geno3141

Solve by elimination:

x + y + z  =  10

-x + y - z  =  0

-x + z  =  1

Combine the first two equations by addition (adding down the columns):

     x + y + z  =  10

    -x + y - z  =    0

    --------------------

         2y       =  10     --->     y  =  5

Replace y with 5 in the first equation:    

     x + y + z  =  10     --->      x + 5 + z  =  10     --->     x + z  =  5

Combine this equation with the third of the original equations by addition:

     x + z  =  5

   -x + z  =   1

   ---------------

         2z  =   6

           z  =  3

Since  y = 5  and  z = 5.5,  substitute these values into the first equation:

     x + y + z  =  10     --->      x + 5 + 3 =  10     --->     x + 8  =  10     --->      x  =  2

These are combinations problems.

Assuming independence, a person who makes 70% of his shots:

Probability of making exactly 3 of 5 shots:  ₅C₃·(0.70)³·(0.30)²  =  10·(0.70)³·(0.30)²  =  0.3087

Probability of making exactly 4 of 5 shots:  ₅C₄·(0.70)⁴·(0.30)¹  =  5·(0.70)⁴·(0.30)¹  =  0.36015

Probability of making all 5 shots:  (0.70)⁵  =  0.16807

Probability of making at least 3 shots:  0.3087 + 0.36015 + 0.16807  =  0.83692

-4 + 3(h + 1)  =  3(h + 1) + h

Notice that you have the "3(h + 1)" term on both sides. Subtract it from both sides:

        -4 + 3(h + 1)  =  3(h + 1) + h

             - 3(h + 1)  =  -3(h + 1)

--->   -4                 =                   h

So your answer is:  h = -4

A postulate is a statement that is assumed to be true.

A theorem is a statement that is proven to be true.

A property is a statement that is true; it can be either a postulate or a theorem.

A conjecture is a statement that is believed (by someone) to be true; it isn't a postulate and it isn't a theorem until it's proven true. It might never become a theorem because it may be false.

The null hypothesis (meaning that there is no difference between the two items being tested) is rejected when the p-value is small because the p-value represents the chance that the probability that you are rejecting the hypothesis because of accidental circumstances is small,

As an example:  you are testing a medicine versus a placebo.

The null hypothesis is that there is no difference between the medicine and the placebo. (Obviously, if you developed the medicine, you want the medicine to work better than the placebo:  you want a difference between the medicine and the placebo.)

The test indicates that there is a difference between the medicine and the placebo. Sometimes the difference is due to the effectiveness of the medicine but sometimes it's just due to pure luck.

The p-value indicates the probability that the difference between the medicine and the placebo was due to pure luck. If the p-value is high, this means that the difference could easily be due to pure luck; if the p-value is small, this (very likely) means that the difference between the medicine and the placebo was due to the medicine.

Why does  a³ - b³  factor into  (a - b)(a² + ab + b²) ?

[You orginally wrote  b³  instead of  b²  --  I'm going to assume that's a typo, because  b²  is correct,  b³  isn't.]

You can show that factoring is correct by multiplying the factors to get the original problem.

 (a - b)(a² + ab + b²)  =  a(a² + ab + b²) - b(a² + ab + b²)

                                  =  a³ + a²b + ab² - a²b - ab² - b³

                                  =  a³ - b³     (after simplifying the like terms)

Since it multiplies correctly, a³ - b³  factors into  (a - b)(a² + ab + b²)

a2 = 1     a3 = 1     a4 = 0     a5 = 0     a6 = 34     a7 = 2

5/7  =  a2 / 2!  +  a3 / 3!  +  a4 / 4!  +  a5 / 5!  +  a6 / 6!  +  a7 / 7!

Start by writing each term with the common denominator 7!;

Left-hand side:  (5/7)·[ (6·5·4·3·2·1) / (7·6·5·4·3·2·1) ]  =  3600 / 7!

Right-hand side:  [ 7·6·5·4·3·a2  +   7·6·5·4·a3  +  7·6·5·a4  +  7·6·a5  +  7·a6  +  a7 ] / 7!

Cancel the denominators of 7!:    3600  =  7·6·5·4·3·a2  +   7·6·5·4·a3  +  7·6·5·a4  +  7·6·a5  +  7·a6  +  a7 

Now, start factoring the right-hand side:

     3600  =  7( 6·5·4·3·a2  +   6·5·4·a3  +  6·5·a4  +  6·a5  +  a6 )  +  a7 

     3600  =  7( 6(  5·4·3·a2  +   5·4·a3  +  5·a4  +  a5  ) +  a6 )  +  a7  

     3600  =  7( 6( 5( 4·3·a2  +  4·a3  +  a4 )  +  a5 )  +  a6 )  +  a7

     3600  =  7( 6( 5( 4( 3a2  +  a3 )  +  a4 )  +  a5 )  +  a6 )  +  a7

{More later, I have to go to supper ...}

Continuing:

3600 / 7  =  514 , Remainder = 2

So, 3600  =  514(7) + 2  =  7( 6( 5( 4( 3a2  +  a3 )  +  a4 )  +  a5 )  +  a6 )  +  a7 

Let  2 = a7  and subtract from both sides  --->     514(7)  =  7( 6( 5( 4( 3a2  +  a3 )  +  a4 )  +  a5 )  +  a6 ) 

Divide both sides by 7     --->     514  =  6( 5( 4( 3a2  +  a3 )  +  a4 )  +  a5 )  +  a6

514/6  =  85, Remainder = 4

85(6) + 4  =   6( 5( 4( 3a2  +  a3 )  +  a4 )  +  a5 )  +  a6

Let 4 = a6  and subtract from both sides  --->   85(6)  =   6( 5( 4( 3a2  +  a3 )  +  a4 )  +  a5 ) 

Divide both sides by 6     --->     85  =  5( 4( 3a2  +  a3 )  +  a4 )  +  a5

85/5  =  17, Remainder = 0

17(5) + 0  =     5( 4( 3a2  +  a3 )  +  a4 )  +  a5

Let 0 = a5 and subtract from both sides  --->   17(5)  =   5( 4( 3a2  +  a3 )  +  a4 )

Divide both sides by 5     --->     17  =   4( 3a2  +  a3 )  +  a4

17/4  =  4, Remainder = 1

4(4) + 1  =  4( 3a2  +  a3 )  +  a4

Let  1 = a4  and subtract from both sides  --->   4(4)  =  4( 3a2  +  a3 )

Divide both sides by 4     --->     4  =  3a2 + a3

This process can be continued one more step, but since a2 and a3 must be whole numbers

   --->   a3 = 1    

   --->   a2  =  1

In summary:  a7 = 2     a6 = 4     a5 = 0     4 = 1     a3 = 1     a2 = 1

Which is a wholy different answer than I had before!  But, they both work!

(I have no idea how I got a6 = 34....)

a₂ = 1     a₃ = 1     a₄ = 0     a₅ = 0     a₆ = 34     a₇ = 2

5/7  =  a₂ / 2!  +  a₃ / 3!  +  a₄ / 4!  +  a₅ / 5!  +  a₆ / 6!  +  a₇ / 7!

Start by writing each term with the common denominator 7!;

Left-hand side:  (5/7)·[ (6·5·4·3·2·1) / (7·6·5·4·3·2·1) ]  =  3600 / 7!

Right-hand side:  [ 7·6·5·4·3·a₂  +   7·6·5·4·a₃  +  7·6·5·a₄  +  7·6·a₅  +  7·a₆  +  a₇ ] / 7!

Cancel the denominators of 7!:    3600  =  7·6·5·4·3·a₂  +   7·6·5·4·a₃  +  7·6·5·a₄  +  7·6·a₅  +  7·a₆  +  a₇ 

Now, start factoring the right-hand side:

     3600  =  7( 6·5·4·3·a₂  +   6·5·4·a₃  +  6·5·a₄  +  6·a₅  +  a₆ )  +  a₇ 

     3600  =  7( 6(  5·4·3·a₂  +   5·4·a₃  +  5·a₄  +  a₅  ) +  a₆ )  +  a₇  

{More later, I have to go to supper ...}

I agree with guest's answer of:  x = 5, y = 49, and z = 97; this answer checks!

My analysis is this (but, I have no idea whether or not it is a correct procedure or only a lucky procedure):

1/x - 1/(xy) - 1/(xyz)  =  19/97     with x < y < z

First find the sum of the left-side using the common denominator of xyz:

1(yz) / (xyz) - 1(z) / (xyz) - 1 / (xyz)  =  19 / 97   --->   (yz - z - 1) / (xyz)  =  19/97

The denominator of the left side has three factors; the denominator of the right side has only one factor, the prime number 97.  It needs two more factors. Since z is the largest of the three factors, claim that  = 97 and multiply the right hand side by xy/xy     --->     (yz - z - 1) / (xyz)  =  19xy / (97xy)

Claim that the numerators are equal:     --->     yz - z - 1  =  19xy

and claim that the denominators are equal:     --->     xyz  =  97xy     with z = 97.

Since z = 97     --->     yz - z - 1  =  19xy     --->     97y - 97 - 1  =  19xy

                                                                   --->     97y - 98  =  19xy

                                                                   --->     97y - 19xy  =  98

                                                                   --->     y(97 - 19x)  =  98

Since the factors of 98 are 2 and 49:         --->     y(97 - 19x)  =  2·49

So, if we are dealing with whole numbers,

either y = 2 or y = 49 (2 won't work           --->     49(97 - 19x)  =  2·49

because y > x  and x = 1 doesn't               --->          97 - 19x  =  2

work).                                                         --->                -19x  =  -95 

                                                                  --->                     x  =  5

If        an = 179   and   an + 4m  =  663

then:                          179 + 4m  =  663

and:                                     4m  =  484

and:                                       m  =  121

Thus:  an + 3m  =  179 + 3·121  =  179 + 363  =  542