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3 years ago

WILL GIVE BRAINLIEST

Mathematics

2 answers:

Gelneren [198K]3 years ago

5 0

Answer:

the answer it's B we got the same test

PUT THE B

irina1246 [14]3 years ago

5 0

Answer:

mm option a is right

Step-by-step explanation:

bcz there is only 1 zero less in weight of elephant as compared to whale hope it may help u

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Write the equation of the line that passes through the points (4, -6) and (-2,3).

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In point slope form, the answer would be y = - 3/2 x it’s one fraction

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3 years ago

Read 2 more answers

FIND AC! PLS HELP!!

Anna007 [38]

Answer:

AC = 7.3 in

Step-by-step explanation:

tan(51°) = 9/AC

1.24 = 9/AC

AC = 9/1.24

AC = 7.26

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3 years ago

Jailed has 60 dimes and 60 nickels

Ivenika [448]

Answer:

75 cents

Step-by-step explanation:

60 dimes is 600 cents or $6, 60 nickels is 300 cents or $3. when we add them together we have $9 and we need 9.75 so we need 75 cents more.

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4 years ago

Find number a and k so that x-2 is afactor<br>OF F(x)=x4-2ax tax-<br>X+k and F(-1)=3

Slav-nsk [51]

Question:

Find numbers a and k so that x-2 is a factor of

$f(x)=x^4-2ax^3+ax^2- x+k$

and

$f(-1)=3$

Answer:

$k = -2$ and $a=1$

Step-by-step explanation:

Given

$f(x)=x^4-2ax^3+ax^2- x+k$

$Factor:\ x - 4$

$f(-1)=3$

Required

Find a and k

For $f(-1)=3$

Substitute -1 for x

$f(x)=x^4-2ax^3+ax^2- x+k$

$f(-1) = (-1)^4 - 2a *(-1)^3 + a*(-1)^2 - (-1) + k$

$f(-1) = 1 - 2a *-1 + a*1 +1 + k$

$f(-1) = 1 +2a + a +1 + k$

$f(-1) = 2 +3a + k$

Substitute 3 for f(-1)

$3 = 2 +3a + k$

Collect Like Terms

$3 - 2 = 3a + k$

$1 = 3a + k$

Also:

If $x - 2$ is a factor, then

$f(2) = 0$

Substitute 2 for x and 0 for f(x)

$f(x)=x^4-2ax^3+ax^2- x+k$

$0 = 2^4 - 2a * 2^3 + a * 2^2 - 2 + k$

$0 = 16 - 2a * 8 + a * 4 - 2 + k$

$0 = 16 - 16a + 4a - 2 + k$

$0 = 16 - 12a - 2 + k$

Collect Like Terms

$2 - 16 = - 12a + k$

$-14 = k - 12a$

$k = 12a - 14$

Substitute 12a - 14 for k in $1 = 3a + k$

$1 = 3a + 12a - 14$

$1 = 15a - 14$

Collect Like Terms

$15a = 1 + 14$

$15a = 15$

Solve for a

$a = \frac{15}{15}$

$a=1$

Substitute 1 for a in $k = 12a - 14$

$k = 12 * 1 - 14$

$k = 12 - 14$

$k = -2$

3 0

3 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