All Answers 355970 Answers

They're good

The probability of selecting each club is 1/3. Let M stand for normal member and let C stand for co-president. (these are distinguishable)

Case 1: first club of 5 people is selected (MMMCC)

There are 5 choose 4 = 5 ways to select the four members, and we must get a group of MMCC in our choice. There are thus 3 choose 2 = 3 ways to choose MMCC out of MMMCC. The probability here is 3/5, and the total probability of this case happening and the co-president restriction is 1/3 * 3/5 = 1/5.

Case 2: second club of 7 people is selected (MMMMMCC)

There are 7 choose 4 = 35 ways to select the four members, and we must get a group of MMCC in our choice. There are thus 5 choose 2 = 10 ways to choose MMCC out of MMMMMCC. The probability here is 10/35 = 2/7, and the total probability of this case happening and the co-president restriction is 1/3 * 2/7 = 2/21.

Case 3: third club of 8 people is selected (MMMMMMCC)

There are 8 choose 4 = 70 ways to select the four members, and we must get a group of MMCC in our choice. There are thus 6 choose 2 = 15 ways to choose MMCC out of MMMMMMCC. The probability here is 15/70 = 3/14, and the total probability of this case happening and the co-president restriction is 1/3 * 3/14 = 1/14. 

1/5 + 2/21 + 1/14 = 11/30

See https://web2.0calc.com/questions/combination-probability

< This led me to think that if he had started to save these coins he probably had at least 1 of each. > 

That's a good point, and a decent inference, and its opposite is equally valid.  It becomes a math problem with a nested language problem, namely, what does "started" imply.  Does it incorporate intent?  Let's think about an analog (pun intended).  How about:  I've started dieting but I haven't lost any weight yet.  I've started collecting nickels but I don't have any yet.  

If you want to require at least one nickel, then you can't have 2 dimes.  

That would leave you with 1 dime, and then you could have 2 nickels.  

13 quarters     3.25  

  1 dime           0.10  

  2 nickels        0.10  

  4 pennies      0.04  

total                 3.49  

_.

Joe, Alex and Sean shared some stamps in the ratio 3:8:5. Sean kept 

1/5 of his stamps for himself and gave the rest of his stamps to Joe and

Alex in the ratio 3:5. As a result, Alex had 96 stamps more than Joe.

How many stamps did Joe have in the end?

Let's say they all have x stamps.

Joe has x * 3/16 stamps, Alex with x * 1/2, and Sean with x * 5/16.

After giving away his stamps, Sean has 1/16 * x stamps. 

Joe got (1/4 * x) * 3/8 = 3/32 * x stamps. Joe now has (3/32 + 3/16)x = 9/32 x stamps.

Alex got (1/4 * x) * 5/8 = 5/32 * x stamps. Alex now has (5/32 + 1/2)x = 21/32 x stamps

21/32 x = 9/32 x + 96

12/32 x = 3/8 x = 96

x = 256.

256 * 9/32 = 72.

The guests apparently gave you correct answers. Is there a mistake/misinterpretation anywhere?

(in case of a mistake, let's check)

256 stamps means Joe has 48 stamps, Alex has 128 stamps, and Sean has 80 stamps. 

Sean now has 16 stamps, and 64 stamps were given away in the ratio 3/5.

24 stamps given to Joe, 40 stamps given to Alex.

Joe has 72 stamps, Alex has 168.

72+168+16 = 256

168-72 = 96

I don't see a problem...? 

Let B be the fully charged battery capacity.

Talking continuously for 1 hour drains B/3 capacity.

Standby for 1 hour drains B/72 capacity.

The standby has drained 27 * B/72 = 3B/8 capacity. 

The conversation drained B/3 * 45/60 = B/4 capacity.

In total, we've drained (3/8 + 1/4) = 5/8 B capacity.

We can talk for 3/8 B more capacity.

How many hours worth is 3/8 B on conversation mode?

B/hr = 3

3B/8hr = 3*3/8 = 9/8 hr.

9/8 hr = 67.5 min

Amount of money Jonathan receive in dontain

from the community = 

Total worth of 7 Easter baskets + Money left overs

at the end 1 ten dollars and twenty-five cents

= 7 * worth of each basket + ten dollars and twenty-five cents

= 7 * Twelve dollars and fifty-five cents + ten dollars and twenty-five cent

= 84 dollars 385 cents + 10 dollars 85 cents 

= 94 dollars 410 cents

As we know that 1 dollars = 100 cents

410 cents = 4 dollars and 10 cnets

Now,

Total amount of money Jonathan receive in donation from community

= Ninety eight dollars and 10 cents

= $98.10

Let sum of prize money be $p.

--> 1/8P is divided equally for 4 individual player

So, 7P/8 is divided to group player

7P/8 * 1/4 = 20

P = 20 * 2 * 8 

P = $320

1/8P = 320/8 = $40

Now, we have to divide $40 equally among 4

individual player 

* 40/4 = $10 each

Individual player receive $10 each.

Without loss of generality, assume they played 250 games.

Then, they've already played 200 games, and won 48/100 * 200 = 96 games.

They need to win 125 games to have 50% wins, which means they need to win 125 - 96 = 29 more games.

They have 50 more games to play, so they must win 29/50 of their games, or 58%. 

To check, take the weighted average of the two values. 48 * 80% + 58 * 20% = 50. 

This works, as we needed 50% wins/losses.

\(\frac{r-3}{\sqrt{r}} = -2 \sqrt{r}\)

Since the denominator of the left side cannot be 0, √r ≠ 0, and r≠0. 

Multiplying both sides by √r, we have r-3 = -2r.

Thus, 3r = 3 and r=1.

$12.55 * 7 = $87.85

That's how much community money he spent on the baskets.

$87.85 + $10.25 = $98.10

Money spent + leftover money

Perhaps Joseph had the intention of saving nickels, but he didn't find any nickels to save.

It's safe to assume that Joseph can have 0 of one of the coins, since it isn't specified in the problem.

The remainder of the money is 14*20 = $280.

This is also equal to 7/8 of the prize money. If 280 is 7/8, and we must find 1/8, we divide 280/7 = 40. 

Each individual player received $10.

1-9             1 digit each      9

10 - 99       2 digits each    180

100 - 488   3 digits each    1167  digits        total  1356

Write equations!

Let S be Aaron's regular salary.

1/4(2/3S - 350)-60 = 240.

1/4(2/3S - 350) = 300

2/3S-350 = 1200

2/3S = 1550

S = 2325

$36 000 / 24 000 eggs  * (36000 * 1.05) * 1.20 = 68040 dollars  

     for 36000 * 1.05 = 37 800 dozen eggs

Hello and thank you very much, I appreciate your help

The only thing that confused me was the fact that it said Josef has started saving quarters nickels, dimes, and pennies. 

This led me to think that if he had started to save these coins he probably had at least 1 of each. 

Looking at it from your perspective it totally makes sense. 

If you can answer it, can you tell me why, even though it says he had already started saving these coins it meant he could also have 0?

Can you explain a bit?  

No, the first guest can not because the answer is incorrect. 

But I'll try to explain.  

If you have 14 quarters, you have 3.50 in quarters.  This would pay for the 

hamburger, so take away one quarter, leaving 13 quarters.  13 quarters totals 3.25.  

If you have 5 dimes, you could use them along with 3.00 in quarters to pay for 

the hamburger, so take away one dime, leaving 4 dimes.  4 dimes totals 0.40.  

If you have 1 nickel, you could use it along with 2 dimes and 13 quarters to pay    

for the hamburger, so you cannot have a nickel at all.   

You can have 4 pennies, because if you have 5 you could use them in place of the

nickel you can't have.  4 pennies totals 0.04.   

Adding up all the coins, the total amount you can have is $3.69.   

_.

بصثب

The remainder is -17.

The remainder is 7x + 3.

Hello and thanks for answering. Can you explain a bit?

The most amount of money is $3.67.

oops...here's the pic of the prob:

it was wrong

There are 40 terms that are divisible by 5.