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975   

Does anybody else think this new AI interface is awful?   

I think it is, too.  I have a few math-related answers that   

have been waiting moderation/publishing for many days.   

But then, sometimes my answer goes up immediately.   

I think that the A.I. is a little short on the "I" part.  

Of course, you'll never see this comment if it languishes   

awaiting A.I. moderation for endless days.  :-)  

_.

How many integers from 1 to 9 are divisors of the five-digit number 24,551?   

This again?  The literal answer is that all of them are.   

But, of course, I think you mean which are evenly divisible into 24,551. 

This is easy to find out.  You just try all the digits that might qualify.   

It ends with an odd number, so you don't have to try any of the even numbers.   

It doesn't end with a 5 or a zero, so you don't have to try 5. 

So that leaves 3, 7, and 9 to try.  That shouldn't be too hard to do.  

edited to add:  Haha, I fell for it.  24,551 is a prime number; the answer is none.    

_.

Don't worry. As AI advances, moderation will improve significantly each day. In a few years, it might even replace human moderation effectively. Time will tell

1/5 < M/56 < 5/4

1/5 < M /56                M/56  < 5/4

56/5 < M                   M < 56 * 5 / 4

11.2 <  M                  M < 70

The integer values range from  12  to  69  inclusive

The average of these =   sum of the  terms  /  number of terms

Number of terms =  69 -12 + 1 =  58

Sum =  (12 + 69) * (58/2) = 2349

Average  =   2349 / 58  =  40.5

Perimeter = 4 * 5  + pi * 5 =  5 ( 4 + pi)  ≈  35.7

Find the number of integers \(n\) that satisfy both of the inequalities:

1. \(4n + 3 < 253\)
2. \(-7n + 5 < 24\)

Solution:

1. For the first inequality, solve for \(n\):
   \[4n + 3 < 253\]

   Subtract 3 from both sides:
   \[4n < 250\]

   Divide both sides by 4:
   \[n < \frac{250}{4} = 62.5\]

2. For the second inequality, solve for \(n\):
   \[-7n + 5 < 24\]

   Subtract 5 from both sides:
   \[-7n < 19\]

   Divide both sides by -7 (and reverse the inequality sign):
   \[n > \frac{-19}{7}\]

3. To satisfy both inequalities, \(n\) must be an integer between \(-19/7\) and 62.5.

4. Count the integers within this range:
   -3, -2, -1, 0, 1, 2, 3, ..., 61, 62

5. There are 66 integers within this range that satisfy both inequalities.

So, the number of integers \(n\) that satisfy both inequalities is 66.

Simplified Expression:
((mn^3)^2) * (((2^-2)(m^-1)(n^3))^(-2))

1. Simplify each part separately:
   - (mn^3)^2 becomes m^2n^6.
   - (((2^-2)(m^-1)(n^3))^(-2)) becomes (2^2)(m^1)(n^3).

2. Apply the negative exponent rule:
   - 2^-2 becomes 2^2.
   - m^-1 becomes m^1.
   - n^3 remains n^3.

3. Simplify the remaining exponents:
   - 2^2 is 4.
   - m^1 is m.
   - n^3 remains n^3.

4. Multiply all the simplified terms together:
   4 * m * n^3 * m^2 * n^6.

5. Combine like terms:
   4m^3n^9.

So, the simplified expression is 4m^3n^9.

| Simplified Expression |105p^6 --- 21p^5 Divide the coefficients: 105 ÷ 21 = 5 Divide the variables (p^6 ÷ p^5 = p^(6-5) = p^1 = p) The simplified expression is: 5p

4 add'l pails would be 1/2 more of the 8 already in the tank then the tank will be      7/12  +  7/12 * 1/2   =  7/8  full 

   then 1/8 th more of the tank requires 3 jugs  to completely fill it.....       wellll then   8 / 8 of the tank will require   8 * 3 = 24 jugs 

Daniel filled 7/12 of a tank with 8 pails of water. Another 4 pails and 3 jugs are needed to fill the tank completely. How many jugs of water can the tank hold?  

8 pails of water brings the tank level up to 7/12.   

4 pails of water is half of 8 pails, so that raises the tank   

level up an additional half of 7/12 which is 7/24.   

Those 12 pails have taken the tank level to 7/12 + 7/24 = 21/24..   

The last 3 jugs of water adds the last 3/24 to completely fill the tank,   

therefore each jug is 1/24 of a tank.   

At 1/24 tank per jug, it would take 24 jugs to fill the tank.   

_.

Let the capacity of the tank be x jugs. 8 pails fill 7/12 of the tank, so 1 pail fills 7/12 * 1/8 = 7/96 of the tank. 4 more pails and 3 jugs fill the remaining 1/12 of the tank, so 1 jug fills 1/12 - 4/96 = 1/32 of the tank. x * 7/96 = 3 * 1/32 x = 3 * 32/7 = 12.86 jugs. Since the tank can not hold 0.86 jugs, it can hold 12 jugs of water. So the answer is 12

Since (-2, 3) is on the graph of y = f(x), we know that f(-2) = 3.

Let (a, b) be a point on the graph of y = f(2x+1)+3. This means that f(2a+1) + 3 = b.

Substituting f(-2) = 3 into the equation f(2a+1) + 3 = b, we get 3 + 3 = b .

Therefore, b = 6 and the point (a, b) = (a, 6) must be on the graph of y = f(2x+1)+3.

Note: We can also find the value of a by substituting b = 6 into the equation f(2a+1) + 3 = b. This gives us f(2a+1) = 3, so f(2a+1) = f(-2). Since (-2, 3) is on the graph of y = f(x), we know that f(-2) = 3, so f(2a+1) = 3. This means that 2a+1 = -2, so a = -3. Therefore, the point (a, b) = (-3, 6) must be on the graph of y = f(2x+1)+3.

There are two ways to solve this problem.

Method 1: Using complementary counting

We can solve this problem by finding the number of ways in which no flavor is selected by exactly three children and subtracting from the total number of ways in which the children can choose their ice cream flavors.

The total number of ways in which the children can choose their ice cream flavors is 36=729.

There are three ways to choose which flavor will be selected by exactly three children. Once we have chosen the flavor, we can assign it to any three children in (36​)=20 ways.

The remaining three children will each have 2 choices for their ice cream flavor. So the total number of ways in which no flavor is selected by exactly three children is 3×20×23=540.

Therefore, the number of ways in which some flavor of ice cream is selected by exactly three children is 729−540=189​.

Method 2: Using casework

We can also solve this problem by considering the three cases where the flavor selected by exactly three children is:

The chocolate flavor.

The vanilla flavor.

The strawberry flavor.

In each case, we can count the number of ways to choose the three children who will get that flavor and the number of ways to choose the flavors for the remaining three children.

The total number of ways is then the sum of the number of ways in each case.

The casework is as follows:

Case 1: The chocolate flavor.

There are (36​)=20 ways to choose the three children who will get chocolate. The remaining three children can each choose from the two remaining flavors, so there are 23=8 ways to choose their flavors.

Case 2: The vanilla flavor.

The number of ways is the same as in Case 1.

Case 3: The strawberry flavor.

The number of ways is the same as in Case 1.

Therefore, the total number of ways is 20×8+20×8+20×8=189.

Call the side of of one of the smaller shaded squares = 1

The  total shaded area = 8(1^2) + 4 (pi) (1^2) / 4  =  8 + pi

The diameter of the  circle =   1 + 1 + 2sqrt(2)  =  2 +2 sqrt (2)

Then the radius = 1 + sqrt (2)

Area =  pi ( 1 + sqrt (2))^2  =   pi ( 3 + 2sqrt (2))

Ratio of shaded area  / Area of circle  =  [ 8 + pi ] /  [ pi ( 3 + 2sqrt (2)) ] ≈  .61

Call the original mass in A = A

Call the original mass in B = B

When (4/5)  of the mass in A is  poured into B

A now contains (1/5)A

B contains B + (4/5)A

When (1/2)  of the mass of B is then poured into A

A  now contains (1/5)A  + [ B + (4/5)A ] / 2  =   (3/5)A + (1/2)B

B now cointains (B + 4/5)A)  / 2  =  (1/2)B + ( 2/5)A

And we have that 

[ (3/5)A + (1/2)B ]               7

______________  =        __              cross-multiply

[ (1/2)B + (2/5)A ]              5

5 [ (3/5)A + (1/2)B ] =  7 [ (1/2)B + (2/5)A ]

3A + (5/2)B =  (7/2)B + (14/5)A

3A - (14/5)A = (7/2)B - (5/2)B

(1/5)A = B

A / B  =   5 / 1    (the original ratio of their masses )

Call the midpoint of MO =  Q

Call the altitude A

NP =  (2/3)(25) =  50/3

PM =  25/3

QN^2 =  [ A^2 - (50/3)^2 ]

QN^2  = [ MQ^2  - (25/3)^2 ]

A^2  - (50/3)^2  = MQ^2 - (25/3)^2

A^2 - 2500/9  = MQ^2 - 625/9

MQ^2 = A^2 - 1875/9

MN ^2  =  MQ^2  + A^2

25^2  =   A^2 - 1875/9 + A^2

625 = 2A^2 - 1875/9

5625  = 18A^2 - 1875

18A^2 = 5625 + 1875

18A^2 = 7500

A^2  = (7500 / 18)  = ( 1250/3)

A = sqrt (1250/3)  =  sqrt ( 625 * 2 / 3)  =  25 sqrt (2/3)  =  25sqrt (6) / 3

w = 12

tan CBQ =  (1/5)

sin CBQ = sin PBQ =  1 / sqrt [ 1^2 + 5^2 ] = 1  / sqrt (26)  = sqrt (26) / 26

Let the width = W

Let X = (W, W)

Let V = ( (1/2)W , 0)

VX =   sqrt  ( (W - (1/2)W)^2  + W^2  =  sqrt [ W/4 + W^2 ]  = Wsqrt (5) / 2

XY =  W sqrt (2)

XY / XV =  W sqrt (2)  / ( (W sqrt (5) / 2)  =  2sqrt (2) / sqrt (5)  =  sqrt (8) / sqrt (5)  = sqrt (8/5) ≈ 1.3

Law of Cosines

BC^2 = BD^2  + DC^2  - 2(BD * DC)cos (60°)

BC^2  = 3^2 + 5^2  - 2 (3*5)(1/2)

BC^2  = 34 - 15

BC^2 = 19

BC = sqrt (19)

BD^2  = BC^2  + DC^2 - 2(BC * DC)cos ( BCD)

3^2  = 19 + 5^2 - 2 ( sqrt (19)*5) cos (BCD)

35 / ( 10sqrt (19)  = cos BCD

7 / (2sqrt 19)  =  3.5 / sqrt (19)  = cos BCD  = sin ACE

arcsin (3.5 / sqrt 19)  ≈ 53.41°

Angle CAE =  120 - 53.41  ≈ 66.59° 

Law of Sines

CE  /sin (CAE) = AC / sin (AEC) 

CE  /sin 66.59 = AC  /sin 60

 3  /sin 66.59 = AC/sin 60

AC = 3 sin 60 / sin 66.59   ≈ 2.83

AE / sin (ACE)  = AC / sin (AEC)

AE / sin (53.41) = 2.83 /  sin (60)

AE = 2.83 sin (53.41) / sin (60) ≈  2.62

I need help with these problems too.  Please solve these!

1. You draw one card from a standard deck of playing cards. If you pick a heart, you will win $10. If you pick a face card, which is not a heart, you win $8. If you pick any other card, you lose $6. Do you want to play? Explain.

2. The world famous gambler from Philadelphia, Señor Rick, proposes the following game of chance. You roll a fair die. If you roll a 1, then Señor Rick pays you $25. If you roll a 2, Señor Rick pays you $5. If you roll a 3, you win nothing. If you roll a 4 or a 5, you must pay Señor Rick $10, and if you roll a 6, you must pay Señor Rick $15. Is Señor Rick loco for proposing such a game? Explain.

3. You pay $10 to play the following game of chance. There is a bag containing 12 balls, five are red, three are green and the rest are yellow. You are to draw one ball from the bag. You will win $14 if you draw a red ball and you will win $12 is you draw a yellow ball. How much do you expect to win or loss if you play this game 100 times?

4. A detective figures that he has a one in nine chance of recovering stolen property. His outof-pockets expenses for the investigation are $9,000. If he is paid his fee only if he recovers the stolen property, what should he charge clients in order to breakeven?

5 At Tucson Raceway Park, your horse, Soon-to-be-Glue, has a probability of 1/20 of coming in first place, a probability of 1/10 of coming in second place, and a probability of ¼ of coming in third place. First place pays $4,500 to the winner, second place $3,500 and third place $1,500. Is it worthwhile to enter the race if it costs $1,000?

6. Your company plans to invest in a particular project. There is a 35% chance that you will lose $30,000, a 40% chance that you will break even, and a 25% chance that you will make $55,000. Based solely on this information, what should you do?

7. A manufacturer is considering the manufacture of a new and better mousetrap. She estimates the probability that the new mousetrap is successful is 3 4 . If it is successful it would generate profits of $120,000. The development costs for the mousetrap are $98,000. Should the manufacturer proceed with plans for the new mousetrap? Why or why not?

8. A grab bag contains 12 packages worth 80 cents apiece, 15 packages worth 40 cents apiece and 25 packages worth 30 cents apiece. Is it worthwhile to pay 50 cents for the privilege of picking one of the packages at random?