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3 years ago

The radius of a circle is 11 ft. Find the circumference of the circle.

Mathematics

1 answer:

lapo4ka [179]3 years ago

8 0

Answer:

69.12 ft

Step-by-step explanation:

The formula for circumference of a circle is 2πr

= 2 × 3.142× 11

= 69.12

Hence the circumference of the circle is 69.12 ft

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. The time of a clock as seen in a mirror is 3:15. The correct time is

Oksana_A [137]

Answer:

3:15 cause a mirror cant change time

Step-by-step explanation:

go in ur mirror and look at the time did it somehow get darker or lighter???

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3 years ago

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Solve and in standard form please. you must add. this is from khan academy

Sindrei [870]

We are asked to sum the two trinomials $(-4f^3-5f+16)+(4f^2-f+9)$ . We can start by using the distributive property to remove the parenthesis, giving us $-4f^3-5f+16+4f^2-f+9$ . Next, when we combine like terms, we get $\boxed{-4f^3+4f^2-6f+25}$ . Hope this helped!

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3 years ago

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Whats 937X928(82|23) like what is this-

Zepler [3.9K]

Answer:

(71 301 952, 19 999 328)

Step-by-step explanation:

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3 years ago

5. A farmer puts apples into 36 crates. Each crate has 25 kilograms of apples in it. She sells 486,235 grams of apples. How many

Lynna [10]

There would be 413,765 kilograms after selling 486,235 grams of apples

How many grams make a kilogram?

In each kilograms, there are 1,000 grams, which means we need to first of all convert the total number of apples in kilograms to grams.

Each crate has 25 kilograms of apples in it

1 crate =25 kilograms

36 crates=36*25 kilograms

36 crates=900 kilograms

1 kilogram=1000 grams

900kilograms=1000*900

900 kilograms=900,000 grams

The number of grams left after having sold 486,235 grams of apples, in essence, the quantity of apples left thereafter is 900,000 grams minus the 486,235 grams of apples sold

apples left=900,000-486,235

apples left= 413,765 kilograms

Find out more about conversion of kilograms to grams on:brainly.com/question/9301317

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2 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

3 years ago