All Answers 352732 Answers

\(P[\text{5 games}] = P[\text{5 games | A wins}]+P[\text{5 games | B wins}]\)

\(P[\text{5 games | A wins}]=p^4 (1-p)\\ P[\text{5 games | B wins}]=p (1-p)^4\\ P\text{5 games}=p^4(1-p)+p(1-p)^4 = \\ p(1-p)[p^3+(1-p)^3]\)

I don't think the last answer in #1 should be checked.  We're given no information about a probability of vaccinating a cat.

One would think the vet would know how many cats they vaccinated exactly.  After all someone has to pay the bill!

The rest is correct.

In #2 the first two are indeed likely to be normally distributed, if for no other reason than that there are so many people entering the store the law of large numbers kicks in.

The rest of #2 you have correct.

\(\text{The math books can be either to the left or right of the english books. That's 2 choices}\\ \text{Within the math books there are $3!$ arrangements}\\ \text{Within the english books there are $5!$ arrangements}\\ \text{The total number of arrangements is thus $2 \cdot 3! \cdot 5! = 1440$}\)

is 0 greater than or equal to negative 12   

This is a semantics issue.  

If "greater than or equal to" means mathematically > then yes. 

If "greater than or equal to" means which one, then it's greater than. 

0 is greater than –12 but 0 is not equal to –12

_.

The intersections are (12/13*(3*sqrt(3) - 1),6/13*(3 + 4*sqrt(3)) and (-12/13*(1 + 3sqrt(3)), -6/13*(4*sqrt(3) - 3), so the length of the chord is sqrt((12/13*(3*sqrt(3) - 1 + 12/13*(1 + 3*sqrt(3))^2 + (6/13*(3 + 4*sqrt(3)) + 6/13*(4*sqrt(3) - 3))^2) = 24*sqrt(39)/13.

12 =2 x 2 x 4, or:

12 =3 x 4, or:

12 =2 x 6

Note: I don't think that you can have 4 DISTINCT digits whose product = 12. There must be "repeats" !!

1126  1134  1143  1162  1216  1223  1232  1261  1314  1322  1341  1413  1431  1612  1621  2116  2123  2132  2161  2213  2231  2312  2321  2611  3114  3122  3141  3212  3221  3411  4113  4131  4311  6112  6121  6211  Total =  36

Note that all of the answers are the sam except for the shift part

-pi/4   will shift it appropriately .25 units  to the right

(1)   n=50; p = 1 - ((365 nPr n) / (365^n))=97.03735796%

(2)  n=51; p =  ((365 nPr n) / (365^n))=2.55680067%

(3) [50C2 x 365!] / [365^50 x (365 - 48 +1)!=0.00011393 09727%

the 9 x   part compresses the sinusoid wave from   2 pi   to    2pi/9 sec  = ~~.70 sec

thx so much 

Zero is greater than every negative number, so zero is greater than -12.

It is also correct to say that "0 is greater than or equal to -12".

A = anne now      4 years ago Anne was A-4

K = Katie now        four years ago Katie was    K-4  and this = 2(A-4)

In SIX years   A+6 = K

Two equations, two unknowns....solve for A  and K

K-4 = 2(A-4)

(A+6)-4 = 2A-8

10 = A                         K = A+6 = 16 y/o

rate * TIME = DISTANCE

DISTANCE/TIME =rate

18  /  9/8   =  s + c             s = boat speed   c = current

18  /   9/5  =  s - c             add the two equations together to solve for 's'    then you can compute    'c'  (if ya want to)

(EDITED)

Use law of cosines

a^2 + b^2 - 2ab cos (theta)     = c^2          c is the distance she will have to hike       a = 7.4 m  b = 5 m   theta =38 degrees    

Assuming only 365 days in a year --

1)  Jack's birthday can be any day:  365/365

      Mike's birthday must be the same day:  1/365

      Final probability:  (365/365) x (1/365)

2)  The professor's birthday can be any day.

      The first student can have a birthday any day except for the professor's:  364/365

      The next student cannot match either of the two previous persons:  363/365

       Etc.

       Final probability:  (364/365) x (363/365) x (362/365) x ... x (315/365)

3)  This one is a guess:

      The first person's birthday can be any day.

      To have only two persons with the same birthday, 48 persons cannot match:

            (364/365) x (363/365) x ... x (317/365)

       The one that matches:  (49/365)

       and this person can be any of the 49:

       Final probability:   (364/365) x (363/365) x ... x (317/365) x (49/365) x 49

       But this is just a guess.

I assume you mean:

\(2x-3y=40\)  (1)

\(x+5y=7\) (2)

Now let's multiply (2) with -2 

\(2x-3y=40\)

\(-2x-10y=-14\) 

Add these two equations

\(-13y=26\)

\(y=-2\)

Now to find x you just substitute  the value of y in either of the two equations

\(x+5y=7 ,x+5(-2)=7, x=7+10=17\)

So,

x=17

y=-2

Thank you Alan!

As follows:

The specific coordinates of P and Q I should just have labelled x_P, y_P, x_Q and y_Q rather than x_OP etc. but, hopefully, you will get the idea!

l = large cups       amount of large cups = 16 x  ounces

small cups = 2x     amount of small cups = 8 (2x)      summed = 320 ounces

16x      +     8 (2x)       =    320        solve for 'x' the number of 16 ounce cups, the 8 ounce cups = 2x

  If you haven't signed on for a while (I do not know the length of time involved) , you get removed from the users list......you will re-appear the enext time you sign on.

how do you solve 6x+3y=14 and 5x-2y=8.

Here's another way of looking at the problem:

Probability he will roll no 1's = (5/6)³. 

Probability he will roll exactly one 1 = 3*(1/6)*(5/6)².

Hence probability he will roll at least two 1's = 1 - ((5/6)³ +  3*(1/6)*(5/6)²) = 2/27