All Answers 355923 Answers

We have the following  exponential function

N = 285 (1.75)^x    where N is the number of subscribers after x years

Since 1994 is 9 years after 1985, we have

N  = 285 (1.75)⁹    ≈  43,872  subscribers

 ( That's a pretty big increase  !!!  }

A pic for #2  might be helpful......

Here's the first one

Call the height of the water in the tank, H

And ......using similar triangles, we can express the radius, R, at this point thusly :

16 / 96  = R / H

1/6 = R/H

H/6  = R

Now  when the tank is 25% full, it will hold

(1/4) (1/3) pi 16^2 * 96   =   2048 pi  ft^3

So....to find the height, we have

2048 pi   =  (1/3) pi  (H/6)^2 * H     divide out pi, multiply both sides by 3

6144  =  H^3 / 36       multiply both sides by 36

221184  =  H^3

Take the cube root of both sides

48³√2     =   H

So

a +  b   =   48 +  2   = 50

There are several numbers missing from the official solution you pasted.  When you paste data, you should check to see that it is all there.

You submitted your solution to AoPS and the reviewer said your solution (1/4) is correct.  Well, assuming you are not telling us a big fib (that means lying), the reviewer is wrong.  I’m sure the official answer/solution goes well into the hierarchy, where a Ph.D. approved it, without a proper review. The wrong answers that often appear in the back of math books also receive approval. Once a solution is believed correct, the analysis ends, and that’s the end of it.  

This question has some powerful juju, Jojo. (Juju means magical karma.)  Despite the impeccable credentials of AoPS Professors, teachers, and reviewers, (1/4) is the WRONG answer.  The answer is wrong is because the solution method is wrong for this question. It should be obvious that Grogg does not arbitrarily choose whether to color a number blue; if he did then that would be the correct solution.

Grogg randomly chooses a set of numbers from zero to six, then he randomly colors numbers blue to match the count of the set. If Grogg chooses three numbers then he randomly colors three of the six numbers blue.  If he chooses zero numbers the he colors none, if he chooses 6 then he colors all six.  Sets of zero, one, two, three, etc, all have an equal (1/7) probability of selection. Any of the six numbers also have an equal probability of selection in each set. 

However, if Grogg considers each of the six numbers individually, and randomly chooses to color it or not, then the “set counts” become a binomial distribution: 1,6,15,20,15,6,1.

The sum of this binomial distribution is 64 and the probabilities of Grogg selecting each set are 1.5625%, 9.375%, 23.4375%, 31.25%, 23.4375%, 9.375%, and 1.5625%.

As you can see this distribution is very different from the uniform probability of 14.285% for each set, where Grogg chooses each set independently. 

In any case, good for you, for getting the correct answer. One lesson I never learned very well in the lower grades is, “the ‘correct’ answer is the one the teacher says it is.” This is good advice because the teacher usually is correct. Personally, though, if there is a conflict, I’d rather have a lower grade and the correct solution. I definitely got the lower grade, but I was less sure about the correct solution. 

In this case, I’m very sure of the correct solution. 

^{And it is not nice to go around media like this with an attitude like that.}

Really? What kind of media would make it nice?

GA

Nice, Melody   !!!!

Look at the graph here :

https://www.desmos.com/calculator/jj4kxyynwi

The first statement is true

The second is true, as well  ⇒ year 5   = 8.8 ft   ,  year 7 = 15.9 ft

The third statement is false.....the initial  height is 1 ft

The fourth statement is also true.....when x  = 15, the tree has almost reached its max height  of about 30 ft

Three consecutive positive odd integers $a$, $b$ and $c$ satisfy $b^2 - a^2 = 344$ and $c^2 - b^2 > 0$. What is the value of $c^2 - b^2$?

b-2, b,  b+2        where a is odd

\(b^2-(b-2)^2=344\\ b^2-b^2+4b-4=344\\ 4b-4=344\\ b-1=86\\ b=87 \)

a=85

b=87

c=89

89^2-87^2   = 352

A circle has a sector with an area of 33\pi  and central angle of (11)/(6)\pi  radians. what is the area of the circle?

Note that (11/6) pi  rads  = 330°  =  330/360 =  11/12  of the whole area of the circle

So....to find the whole area we have

33pi  (12/11)  =

(33/11) * 12 * pi  =

 3 * 12  pi  =

36 pi

f(x)  = cx^2 + 2x    x < 2

f(x) = x^3 - cx        x ≥ 2

Let's just substitute 2  into each function for x

c(2)^2 + 2(2)  =   4c + 4

(2)^3 - c(2)  =  8 - 2c

Set these results equal  and solve for c

4c + 4  = 8 - 2c      add 2c to both sides, subtract 4 from both sides

6c  = 4      divide both sides by  6

c  =2/3

See the graph here that confirms the result :

https://www.desmos.com/calculator/ahns4csjpv

These  numbers  are very familiar

The  three known text sizes  [ from least to greatest ]  are

Phi^-1

Phi⁰

Phi²

The unknown text size  is Phi¹  = Phi  =  1.618

Note that the point  (1,0)  is the first point on the graph

And the point  (5,2)  is also on the graph

So.....the top  right graph is correct

Note that the bottom right graph couldn't be correct becuse no negative values for x  are allowed......these would make the quantity under the radical negative  which would not be a real number

Domain

x  cannot be any value < -5   because we would be taking the root of a negative number which isn't real

So....the domain  is   [-5, infinity )

The range is trickier

As small as the quantity under the radical can be is 0  [  when x  = -5 ]

So  1/2  * 0   =   0

And as x grows larger than -5, the function keeps increasing to infinity

So....the range is  [ 0, infinity )

The first answer is correct

The answer is 4.

f(x)  =  l x - 3 l  + 2

f(x + 2)  =   l (x + 2) -3 l + 2  =  l x - 1 l + 2

This shifts the graph 2 units to the left

Look at both graphs here :

https://www.desmos.com/calculator/8m6gjryhfu

                      O     A      B    AB    Total

Male (M)      210  174    74    42     500

Female (F)   315  261  111    63     750

Total             525  435  185   105  1250

P (F l A )   =  261/435  =   3/5

P (F ∩ A ) =  261 / 1250

P(F) * P(A)  =  750/1250 * (435/1250)  = 261 / 1250

So....the events are independent

Answer   "C"

Your question doesn't make sense!

If you were thinking of the LCM[10,700], it doesn't equal 7,000, but 700!.

Well, the next smaller scale size to 2.863 will be:

2.863 / 1.42 =2.016

There is a very simple formula for this kind of problem:

D = ~Sqrt[1.5 x H], where D =distance in miles, H=Height in feet.

D =~sqrt[1.5 x 2.7 miles x 5,280 feet]

D =~Sqrt[ 21,384]

D=~ 146.23 miles - distance to the horizon.

First one

The complement of set P  are all the elements is the universal set U which are not in set P

So we have  {2, 4, 6 }  =  "C"

volume in question   =   volume of cylinder - volume of sphere

volume of cylinder   =   ( area of base )( height )

volume of cylinder   =   ( π r² )( height )

volume of cylinder   =   ( π(3/2)² )( 8 )     cubic inches

volume of cylinder   =   ( π(9/4) )( 8 )   cubic inches

volume of cylinder   =   18π                 cubic inches

volume of sphere   =   (4/3)π r³      

volume of sphere   =   (4/3)π(2/2)²  cubic inches

volume of sphere   =   (4/3)π           cubic inches

volume in question   =   18π - (4/3)π   cubic inches

volume in question   =   (50/3)π         cubic inches

volume in question   ≈   52             cubic inches

v=4/3*π*5^3

v=4/3*π*125

v=166.7πcm^3

v=4/3*π*7^3

v=4/3*π*343

v=457.3πcm^3

457.3πcm^3-166.7πcm^3=290 2/3π

Ans=D. 290 2/3π   

omg yes thankyou so much!

If you have a bag of 10 marbles, where 3 of them are blue and 7 of them are green...

probability that a randomly selected marble is blue  =  \(\frac{\text{num of } \underline{\text{marbles}} \text{ that are blue}}{\text{total num of }\underline{\text{marbles}}}\)

Likewise....

probability that a randomly selected student with a part time job is on the honor roll  =

\(\frac{\text{num of } \underline{\text{students with a part time job}} \text{ that are on the honor roll}}{\text{total num of }\underline{\text{students with a part time job}}}\)

num of students with a part time job that are on the honor roll  =  22% of 1200

num of students with a part time job that are on the honor roll  =  0.22(1200)

num of students with a part time job  =  60% of 1200

num of students with a part time job  =  0.6(1200)

probability that a randomly selected student with a part time job is on the honor roll  =

\(\frac{0.22(1200)}{0.6(1200)}\,=\\~\\ \frac{0.22}{0.6}\,=\\~\\ \frac{22}{60}\,\approx\\~\\ 37\%\)

I hope that made some sort of sense....

Awesome Cphill!