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3 years ago

15

YALL HELP plz give me a fraction problem that contains division and multiplication !!

1 answer:

Luba_88 [7]3 years ago

4 0

Answer:

This one is a bit easy, so if you need a harder one, just let me know.

((32/48) x (4/6)) / (2/5)

Step-by-step explanation:

1) You can simplify both (32/48) and (4/6) to (2/3).

2) Multiply (2/3) by itself, giving (4/9).

3) Dividing by a fraction is equivalent to multiplying by its reciprocal, or opposite. In this example, the reciprocal of (2/5) is (5/2). Thus, (4/9) x (5/2) = (20/18).

4) Simplify your answer, to (10/9) or 1 and 1/9 (as a mixed fraction).

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Help me pleaseeeeeeeeeeeeeee

dangina [55]

Answer:

W=1000C/tc

Step-by-step explanation:

First multiply both side 1000: 1000C=Wtc

Divide both side by tc: 1000C/tc=W (t, c ≠0)

7 0

3 years ago

Define an action adder :: io () that reads a given number of integers from the key board, one per line, and displays their sum.

aleksandrvk [35]

The program is an illustration of loops

<h3>What are loops?</h3>

Loops are program statements that are used to perform repetition

<h3>The main program</h3>

The program written in Python, where comments are used to explain each line is as follows:

#This initializes sum to 0

summ = 0

#This gets input for the first number

num = int(input())

#This is repeated while num is not -1

while num!= -1:

#This calculates the sum

summ+=num

#This gets input for the num

num = int(input())

#This prints the sum

print(summ)

Read more about loops at:

brainly.com/question/16397886

4 0

2 years ago

Maclaurin series of sinx

olga nikolaevna [1]

Answer:

cos(u) and sin(u) can be expanded in with a Maclaurin series, and cos(c) and sin(c) are constants.

Step-by-step explanation:

thats it

7 0

3 years ago

Candice spent 5 1/4 hours doing her homework. Her brother, Ronald, spent 1/2 that number of hours doing his homework. How many h

Elden [556K]

Let

x-----> the number of hours that Candice spent doing her homework

y-----> the number of hours that Ronald spent doing his homework

we know that

$x=5\frac{1}{4}\ hours$

convert mixed number to an improper fraction

$5\frac{1}{4}\ hours=\frac{5*4+1}{4}= \frac{21}{4}\ hours$

so

$x=\frac{21}{4}\ hours$ -------> equation A

$y=\frac{1}{2}x$ ------> equation B

Substitute equation A in equation B

$y=\frac{1}{2}*\frac{21}{4}=\frac{21}{8}\ hours$

convert an improper fraction to a mixed number

$\frac{21}{8}\ hours=2.625=2+0.625=2+\frac{5}{8}=2\frac{5}{8}\ hours$

therefore

<u>the answer is the option B</u>

$2\frac{5}{8}\ hours$

5 0

3 years ago

Read 2 more answers

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

3 years ago