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3 years ago

What’s the least common multiple of the denominators of 9+3/4x=7/8x-10

Mathematics

1 answer:

Alona [7]3 years ago

8 0

Answer:

x = 72

Step-by-step explanation:

$9 + \frac{3}{4} x = \frac{7}{8} x - 10 \\ \frac{3}{4}x - \frac{7}{8} x = - 10 - 9 \\ \frac{6}{8} x - \frac{7}{8} x = - 19 \\ - \frac{1}{8}x = - 19 \\ x = \frac{ - 9}{ - \frac{1}{8} } \\ x = 72$

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One night Puvi spent 1/3 of his money on dinner and then 1/4 of the remaining money on dessert. He then had just enough left so

arsen [322]

Answer:

Step-by-step explanation:

Cost of two tickets = 2 * 30 = £60

Let the amount with Puvi = £x

Money spent on dinner = (1/3)*x

$= \dfrac{1}{3}x$

$Remaining \ money =x - \dfrac{1}{3}x =\dfrac{3}{3}x-\dfrac{1}{3}x=\dfrac{2}{3}x$

$Money \ spent \ on \ dessert = \dfrac{1}{4} *\dfrac{2}{3}x =\dfrac{1}{6}x\\\\\\$

Total money - money spent on dinner - money spent on dessert = 60

$x -\dfrac{1}{3}x -\dfrac{1}{6}x=60\\\\\\\dfrac{6}{6}x-\dfrac{1*2}{3*2}-\dfrac{1}{6}x =60\\\\\\\dfrac{6}{6}x-\dfrac{2}{6}x-\dfrac{1}{6}x=60\\\\\\\dfrac{6-2-1}{12}x=60\\\\\\\dfrac{3}{12}x=60\\\\\\x=60*\dfrac{12}{3}=60*4=240$

Money that Puvi had at the start of the night = £ 240

8 0

2 years ago

The product od 2 and y is at least 10

alexgriva [62]

Hello There!

It could be represented by this:
$xy \geq 10$

Hope This Helps You!
Good Luck :)

- Hannah ❤

8 0

3 years ago

Can someone help me please I need the mode, median, range and the mean.?

Neko [114]

Answer:

First the mode. Since 5 popped up the most, 5 is the mode.

Next is the median. I crossed 1 dot from each side until it shows the last dot, and 5 was the last one.

After that the range. 9-2=7

Finally the worst, the mean... 2+2+3+3+3+4+5+5+5+5+5+6+6+6+8+9+9+9+9

=104/19=5.47

SO, Mode=5 Median=5 Range=7, and the mean is 5.47 (rounded nearest hundred)

7 0

3 years ago

If sin(2x)=cos (x+30°), what is the value of x?

Murljashka [212]

Answer:

x=20

Step-by-step explanation:

Please mark as brainliest

3 0

3 years ago

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