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3 years ago

Which of the following values is a solution of |2-x| less than 4

Mathematics

1 answer:

morpeh [17]3 years ago

4 0

The correct answer is -1.

In order to solve this, we need to split into the positive and negative version of the answers. Let's start with the positive version.

2 - x < 4

-x < 2

x > -2 ----> NOTE: When we divide by -1, we have to change the direction of the sign.

Now we'll do the negative version.

2 - x > -4

-x > -6

x < 6

So we know the number must be greater than -2, but less than 6. The only number on this list that fits that is -1.

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Use the diagram. AB is a diameter, and AB is perpendicular to CD. The figure is not drawn to scale.

Westkost [7]

Answer:

The measure of arc BD is 147°

Step-by-step explanation:

we know that

If segment AB is perpendicular to segment CD

then

The measure of the inner angle CPA is a right angle (90 degrees)

Remember that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

m∠CPA=(1/2)[arc AC+arc BD]

substitute the given values

90°=(1/2)[33°+arc BD]

180°=33°+arc BD

arc BD=180°-33°=147°

7 0

3 years ago

Below is an equation being solved.

Fudgin [204]

Answer:

There are no errors

Step-by-step explanation:

The math is correct

5 0

3 years ago

Work out the sum of

Sladkaya [172]

Answer: 1 and 3/70

Explanation: Find the least common denominator and then combine like terms. The LCM is 70.

8 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

Which of the trigonometric ratios has a value that is undefined?

FinnZ [79.3K]

Answer:

$\tan \left(\frac{\pi }{2}\right)$

Step-by-step explanation:

We know that tan(x) is defined as sin(x)/cos(x):

$\tan \left(\frac{\pi }{2}\right),\\\sin \left(\frac{\pi }{2}\right)=1,\\\cos \left(\frac{\pi }{2}\right)=0,\\\\$

$\tan \left(\frac{\pi }{2}\right) = 1/0 \\$

1/0 is undefined as 0 is in the denominator

6 0

3 years ago