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2 years ago

the rate of change of sales tax on trousers is 10% one can buy a pair of trousers at$660. What is the marked price?

Mathematics

1 answer:

Anika [276]2 years ago

8 0

Step-by-step explanation:

Total price = Marked price + Sales tax

= Marked price + (0.1 * Marked price)

= 1.1 * Marked price.

Since the total price is $660, the marked price is $660 / 1.1 = $600.

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The product of 3 and a number x is 23 . What is the value of x?

Scrat [10]

I believed there's a typo in your question
<span>The product of 3 and a number x is 2/3
if so
</span>3x = 2/3
x = 2/3 * 1/3
x = 2/9

answer
x = 2/9

8 0

3 years ago

Solve for x. −1.9=x/0.5

Feliz [49]

-1.9 = x/.5 given equation

multiply both sides by .5

.5(-1.9) = x/(.5) *(.5)

-.95 = x

check: -1.9 = -.95/ .5

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3 years ago

[REALLY EASY QUESTION]

AVprozaik [17]

Answer:

A. h(x)=4x+2

Step-by-step explanation:

Because it's shifted to the left, you would subtract three units from 5, resulting in 4x+2. Hope this helps, have a nice day! :)

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2 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

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3 years ago

Goran is participating in a 6-day cross-country biking challenge. He biked for 51, 65, 62, 46, and 62 miles on the first five

zimovet [89]

Answer:

i dont understand?

Step-by-step explanation:

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3 years ago