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3 years ago

There are 203 apples in a basket. How many children can share these apples equally? How many apples would each child get?

2 answers:

7 0

You can divide the number 203 into 1, 7, 29 or 203 parts, therefore:

<span>All possible numbers of children in ascending order are: 1, 7, 29, 203

</span><span>All possible numbers of apples in descending order are: 203, 29, 7, 1</span>
Нope this helps

tangare [24]3 years ago

6 0

I think this is factorials
so an example would b....... 1*2*3*4*5.... ect

hope this helps

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The greater of two consecutive integers is 15 more than twice the smaller. Find the integers.

Vlad [161]

Well, we know that the two consecutive numbers have to be negative, because if it was positive, the smaller number multiplied by two would be much greater. (unless 1 and 2). So, by guessing and checking, we will get -13 and -14 :)

3 0

3 years ago

A population p of migrating butterflies changes over time it is represented by the equation p=100,000*4/5 where w is the number

Trava [24]

Given that,

A population p of migrating butterflies changes over time it is represented by the equation

$p=100,000\times (\dfrac{4}{5})^w$ Where w is number of weeks.

To find,

The population after 2 weeks.

Solution,

We have,

$p=100,000\times (\dfrac{4}{5})^w$

Put w = 2 in the above equation.

$p=100,000\times (\dfrac{4}{5})^2\\\\=100,000\times \dfrac{16}{25}\\\\=64000$

So, the population after 2 weeks is 64000 .

7 0

3 years ago

A number is chosen at random from 1 to 50 find the probability of selecting composite numbers

Anit [1.1K]

34/50= .68 or 68%

hope it helps

3 0

3 years ago

If (x + yi) + (4(x + yi) + 6i) = 10, what is x + yi?

Verizon [17]

Answer: (10 - 6i)/5

(x + yi) + (4(x + yi) + 6i) = 10

⇔ 5(x + yi) + 6i = 10

⇔ 5(x + yi) = 10 - 6i

⇔ x + yi = (10 - 6i)/5

Step-by-step explanation:

8 0

3 years ago

The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

7 0

2 years ago