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The mean of a group of numbers is what everyone usually calls the average. Add up all your numbers and then divide that answer by the number of terms (13 in this case).
To find the median, rearrange your numbers from smallest to biggest. The median will be the one in the middle. You have 13 numbers to sort so one of them will be in the exact middle. If you have an even number of terms, I think you take the mean (average) of the 2 middle terms. For example: 2, 3, 5, 7, 8, 10 The two middle terms are 5 and 7 so the median is 6.
The mode is the number(s) which appears the most time. In your group, it looks like 20 is the only one that appears more than once, so it's the mode.
Edited: Once again, I'm just a little bit late. Good work, Ganesh.
139. If
136. A certain sum of money doubles in 10 years at simple
interest. What is the rate of interest?
Ganesh - I agree with kylekatarn. Excellent work. Are you keeping track of which ones haven't been successfully answered? If so, maybe publish that list occasionally.
Are you sure about #132?
You need to break the equations into factors. From experience, I know that the solution will probably involve breaking down the equation into the form (x+a) (x+b) = 0. Let's see what happens when you multiply that equation out:
Your initial equation looks like that. In this case, a+b = 5 and ab=-24. Keep in mind that either or both a and b can be less than 0. What 2 numbers multplied together equal 24? There's the following pairs of numbers: 1 and -24, -1 and 24, 2 and -12, -2 and 12, 3 and -8,
-3 and 8, 4 and -6, -6 and 4.
Now of those groups which, if any, of those pairs when added together equal 5? Just one of them and it's -3 and 8. So your equation can be factored into (x-3)(x+8)=0 . The only ways that equation can be equal to 0 is when x-3=0 or x+8=0 That happens when x is equal to 3 or -8. So that's your answer. Plug those values in for x to double-check your solution.
Both solutions check out okay.
Equations don't often factor out as easily as this one did. Your book is giving you these questions because they do factor cleanly. Later I'm sure you'll get equations that are as easy and that's when the quadratic formula comes in handy.
Let's look at the second problem. x²-3x-28=0. The same method can be used. You want to find a and b where ab=-28 and a+b=-3. You'll find out that a and b work out to -7 and 4 so that you get (x-7)(x+4)=0. Again, the only way that equation can be equal to 0 is if one of the 2 equations inside the parenthesis is equal to 0. So x is equal to 7 or -4. These types of problems will get easier with experience and practice.
The third problem is a little different. 4x²=3x
Same prinicipal now. For that equation to be equal to 0, either 4x-3=0 or x=0. For 4x-3=0, x=3/4.
Edited - Ganesh beat me to the punch. While the quadratic equation may sometimes be simpler and always work, I'm sure your teacher is looking for you to factor out these equations. Personally I always took a quick look to see if it can be easily factored. If so, I'm done. If not, then I resort to the quadratic equation.
5.The prime numbers p and q are called twin primes if |p-q|=2, Prove that if p and q are twin primes greater than 3 then q+p is divisible by 12
Logic similar to the that used to solve #4 can be used here. I don't know if you would call this a "proof" though. More of a demonstation.
If q+p is divisible by 12, it has to be divisble by 3 and 4. More importantly, if a number is divisible by 3 and by 4, it's divisible by 12 (if your teacher is a stickler for details, you may have to prove that also).
Let's make n equal to the smaller of p and q. That makes the other number equal to n+2. Let's first show that n + (n+2) is divisible by 4.
n+(n+2) = 2n + 2
=2(n+1)
We factored out a 2, so we know p+q is divisble by 2. Add since n is odd, n+1 is even and therefore also divisible by 2. So p+q is divisble by 4.
Now for divisble by 3. Just like in #4, for every (odd) number, one and only one of the following will be true:
Case 1: n mod 3 = 0;
Case 2: n mod 3 = 1; or,
Case 3: n mod 3 = 2
Keep in mind that n is a prime and also equal to the smaller of p and q. Since n is prime,
Case 1 can't be true. If Case 2 is true, then ((n+2) mod 3) = 0. That makes n+2 non-prime (divisble by 3). So Case 2 is also ruled out. That means Case 3 is true. Therefore,
((n+2) mod 3) = 1 and ((n + (n+2) mod 3) = 0. So, n + (n+2) is divisble by 3.
It now has been shown that n + (n+2) is divisble by 3 and by 4. Therefore it is also divisble by 12.
#4. In another topic, it was shown that x, x+4 and x+8 can't all be primes. The same logic works for x, x+2 and x+4. Simply put, one of those 3 numbers has to be divisble by 3.
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.
Another way to think about it is to remember what the definition of Y intercept is. It is the value of Y when X=0. Substitute 0 for x in your original equation and you'll find Y=-7.
4x - 10y = 11
-10y = 11 - 4x (subtract 4x from each side)
y = -11/10 + 4x/10 (divide each side by -10)
y = (2/5)x -11/10 (rearrange and simplify)
So the answer is C
It doesn't matter whether you solve for x or for n first. Once you find one value it's easy to solve for the other. You have the following 2 equations:
n=x-12
n+2x=39
Substitute x-12 for n in the second equation:
x-12+2x=39
3x=51
x=17
Now solve for n:
n=x-12
n=17-12
n=5
For me, that was the easiest way to do it. The approach you were taking was fine too:
n+2x=39
2x=39-n
x=(39-n)/2
Now use this value for x and substitute into your other equation: n = x-12
n=x-12
n=(39-n)/2 - 12
2n = 39-n -24 (multiply each side by 2 to get rid of the fractions)
3n = 15
n =5
Same answer, just not quite as easy.
Plugging 12 in for x:
12(12-12) = 2(12-12)
12(0) = 2(0)
0 = 0
Plugging 2 in for x:
2(2-12) = 2(2-12)
-20 = -20
The one car travel 18 miles more than the other.
The speed of one car is 12 mph faster than the other. To get 18 miles ahead of the other car, they must have been driving for 1 1/2 hours.
One car went 93 miles in an hour and a half:
93 / (3/2) = 93 * 2/3 = 62 mph
The other car went 111 miles:
111 / (3/2) = 111 * 2/3 = 74 mph.
I just realized that all my units should have been km rather than miles. Same numbers, different units.
x+y = 100 x is the $9 coffee, Y is the 12.
9x + 12y = $11.25 (100)
9x+12y = 1125
now substitue 100-y for x
9(100-y) + 12y = 1125
900 - 9y + 12y = 1125
3y = 225
y =75
And x = 100-y=25
Doublecheck: $9/pound (25 pounds) + $12/pound (75pounds)= $225 + $900 = $1125
How about $10 and $50. That would be all of the US bill denominations in ascending order.
You're doing great. X is your width so your rectangle has a width of 5 and a lenght of 2 + 2X = 12. To doublecheck: 2 times the width + 2 times the lenght = 2*5 + 2*12 = 10+24 =34.
By the way, there's one mistake in your equations. Since you came up with the right answer, it must have been a typo. 2(2+2x)+x=34 should be 2(2+2x)+2x=34
The 2 equations are essentially the same. One of them is just the other multiplied by -2. So you won't be able to solve for X and Y. For that, you would need 2 different equations. So either it's a trick question or there is a typo.
16/25, 7/100, 9/50, 11/100
I'm not sure what you're asking but I think we want to add this four numbers together.
16/25 = 64/100 (multiplied numerator and denominator by 4)
7/100 = 7/100
9/50 = 18/100 (multiplied by 2)
11/100 = 11/100
Add all those together: 64/100 + 7/100 + 18/100 + 11/100 = (64+7+18+11)/100 = 100/100 = 1
Maybe the 4 numbers represents fractions of the total mail. Each number could represent a percentage of the whole mail system. Whatever the first type of mail is, it makes up 64% of the mail. The second group is 7%, the third is 18% and the fourth is 11%. Together they add up to 100%.
I agree. Your answer differs from the text book because you're using different ranges of r. The text book uses r = 2 to 10 and you use r = 1 to 9. The following would also be a valid (and equal) answer:
Since each spin is independent of all other spins, it doesn't matter that a 0 just came up. The odds of spinning a 0 are 1 in 37 no matter whether the last 33 spins were all 0 or if a 0 hasn't come up in the last 1000 spins (I believe you when you say the wheel was unbiased!). I think the answer would be the same if a 16 just came up or if you asked how many non-32 spins come up before you get a 32.
Randomly walk up to a roulette table and count how many spins in takes to get a zero. Once it hits, walk to another table and do the same over and over again. It may hit on the first spin, it may hit on the 50th spin. I'm going to say on average that you would have to wait 18 or 19 spins (37/2).
the value of (-5) - (-48) / (-16) + (-2) x 6 is:
Do the division and multiplications first.
-48 / -16 = 3 and -2 x 6 = -12
Substitute those values back in your original equation:
-5 - (3) + (-12)
-8 - 12 = -20
I agree with you George. I was actually just trying to get Mr Brown riled up with the whole .9999... = 1 fiasco! For those who don't know what I'm talking about, see the threads dealing with .9 in the "This is Cool" forum.