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3 years ago

3 friends by pizza they split the cost 3 ways . the pizza cost 15 dollars amd salad is 6 dollars how much will each girl pay

Mathematics

2 answers:

Galina-37 [17]3 years ago

4 0

7 dollars friend 7 dollars

Natalija [7]3 years ago

3 0

The total cost is $21. If you divide 21 by 3 you get $7 each person.

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Help me plz I do not understand

9966 [12]

It is quite simple actually since it is simple division with the negative rules.
Just simplify the problem and use the negatives to determine if it is positive or negative. (two negatives equal a positive and one negative equals a negative)

5 0

3 years ago

Find the point on the line y=2x that is closest to the point (4, 5) .

alexgriva [62]

That point would be : (2.5,5)

7 0

2 years ago

Find the product of this please thank you

iVinArrow [24]

You combine the like terms to get
2s^2y^2-4a^2sy+2a^4

7 0

3 years ago

The mold for a cone has a diameter of 4 inches in 6 inches tall what is the volume of the cone mold to the nearest 10th

anygoal [31]

Hey there!

The formula for the volume of a cone is $V=\pi r^{2} \frac{h}{3}$ .

In this cone, we have the given information:

Its diameter is 4 inches
Its height is 6 inches

The information we need for this formula is the radius and height.

We already have the height, but not the radius.

Fortunately, we can easily get it by dividing the diameter by 2.

$4 \div 2 = 2$

So, the radius is 2.

Let 3.14 be $\pi$ .

Plug in each value into the equation. $V = 3.14 \times 2^{2} \times \frac{6}{3}$

Simplify the exponent and fraction.

$V = 3.14 \times 4 \times 2$

Multiply the three numbers together.

$V = 25.12$

But wait! We have to round this to the nearest tenth.

When rounding, we look at the place value one less than the one we are rounding to. Since we are rounding to the nearest tenth, we look at the hundredths place, which is the digit 2.

Since 2 is less than 5, we round down to 25.1.

This is the volume of a cube, and it is measured in inches, so the unit we will use is $in^{3}$ .

So, your <u>final answer</u> is $25.1 in^{3}$ !

Hope this helps!

8 0

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

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

5 0

3 years ago