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Remember the rule that  aⁿ x a^m   =  a^m+n

So...... 3⁴ x 3⁴ = 3⁴⁺⁴ = 3⁸

Does that help ??

please can you help me is for my chemistry class

 srry i mean 3^4 x 3k = 3^8, write down the value of k

This is correct, Melody.......!!!!

Another way to look at this is.......call the amount of money the person bets = B...........and call the amount of money he wins  = 3B

So the expected value of any bet is

3B(1/18) - B(17/18) =

3B/18 - (17B)/18  =

-(14B)/18 =

-.7778 B 

In other words......the player stands to lose about 78 cents on each dollar he bets........

(Contrast this to the game of roulette where, on the "American Wheel," the player stands to lose a little more than 5 cents per dollar bet on any spin !!! )

Use the distributive property:

3(2 - 3x)     =     3(2) - (3)(3x)  =  6 - 9x

To find the percent of increase:

     first, find the amount of increase:     164 - 151 = 13 pounds

     then, divide by the original amount:  13 / 151  =  0.0861

     to turn into a percentage, multiply that answer by 100:  0.0861 x 100  =  8.6%

To find the payment:     P  =  i·A / [1 - (1 + i)^(-N) ]

For one year:

     i = interest rate, as a decimal, per period:  0.072 / 3  =  0.024

    A = amount borrowed:  3000

    N = number of payments:  3

    P  =  (0.024)(3000) / [ 1 - (1 + 0.024)^(-3) ]   =  $1048.38

For three years:    If it's three equal installments per year:      N = 9

    P  =  (0.024)(3000) / [ 1 - (1 + 0.024)^(-9) ]   =  $374.60

For five years:       If it's three equal installments per year:      N = 15

    P  =  (0.024)(3000) / [ 1 - (1 + 0.024)^(-15) ]   =  $240.52

I lost the question towards the end ...

Perhaps you mean:   2 x 2 x 2 x 2 x 2 x 2 x 2   =   2 ^ 7

(I got the 7 by counting the number of 2's in the problem.)

Growth and decay problems:  By how much will a population grow or decline? What is the half-life of an element?

Using logarithms to solve problems in which the variable is an exponent.

Solving those problems taht arise using calculus.

Solving problems in probability that use the normal curve of distribution, whose function is described in terms of e.

This is a volume problem:  volume  =  area of base x height  

                  which is also:  volume  =  area of base x depth

(However, the 0.6 sqm of material must actually be 0.6 cubic meters of material.)

volume = 0.6 cubic meters          area of base = 6.0 square meters         depth = unknown

----->     0.6  =  6.0 x depth

Dividing by 6.0:     ----->     0.6 / 6.0  =  depth     ----->     0.1 meters  =  depth

If you want a point that is one-third of the way from (2, 3) to (5, 9), it will have to be one-third of the way in the x-direction and one-third of the way in the y-direction.

In the x-direction:  the distance from 2 to 5 is 3; one-third of 3 is 1; adding 1 to 2 (the beginning x-value), answer = 3.

In the y-direction:  the distance from 3 to 9 is 6; one-third of 6 is 2; adding 2 to 3 (the beginning y-value), answer = 5.

The location of the point on-third of the way from (2, 3) to (5, 9) is (3, 5).

tan(32°)  =  x/6.5

0.6249  =  x/6.5

x  =  -.6249 x 6.5

x  =  4.062

Another way to get CPhill's answer:

Still using his idea of subtracting the probability of getting no kings from 1, the total possible probability.

The only way that you can get no kings is:

to get no king for the first card:  probability = 48/52    

to get no king for the second card:  probability = 47/51

to get no king for the third card:  probability = 46/50

to get no king for the fourth card:  probability = 45/49

and, finally:

to get no king for the fifth, and last, card:  probability = 44/48

Multiplying all these probabilities together:  48/52 x 47/51 x 46/50 x 45/49 x 44/ 48  =  0.6588

Subtracting this from 1.000:  1.000 - 0.6588  =  0.341

@@ End of Day Wrap       Sat 22/11/14       Sydney,   Australia     Time 12:35 am  (Really Sunday)         ♬

Hi everyone,

Today was probably the busiest Saturday that we have ever had.  The forum is growing exponentially.

Anyway there were lots of great answers provided by Alan, CPhill, Geno3141, BabaLlama, SelenaPubblT, Mathematician, Heureka, and Happy7.  Thank you, you are all great.    

There were many questions that did not get answered properly today - many but not of them were probablility.   I collected the address and put them in this thread.  ( I appologise if there is any doubling up)

Sun 23/11/14

Finding percent of a number (or money)

I suggest you look up a table OR here is a site that you could make good use of

Should there be any brackets here?

What is the sum of the first 25 terms of the sequence 3, 17, 31, 45, ..

d=14 a=3 n=25

Sn=n/2(2a+(n-1)d) = 25/2(6+24*14) 

$${\frac{{\mathtt{25}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{14}}\right) = {\mathtt{4\,275}}$$

The random variable X has a binomial distribution with n = 10 and p = 0.01. Determine the following probabilities.

            a. P(X = 5)                  c. P(X ≥ 9)

            b. P(X ≤ 2)                  d.P(3 ≤ X < 5)

a)  10C5*0.01^5*0.99^5

b) P(0 or 1 or 2 successes)=  0.99^10 + 10C1*0.01*0.01^9 + 10C2*0.01^2*0.99^8

c) P(9 or 10 sucesses) = 10C9*0.01^9*0.99 + 0.99^10

d) P(3 or 4 sucesses) = 10C3*0.01^3*0.99^7 + 10C4*0.01^4*0.99^6

I think these are right but it is late - i am tired and probability is not my strong suite.

The phone lines to an airline reservation system are occupied 40% of the time. Assume that the events that the lines are occupied on successive calls are independent. Assume that 10 calls are placed to the airline.

a.  What is the probability that for exactly three calls the lines are occupied?

10C3* 0.4^3*0.6^7

$${\left({\frac{{\mathtt{10}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.6}}}^{{\mathtt{7}}} = {\mathtt{0.214\: \!990\: \!848}}$$

b. What is the probability that for at least one call the lines are not occupied?

1-P(all are cccupied)=$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.4}}}^{{\mathtt{10}}} = {\mathtt{0.999\: \!895\: \!142\: \!4}}$$

The probability that a person carries the gene is 0.1. What is the probability that 4 will have to be tested before 2 with the gene are detected?

I think this really means that there is one sucess in the first 3 and then the forth one is a success.

P(1 succes in first 3)=3C1*(0.1)^1(0.9)^2

$${\mathtt{3}}{\mathtt{\,\times\,}}{\left({\mathtt{0.1}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{0.9}}\right)}^{{\mathtt{2}}} = {\frac{{\mathtt{243}}}{{\mathtt{1\,000}}}} = {\mathtt{0.243}}$$

P(1 succes in first 3 FOLOWED by a success) = 0.243*0.1 = 0.0243

Once again - I think that is correct  

a) P(FFFS)=0.2^3*0.8 

$${{\mathtt{0.2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{0.8}} = {\frac{{\mathtt{4}}}{{\mathtt{625}}}} = {\mathtt{0.006\: \!4}}$$

b) 1-P(FFFF) = 1-0.2^4 

$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.2}}}^{{\mathtt{4}}} = {\frac{{\mathtt{624}}}{{\mathtt{625}}}} = {\mathtt{0.998\: \!4}}$$

c) P(FFF)=0.2^3

$${{\mathtt{0.2}}}^{{\mathtt{3}}} = {\frac{{\mathtt{1}}}{{\mathtt{125}}}} = {\mathtt{0.008}}$$

That is what I think anyway.  :))

Perhaps someone would like to check my answers?

That sounds like it could be right Chris.  

I don't have any better ideas.  3 more points for you   :))

What is the question?

I think that this question is poorly worded.

there is no mention of the probability of a family not visiting either  park.

SO  I have decided to interpret it like this.

Of the families that visit Sand park or Rainbow park,  20% visit both and 80% visit Rainbow

Find the probability that a family visits the Sand Crab Water Park if it is known that they have already visited Rainbow Gardens Amusement Park.

---------------

NOW

20% go to both and 80% go to Rainbow.  This must mean that 60% go to Rainbow but not to Sand

This means that 20/60 = one third of those that go to Rainbow also go to Sand

So the desired probablility is 1/3

I think that is correct.