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The answer 8/15 is incorrect. The correct answer is \(-8/15\)

(a) Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

(b) To choose three numbers, we can choose two groups one group with two numbers and the other group has one number.  The total of n + 2 numbers can be separated into two groups.

In the first way, one group has 2 numbers, and the other group has n numbers.  They form a total of n + 2 numbers.  There are C(2,2) = 1 ways to choose two numbers from the 2 group.  There are C(n,1) = n ways to choose one number from the one group.  This gives us a first term of 1*n.

In the second way, one group has three numbers, and the other group has n - 1 numbers.  They form a total of n - 1 numbers.  There are C(3,2) = 2 ways to choose two numbers from the 2 group.  There are C(n - 1,1) = n - 1 ways to choose one number from the one group.  This gives us a second term of 2*(n - 1).  The pattern will continue until we reach n.  So the two sides are equal.

Let F represent a full tank:
1/5F + 3 =1/4F, solve for F
F = 60 Liters when it is full

Volume of a cylinder   pi r² h

1/5 full   is the   1/5 pi r^2 h

add 3 liters

1/5 pi r^2 h+ 3   =  1/4 pi r^2 h

3 = (1/4 - 1/5 ) pi r^2 h

3 = 1/20 pi r^2 h

60 liters = pi r^2 h = volume when full

a = 5/2,  b = 9/2,  c = 8,

So, c = 8

h(t)=-4.9t^2+14t-0.4       substitute  6 in for h(t) and solve for t

    you should get two VALUES for t   the ball is above 6m for the time in between

6 = -4.9t^2+14t-.4

-4.9t^2 +14t- 6.4 =0    solve for t = 0.571429   and   2.28571    now you calc the time the ball is above 6m

(1) Solving the inequality, the answer is 5 seconds.

(2) Solving the inequality, the answer is (-5,2).

(3) The answer is (0,inf).

Common ratio = 8 / 15

I was thinking there should be a search for users/members of this site, (E.g. "type name" --->User profile appears)     

The list on the right is usually truncated, so click the down arrow at the bottom to expand it, and then use your browser Find feature to search for the name. 

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So basically, simplifying what ElectricPavlov said, the average age should be 825/25 = 33.

Find the common ratio of the infinite geometric series

\(\frac{5}{6}-\frac{4}{9}+\frac{32}{135}-\dots\)

Correct!

Since the do an out and back trip....the up and down portions reverse and the flat remains the same

ANSWER 24 MPH

example:   on the way out   say    60 m up hill   = 3 hours

                                                     flat     48       = 2 hrs'

                                                     down  120    = 4 hrs

on the way back                           up 120          = 6 hrs

                                                     flat 48           = 2

                                                     down 60      = 2                          total miles 456 miles

                                                                                                        total time = 19 hrs                 avg m/h = 456/19 = 24 m/hr

[(10*30) + (15*35) ] / (10+15)= avg age of all of the members

min{(y - x)^2/((y - z) (z - x)) + (z - y)^2/((z - x) (x - y)) + (x - z)^2/((x - y) (y - z))} = 3 

at (x, y, z) = (33/10, 8/5, -23/5)

Find the smallest possible value of

\(\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}, \)

where x, y and z are distinct real numbers.

Thanks Guest,

I just drew it up all 6 possibilities. You are right, only 2 of them crossed.  

So I concur, the prob is 1/3

The probability that when labeling the points the chords intersect is not 1/2, it is 1/3

How many even perfect square factors does 2^4*7^9 have?

Take the exponent of each even prime factor: 2^4 =[4 / 2 + 1] = 3 - it has only 3 EVEN perfect squares as factors.

If you put any 4 distinct points on a circle then label them A1, A2, B1, B2

Then join the 2 A points together and also join the 2 B points together.

Then exactly half the time they weill cross over each other.

Using this logic I think the probability that they wil cross is 50%.

I have not seen this question before. Maybe I am over simplifying it but it seems logical to me.

ok thanks, I shall try and remember.  

since 2*7  and 1*14 are the only integer factors of 14 .

It does follow that 192 must be the smallest one.

Here is how to do it.

Add it all up, put equal to 43.  then solve.

help

\(\begin{array}{lccccccccccc} \text{row }1:& & & & & & 1 \\ \text{row }2:& & & & & 2 & & 2 \\ \text{row }3:& & & & 3 & & 4 & & 3 \\ \text{row }4:& & & 4 & & 7 & & 7 & & 4 \\ \text{row }5:& & 5 & & 11 & & 14 & & 11 & & 5 \\ \text{row }6:& 6 & & 16 & & 25 & & 25 & & 16 & & 6 \end{array} \)

If we form a number triangle as above,

what is the sum of all the numbers in the triangle with 20 rows?

\(\text{Let row $=r$} \\ \text{Let the sum of all the numbers in the rth row $=s_r$ } \\ \text{Let the sum of the numbers in the triangle from 1 to rth row $=s_{1\ldots r}$ }\)

\(\begin{array}{|rcll|} \hline s_1 &=& 1 \\ s_2 &=& 1 +3\cdot( 1 ) \\ s_3 &=& 1 +3\cdot( 1+2 ) \\ s_4 &=& 1 +3\cdot( 1+2+4 ) \\ s_5 &=& 1 +3\cdot( 1+2+4+8 ) \\ s_6 &=& 1 +3\cdot( 1+2+4+8+16 ) \\ \ldots \\ s_r &=& 1 +3\cdot( 1+2^1+2^2+2^3+2^4+\ldots + 2^{r-2} ) \\ &=& 1+ 3\cdot(2^{r-1}-1) \\ \mathbf{s_r} &=& \mathbf{ 3\cdot 2^{r-1} -2 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_{1\ldots r} &=& \sum \limits_{i=1}^{r} \left(3\cdot 2^{i-1} -2 \right) \\ &=& 3\sum \limits_{i=1}^{r} \left(2^{i-1}\right) -\sum \limits_{i=1}^{r} 2 \\ &=& 3\sum \limits_{i=0}^{r-1} \left(2^i\right) -2\sum \limits_{i=1}^{r} 1 \\ &=& 3\sum \limits_{i=0}^{r-1} \left(2^i\right) -2r \\ &=& 3\sum \limits_{i=0}^{r} \left(2^i\right) -3\cdot2^r -2r \quad | \quad \sum \limits_{i=0}^{r} \left(2^i\right) = \left( 2^{r+1}-1 \right) \\ &=& 3\cdot \left( 2^{r+1}-1 \right) -3\cdot2^r -2r \\ &=& 3\cdot 2^{r+1}-1 -3 -3\cdot2^r -2r \\ &=& 3\cdot \left(2^{r}\cdot 2 \right) -3\cdot2^r -2r -3 \\ &=& 2^{r}\left( 6-3\right) -2r -3 \\ \mathbf{ s_{1\ldots r} } &=& \mathbf{ 3\cdot 2^{r} -2r -3 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ s_{1\ldots r} } &=& \mathbf{ 3\cdot 2^{r} -2r -3 } \quad | \quad r = 20 \\\\ s_{1\ldots 20} &=& 3\cdot 2^{20} -2\cdot(20) -3 \\ &=& 3\cdot 2^{20} -2\cdot(20) -3 \\ &=& 3\cdot 2^{20} -43 \\ &=& 3\cdot 1048576 -43 \\ &=& 3\cdot 1048576 -43 \\ &=& 3145728 -43 \\ \mathbf{ s_{1\ldots 20} } &=& \mathbf{ 3145685} \\ \hline \end{array}\)

CU: 1 hour =60 minutes x 60 seconds =3,600 seconds.

15 x 3,600 seconds =54,000 seconds

15 / 54,000 =1 / 3,600

Im sorry calculstoeuser this is incorrect