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consider tossing a biased coin whose probability is of coming up heads is 0.3. if 40 trials are performed, what values of k would have probabilities closest to half the probability of the expected value of k

By the expected value of k, I am going to assume that you mean the expected probability IF the the coin was not biased.

$$\begin{array}{rll}
2\times \binom{40}{k}(0.3)^k(0.7)^{(40-k)}&=&\binom{40}{k}(0.5)^k(0.5)^{(40-k)}\\\\
2\times (0.3)^k(0.7)^{(40-k)}&=&(0.5)^{40}\\\\
2\times \left(\frac{3}{10}\right)^k\left(\frac{7}{10}\right)^{(40-k)}&=&\left(\frac{1}{2}\right)^{40}\\\\
2\times \frac{3^k\times 7^{(40-k)}}{10^k\times {10^{(40-k)}}}&=&\frac{1}{2^{40}}\\\\
\frac{3^k\times 7^{40}}{10^{40}\times 7^k}&=&\frac{1}{2^{41}}\\\\
\frac{3^k}{ 7^k}&=&\frac{10^{40}}{2^{41}\times 7^{40}}\\\\
\left(\frac{3}{ 7}\right)^k&=&\frac{10^{40}}{2^{41}\times 7^{40}}\\\\
log\left[\left(\frac{3}{ 7}\right)^k\right]&=&log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]\\\\
k\times log\left[\left(\frac{3}{ 7}\right)\right]&=&log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]\\\\
k &=&\frac{log\left[\frac{10^{40}}{2^{41}\times 7^{40}}\right]}{log\left(\frac{3}{ 7}\right)}\\\\



\end{array}$$

$$\left({\frac{{log}_{10}\left({\frac{{{\mathtt{10}}}^{{\mathtt{40}}}}{\left({{\mathtt{2}}}^{{\mathtt{41}}}{\mathtt{\,\times\,}}{{\mathtt{7}}}^{{\mathtt{40}}}\right)}}\right)}{{log}_{10}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}}\right) = {\mathtt{16.702\: \!552\: \!085\: \!923\: \!092\: \!1}}$$

The closest one is k=17

Multiply them with (-1)

If you factor 2x² - x - 3  =  0, you get (2x - 3)(x + 1)  =  0. 

Setting each factor equal to zero:  2x - 3  =  0     and     x + 1  =  0

and solving:                                       x  =  3/2     and     x  =  -1

i is the symbol representing √(-1).

Whole numbers and fractions, including irrational numbers like π, are called "real" numbers.

i is called an "imaginary" number, which is a different type of numbers.

All the real numbers and imaginary numbers form the set of complex numbers.

Examples:  √(-9)  =  √(9)√(-1)  =  3√(-1)  =  3i  (another imaginary number)

Complex numbers can be formed by adding and subtracting real numbers with imaginary numbers, such as

5 + 7i, -2 + i, and 7 - 9i.

Ifall the gumballs are either orange, yellow, or purple, then:

orange = 3/4 = 9/12

yellos = 1/6 = 2/12

orange + yellow = 9/12 + 2/12 = 11/12

Then  1 - 11/12 or 1/12 = purple

This means the third root of 64 -- what number used three times in multiplication has an answer of 64?

It's 4, because 4 x 4 x 4 = 64.

csc^2 (x)- csc(x) - 2=0    factor

(cscx + 1) (cscx -2) = 0      set each factor to 0

So either

cscx  = -1      and this happens at  -3pi/2 ± n 2pi     where n is an integer

OR

cscx = 2        and this happens at pi/6 ± n 2pi     and  at  5pi/6  ± n2pi     agian, where n is an integer

Minimum amount of steps E -> F -> G is 8, but 9 steps is required.

I.e. E -> F in 4 steps & F -> G in 5 steps, or E -> F in 5 steps & F -> G in 4 steps.

So is 40 ways to go from E to F to G in 8 steps, or did you somehow take that additional step into account?

(I don't have spare time to calculate, so cannot check this.. so just wondering)

Lol i asked my brother he said it's +1. He said just to put it as -1*-1

 Negative number are hard the square / cube as they change from negative to positive 

like -2^3 = -8

But i dont know why the calc would analyse it like that and answer it as -1

They are the same question...?

But when put differently they give different answers...

I'm puzzled.

Lol

But why is  

  $$\\-1^2=-1 \;\;when\;\; -1*-1=+1\qquad $maybe you didn't know that?$$$

I hate these mind boggling questions with negatives but i will see.

-1^2 would = -1 

Its the same as one just negative

but 1 itself is a positve and sqrt(1) =1

I have something for you to think about  MG

$$\\(-1)^2=1\\
$so would it be correct to say that$\\
\sqrt{1}=-1\;\;\;?\\
also\\
what \;is\;\; -1^2\;\;\;?$$

Thanks :)

I wouldn't have leanrt square and sqare root without your help.

It is good to see you making good use of the web2 calc MG  :)

1+2 =3

Unless you add something or takeaway something.

That was quite a feat Chris,  

I will do it using remainder theorem.

For what value of c will the polynomial P(x) = -2x^3 + cx^2 - 5x + 2 have the same remainder when it is divided by x - 2 and by x + 1?

When P(x) is divided by x-2 the remainder will be     p(2)=-2*8+c*4-10+2 = -16+4c-8 = 4c-24

When P(x) is divided by x+1 the remainder will be    p(-1)=-2*-1+c*1+5+2=2+c+7 = 9+c

If these are the same then 

4c-24=9+c

3c=33

c=11

Adding onto CPhills answer.

Mathmaticians still haven't found the full number or even if it has a full number.

In pi you can find your phone number , your birthday everything!

But it never repeats!

A decimal place?

It's a ' dot '  that simply looks further into a number 

i.e. 1.5 = 1 and half 

1.5+1.5 = 3 

a half add a half = 1 

1+1+1=3

Not only that but a decimal can also be convertd to a fraction or percentage.

0.25 = 25% = 1/4 

$${{\mathtt{1}}}^{{\mathtt{2}}} = {\mathtt{1}}$$

$${\sqrt{{\mathtt{1}}}} = {\mathtt{1}}$$

1.000e+00? means  $${\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{0}}}$$  =   1*1 = 1

Worse than double counting, multiple counting.

To illustrate why this method is wrong, consider an even simpler situation. Suppose that you have just four people and you want to split them into pairs. How many ways ?

If you employ the method used for the eight person problem, I guess that you would say that the number would be 4C2*2C2 = 6.

However, if you call the people A,B,C,D, you can see that there are only three possible pairings (AB and CD), (AC and BD) and (AD and BC). The basic reason why the first answer is wrong is that each grouping has been counted twice. (AB and CD) and (CD and AB), for example, are being thought of as being two different pairings, meaning that that particular pairing has been counted twice, if that makes sense. The 4C2 = 6 contains AB, AC, AD, BC, BD and CD, but both AB and CD produce the same split into two pairs.

Try the eight into four groups of two again.