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there are 9 balls total in the drawer

the probability that the first one is black is 6/9=2/3

the probability that the second is black (with the first one drawn being black also) is (6-1)/(9-1)=5/8

multiply the probabilities:

(2/3)*(5/8)=10/24=5/12

Test post

Still not displaying properly   

move around terms first

\((1.8 \cdot 6.5)(10^3 \cdot 10^{14})\)

\((1.8 \cdot 6.5)(10^{17})\) multiply the powers of ten, adding the exponents by the exponent rules

multiply the numbers in front

\((11.7)(10^{17})\)

almost done, but this is not in correct scientific notation

add one to the exponent so the final answer will have the correct place value

\(1.17 \cdot 10^{18}\)

and that is the final answer

thank you 

Uh I have no idea 

It's let me see :D

Use the calculator on this website

complicated but he right if you want it simple just put it as

25/2113

25/2113 = 0.0118315191670611

Work = change in KE

Initial KE     1/2 m ( 6^2 + (-1)^2)

final KE        1/2 m ( 8^2 + 4^2)

You can choose 2 students out of 25 in: 25C2 = 300 ways - or 300 pairs.

Since each student can be paired with 24 other students and the 4 Smith brothers can form a total of: 24 + 23 + 22 + 21=90 pairs, therefore the probability that one of the pairs will be a Smith is: 90 / 300 =3 / 10

KE = 1/2 mv^2          v = sqrt ( 6^2 + (-1)^2) ) = sqrt37    m= 3.00

6 x    + 10 (3-x)   = 20      solve for 'x' the rain running time in hours

a)   17:25 + 2:00 = 19:25

d) 07:21 - 12:00 = 19:21    (the previous day)

f) 23:00 + 00:45 = 23:45   

   You should try the rest !

maximum value of f(x)=-3(x-4)+5

                                    = -3x+17       maximum value of this line is + inf

How do you type 5% "out" of 40 on the calculator keyboard?

'5'

'2^nd'

'%'

'2^nd'

'x'

'40'

'='

105  112  119  126  133  140  147  154  161  168  175  182  189  196  203  210  217  224  231  238  245  252  259  266  273  280  287  294  301  308  315  322  329  336  343  350  357  364  371  378  385  392  399  406  413  420  427  434  441  448  455  462  469  476  483  490  497  504  511  518  525  532  539  546  553  560  567  574  581  588  595  602  609  616  623  630  637  644  651  658  665  672  679  686  693  700  707  714  721  728  735  742  749  756  763  770  777  784  791  798  805  812  819  826  833  840  847  854  861  868  875  882  889  896  903  910  917  924  931  938  945  952  959  966  973  980  987  994  Total =  128 - all 3-digit multiples of 7.

161  231  301  371  441   511   581  651  721  791  861  931    Total =12 multiples of 7 ending in 1

Probability = 12 / 128 = 3 / 32 

If I understand your question:

1 - One die has 19 numbers on it and the number "20" is missing:

2 - The 2nd die also has 19 numbers on it with number "9" missing:

3 - If you rolled the 2 dice, these are the totals you should get:

(2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 31, 31, 31, 32, 32, 32, 32, 32, 32, 32, 32, 33, 33, 33, 33, 33, 33, 33, 34, 34, 34, 34, 34, 34, 35, 35, 35, 35, 35, 36, 36, 36, 36, 37, 37, 37, 38, 38, 39) Total =>> 361

Therefore, the probability is: 11 / 361

R, R, R, R, R, B, B, B, B. Out of nine dishes, you can pick 3 dishes as follows:

{B, B, B} | {B, B, R} | {B, R, B} | {B, R, R} | {R, B, B} | {R, B, R} | {R, R, B} | {R, R, R} (total: 8)

Probability =2 / 8 = 1 / 4

Suppose 2 distinct integers are chosen from 1 through 9, inclusive. What is the probability that they are both multiples of 3?    

The multiples of 3 are 3, 6, and 9. 

The probability of choosing one of the three of them is 3/9. 

Then, the probability of choosing one of the other two is 2/8.  

                                                                        3           2            6           1

The probability of both things happening is    –—   x   –—   =   –—   =   –—   

                                                                        9           8           72          12  

g(x) = (f(x))^2    is defined over the same interval as f(x)   [-1,9]      interval width is 10

This should help

Be careful at x =1, as it is a strict inequality!

Mean of the set of four numbers:

  (x1 + x2 + x3 + x4)/4 = 18   

Multiply both sides by 4, so

x1 + x2 + x3 + x4 = 4*18   ...(1)

Mean of the five numbers

(x1 + x2 + x3 + x4 + x5)/5 = 16

Multiply both sides by 5, so

x1 + x2 + x3 + x4 + x5 = 5*16   ...(2)

Replace x1 + x2 + x3 + x4 in (2) by using (1)

4*18 + x5 = 5*16

Subtract 4*18 from both sides

x5 = 5*16 - 4*18 

x5 = 80 - 72

x5 = 8