranged 720 ways (6!).
A six letter word with one letter repeated (AABCDE) can be arranged 360 ways (6! / 2!)
A six letter word with 2 letters repeated (AABBCD) can be arranged 180 ways (6! / (2!*2!)).
So the answer would be the six letter word which has 2 letters that are repeated. The word COFFEE would be an example.
1. No doublet or 7's are thrown. There are 36 (6*6) different possible throws. There are 6 doubles (11, 22, 33, 44, 55, 66) and 6 ways to roll a 7 (16, 25, 34, 43, 52, 61). None of them are duplicated, so there are 24 ways (36-6-6) out of the 36 ways to not throw a doublet or 7. Thats .666 or 66.6%
I'll use a different approach. How many different sets of 2 balls could you possibly choose? That would be (40 choose 2) ways or 780 ((40 * 39) / (2*1)).
How many different ways involve 2 lucky draws? That would be (10 choose 2) or 45.
How many different ways involve 2 unluck draws? That is (30 choose 2) or 435.
How many for 1 lucky and 1 unlucky? (10 choose 1) * (30 choose 1) or 300.
45 + 435 + 300 = 780. That's a good thing.
So:
a) both lucky: 45/780 = .057692
b) both unlucky: 435 / 780 = .55769
d) one lucky and one unlucky: 300 / 780 = .38461
c) one or more lucky: (300 + 45) / 780 = .44231
I used Excel to look at all of the different possibilities. The number of different possible trips is 2^n. Think about a 1 stage journey starting at A. Two possiblilities - B or C. Now consider a 2 stage journey. You take the 2 possible one stage journeys and continue with them. From B you can go to A or C and from C you can go to A or B. ABA, ABC, ACB, ACA. From each town you can go to 2 other towns.
Leg # possible End at A End at B End at C
1 2 0 1 1
2 4 2 1 1
3 8 2 3 3
4 16 6 5 5
5 32 10 11 11
6 64 22 21 21
So the formula looks like it will be:
Using Excel, I came up with 21 possible routes. The number of different possible 6 leg routes is 64 (2 ^64). I just mapped all of them in Excel and then looked to see how many ended in C. I got 21 for C, 21 for B and 22 for A.
A B A C B A C
A B A B C A C
A B A B A B C
A B A C A B C
A B A B C B C
A B C A B A C
A B C A C A C
A B C B C A C
A B C B A B C
A B C A C B C
A B C B C B C
A C A C B A C
A C A B C A C
A C A B A B C
A C A C A B C
A C A B C B C
A C B A B A C
A C B C B A C
A C B A C A C
A C B C A B C
A C B A C B C
Talvon's answer is correct but let's look at where you went wrong Teha:
Your first line: 14/100 = x/4300
The 14 is the number of women on welfare and 100 is the total women. X in the second fraction also represents the women on welfare but 4300 is the number of women NOT on welfare instead of the TOTAL number of women. It should be: