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the probability of winning on the first ticket is \(2\over 40\)

the probability of wining on the second one is \({2-(1) \over 40-(1)}={1\over39}\)

the probabiity of winning on the both is \(\frac{2}{40}\times\frac{1}{39}=\frac{1}{780}=0.00128205\dots\)

\(\frac{1}{5n}+6=-4\)

\(\frac{1}{5n}+\frac{6}{1}=-4\)

\(\frac{1}{5n}+\frac{6\times5n}{1\times5n}=-4\)

\(\frac{1}{5n}+\frac{30n}{5n}=-4\)

\(\frac{1+30n}{5n}=-4\)

\(\frac{1+30n}{5n}\times5n=-4\times5n\)

\(1+30n=-4\times5n\)

\(1+30n=-20n\)

\(1+30n+20n=-20n+20n\)

\(1+50n=-20n+20n\)

\(1+50n=0n\)

\(1+50n=0\)

\(1+50n-1=0-1\)

\(50n=0-1\)

\(50n=-1\)

\(\frac{50n}{50}=\frac{-1}{50}\)

\(1n=\frac{-1}{50}\)

\(1n=-0.02\)

\(n=-0.02\)

thanks

So you have a 1/20 chance to win one prize, with 2 tickets and 1/40 chance to win one prize with one ticket

Using that same chance, you will have 2/80 or 1/40 chance of winning both with 2 tickets.

55,000 = 25(10)2t?,

Solve for t:

55000 = 500 t Divide both sides by 500

t=110 hours.

it is 18

\(f(x)=5+\frac{1}{x+3}\)

\(f(x)=\frac{5}{1}+\frac{1}{x+3}\)

\(f(x)=\frac{5\times(x+3)}{1\times(x+3)}+\frac{1}{x+3}\)

\(f(x)=\frac{5x+15}{x+3}+\frac{1}{x+3}\)

\(f(x)=\frac{5x+15+1}{x+3}\)

\(f(x)=\frac{5x+16}{x+3}\)

i wrote more

 \(f(x) = 5+{1 \over x+3} \\ f(x) = {5(x+3)\over x+3}+{1 \over x+3} \\ f(x) =\frac{5x+16}{x+3} \text{ same as } \frac{ax+b}{x+3}\\ a=5 \\ b=16\)

got to it faster

it is 00001000

Thank you very much!

50*0.10=5

the awnser is 5

So, it can be either red (only) or white spots (only) or red with white spots.

The probability of being any of the above three is 0.80.

The probability of being red (either only red or red with white spots) is 0.62.

The probability of having white spots (either only white spots or red with white spots) is 0.35.

If you add these two probabilities together, you get 0.62 + 0.35  =  0.97.

But, the total probability of all three is just 0.80.

Subtracting  0.97 - 0.80 means that there is an overlap of 0.17.

The overlap is the probability of getting one that is red with white spots.  Answer:  red with white spots  =  0.17.

Red only will be 0.62 - 0.17  =  0.45.

White spots only will be 0.37 - 0.17  =  0.18.

Checking by adding the three probabilities:  0.17 + 0.45 + 0.18  =  0.80.

okay thank you gisonj338 :)

\(0.1x+0.05y=30\)

\(x+y=500\)

Solve for x in the second equation

\(x+y=500\)

\(x=500-y\)

Subsitute x in th first equation and solve for y

\(0.1\times(500-y)+0.05y=30\)

\(50-0.1y+0.05y=30\)

\(50-0.05y=30\)

\(-0.05y=-20\)

\(y=400\)

Subsitute y in the second equation

\(x+400=500\)

\(x=100\)

There would be \(400\) nickels and \(100\) dimes totaling \(500\) coins with a face value of \($30.\)

Do something like this: 1 2/5

acos(3/4) = 0.722734247813

okay thank you. 

4 

The big sack contained $30 in nickels and dimes. If there were 500 coins in the sack, how many were nickels?

Let the number of nickles=N

Let the number of dimes =D

N + D =500

.05N + .10D =$30

Solve the following system:
{D+N = 500 |     (equation 1)
0.1 D+0.05 N = 30 |     (equation 2)
Subtract 0.05 × (equation 1) from equation 2:
{N+D = 500 |     (equation 1)
0 N+0.05 D = 5 |     (equation 2)
Divide equation 2 by 0.05:
{N+D = 500 |     (equation 1)
0. N+D = 100. |     (equation 2)
Subtract equation 2 from equation 1:
{N+0 D = 400 |     (equation 1)
0 N+D = 100 |     (equation 2)
Collect results:
Answer: |  N = 400               and               D=100
 

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