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2 years ago

Lucy had 24 beanbags and gave 1/3 of them to her friend Shelley. How many beanbags does Lucy have left?

Mathematics

2 answers:

Fantom [35]2 years ago

6 0

Answer:8

1/3 divided by 24 equals 8.

zepelin [54]2 years ago

6 0

8 so the person who got it was correct

1/3 divided by 24 equal

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If 3 pounds of apples cost $1.89 how much do 5 pounds cost

Grace [21]

Answer:

$3.15

Step-by-step explanation:

divided by 3 then multiplied by 5

give brainliest please!

5 0

2 years ago

Read 2 more answers

Help please and thank you

Bingel [31]

Answer:

40° D. is your answer to this question

3 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

What is the secant of angle L? <br> A) 15/12<br> B) 13/5<br> C)12/13<br> D)12/5

Gennadij [26K]

Hey there! I'm happy to help!

The secant is the ratio of the hypotenuse to the side adjacent to the angle. However, we do not have the measure of the hypotenuse given to us. This is a right triangle, so we can use the Pythagorean Theorem to find the hypotenuse. Our sides have lengths of 5 and 12.

5²+12²=c²

25+144=c²

c²=169

We square root both sides

c=13

Now, we take our hypotenuse (13) and we put it over the adjacent side to angle L, which is 5.

Therefore, the secant is B) 13/5.

I hope that this helps! Have a wonderful day!

6 0

2 years ago

Find an equation of the tangent line to f(x) = 5x2 − 2x + 10 at x = 7.

laiz [17]