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4 years ago

What fraction represents 1/2 x 2/3

Mathematics

1 answer:

son4ous [18]4 years ago

8 0

Answer: 1/3

hope this helped!

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Jamal borrowed $316 for nine months. He paid it off by paying $38.27 monthly.

Doss [256]

$28.43 is the answer

7 0

3 years ago

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The distance on a map between a student's house and her job is 6 cm on the same map the distance between her house and her schoo

Montano1993 [528]

Answer:

9 km

Step-by-step explanation:

First I did 12 divided by 8 to get the amount that each centimeter is worth.

I got 1.5

After that all i had to do was multiply 1.5 by 6 to get 9

8 0

3 years ago

Bill washes his mom's car once every 2 weeks and gets paid $20.00 each time. How much does he earn in 22 weeks?

vlada-n [284]

Answer:

$220

Step-by-step explanation:

To find for 1 week

$20.00÷2= $10

For 22 weeks

22×10=220

6 0

3 years ago

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Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I

Hatshy [7]

a. $f_{X,Y}$ is a joint density function if its integral over the given support is 1:

$\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1$

so the answer is yes.

b. We should first find the density of the marginal distribution, $f_Y(y)$ :

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy$

$f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}$

Then

$P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}$

or about 0.2019.

For the other probability, we can use the joint PDF directly:

$P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}$

which is about 0.7326.

c. We already know the PDF for $Y$ , so we just integrate:

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5$

5 0

3 years ago

The aorta is the largest artery in the human body. In the average adult, the aorta attains a maximum diameter of about 1.18 inch

Sholpan [36]

Well we know that 1 inch is 2.54 cm. But how wide is .18? Lets do .18 x 2.54 to get 0.4572. Now lets add 2.54 + 0.4572 to get 2.9972. This means that the maximum width of the aorta is 2.9972 cm. Hope this helped :)

3 0

4 years ago