All Answers 358093 Answers

BuilderBoy:

b)  You forgot one arrangement:

(2, 2, 0, 0, 0) ==5! / 3!2! ==10

So: 60 + 10 ==70 ways.

This is the case of 4 IDENTICAL balls into 5 DISTINCT boxes without restrictions:

Formula: [4 + 5 -1] nCr 4 ==8 nCr 4 ==70 ways

Note: To get the correct distribution, look up this page (at the very bottom of it), where there is a table made up by an expert mathematician by the name of Bogart here in Wikipedia:

https://en.wikipedia.org/wiki/Twelvefold_way

a) 

(n/k) / (n / (k - 1)) ==(n/k) x  (k - 1) / n ==(k - 1) / k

b) See the answer here:  https://web2.0calc.com/questions/counting-solutions

See answer #1 here:  https://web2.0calc.com/questions/probability-question_100

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100000 - 999999 : 112496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

See the answer here:   https://web2.0calc.com/questions/counting-problem_10511#r2 

Just one such number:

4096==64^2

4096==16^3

13 = 20

b) 

0, 0, 0, 0, 4

3, 1, 0, 0, 0

2, 1, 1, 0, 0

1, 1, 1, 1, 0

For the first case, there are \({5 \choose 1} = 5\) cases

For the second one, there are \(5 \times {4 \choose 1}= 20\) cases

For the third one, there are \(5 \times {4 \choose 2} = 30\) cases

For the final one, there are 5 cases. 

So, there are a total of \(5 + 20 + 30 + 5 = \color{brown}\boxed{60}\) cases.

a)

1, 1, 2

4, 0, 0

3, 1, 0

\(\color{brown}\boxed{3}\) cases

The wording of this question strongly suggests that the use of Descartes' Rule of Signs is what's intended.

Suppose that we have a polynomial p(x) with real coefficients.

The rule says that the number of positive real roots is equal to the number of sign changes (as we move left to right along the polynomial) or is less than that number by an even integer.

Replacing x by -x to get the polynomial p(-x), a similar statement can be made about the number of negative real roots.

For example, the polynomial

\(\displaystyle p(x) = 3x^{5}-2x^{4}+7x^{2}-9 \\ p(-x)=-3x^{5}-2x^{4}+7x^{2}-9,\)

will have 3 or 1 positive and 2 or 0 negative real roots.

This graph should help (the other two roots are complex)

1/2

because for every WX > YZ, there is another choice of points where WX < YZ (by symmetry)

First of all p + m = 5 because there is a total of 5 pounds of mixture to be made. Now for there to be 40% of almonds there must be 0.4*5 = 2 pounds almonds So as there are 80% almonds in the mixture the amount of m must be 2 / 0.8 = 2.5 pounds so we have p + 2.5 = 5 so p = 2.5 Answer is 2.5 pounds of peanuts and 2.5 pounds of the mixture.

Probability is:

successes/total

We start by finding the total number of draws that you can do:

6 * 5 * 4 * 3 * 2 * 1

Now, we have to find the probability that you actually draw your pair:

6 * 1 

-> For the first pair:

4 * 1

-> After taking out the second pair:

2 * 1 

-> For the third pair

So, successes are: (6 * 4 * 2)/(6 * 5 * 4 * 3 * 2 * 1)

1/(5 * 3)

1/15

So the probability is 1/15.

I might be incorrect as I'm not the best at this, so if you find it incorrect please tell me.

We have the equation:

X + 1 = 0

Subtract 1 from both sides:

X = -1

What about 1:2:3?  What about 2:4:7?  What about 1:8:40?

I think I figured it out.

Transformations of f(x)=log₁₀(x) -> f(x)=2log₁₀(x-4)+3

Vertical stretch: f(x)=log₁₀(x) ->  f(x)=2log₁₀x

Horizontal translation: f(x)=2log₁₀x -> f(x)=2log₁₀(x-4)

Vertical translation: f(x)=2log₁₀(x-4) -> f(x)=2log₁₀(x-4)+3

In terms of x and y, the vertical stretch and vertical translation would affect y, and the horizontal translation would affect x. Therefore, it would be (x+4, 2y+3).

This can be checked by using the point (1, 0) on :

(1+4, 2(0)+3)

= (5, 3)

(5, 3) is a point on the graph f(x)=2log₁₀(x-4)+3.