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2 years ago

Holly says 4.131131113... is a rational number. Which of the following best describes whether Holly is correct and why?

Mathematics

1 answer:

Dovator [93]2 years ago

5 0

Answer:

its d holly is incorrect the decimal part follows a pattern but does no actully repeat

Step-by-step explanation: Your Welcome

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Find an equation in slope intercept form of a line having slope 6 and y-intercept 2

xz_007 [3.2K]

Answer:

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

We have the slope m = 6, and the y-intercept b = 2.

Substitute:

$y=6x+2$

4 0

2 years ago

The 8th grade is selling tickets to the 8th grade formal.The cost per ticket is $15.The decorations and food cost $600,and each

MArishka [77]

Answer:

Step-by-step explanation:

Each ticket is $15. The number of tickets is what we are trying to solve for. The class spends a certain amount of money to prepare for the formal. They hope that the money they make in ticket sales is MORE than what they spend. The expression that represents the number of tickets at $15 each is 15x, where x is the number of tickets. They hope that the sales are greater than what they spend, so what we have so far is

15x >

Greater than what, though? What do they spend? They spend 600 for the food, so

15x > 600...

but they also have to print a certain, unknown number of tickets at .50 each. The expression that represents the printing of each ticket is .5x (we can drop the 0; it doesn't change the answer or make it wrong if we drop it off). So the cost for this affair is the food + the printing.

15x > 600 + .5x

Solve this inequality for x. Begin by subtracting .5 from both sides to get

14.5x > 600 so

x > 41.3

Because we are not selling (or printing) .3 of a ticket, it's safe to say (and also correct!) that they need to sell (and print) 41 tickets. If they sell 41 tickets, the profit is found by

15(41) > 600 + .5(41)

615 > 600

This means that at 41 tickets, they make a profit. At 40 tickets, the inequality looks like this:

15(40) > 600 + .5(40) and

600 > 620. This is not true, so 40 tickets isn't enough.

8 0

2 years ago

At what values of x does f(x) = x^3 - 2x^2 -4x+1 satisfy the mean value theorwm on [0,1]

denis-greek [22]

Answer:

x=1/3

Step-by-step explanation:

A function f is given as

$f(x) = x^3-2x^2-4x+1$ in the interval [0,1]

This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)

Hence mean value theorem applies for f in the given interval

$f(1) = 1-2-4+1 = -4\\f(0) = 1$

The value

$\frac{f(1)-f(0)}{1-0} =\frac{-4-1}{1} =-5$

Find derivative for f

$f'(x) = 3x^2-4x-4$

Equate this to -5 to check mean value theorem

$3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1$

We find that 1/3 lies inside the interval (0,1)

4 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

Factor completely x^2-9x-20

Nadya [2.5K]

Answer:

(x - 10.844) (x + 1.844)

Step-by-step explanation:

5 0

2 years ago