All categories

4 years ago

Determine whether a triangle can be formed with the given side lengths. 11 ft, 2 ft, 15 ft

Mathematics

2 answers:

dimulka [17.4K]4 years ago

6 0

No it can not be formed because the sum of the two smaller side lengths is not greater than the longest length

larisa86 [58]4 years ago

4 0

No, The one side would be longer than the other.

You might be interested in

The quotient of a number and 3 minus two is at least -12

Amanda [17]

X/3 - 2 greater than or equal to -12
add two
x/3 is greater than or equal to -10
now multiply both sides by 3
x is greater than or equal to -30
Hope this helped! If it did you should select my answer as brainliest!!

6 0

3 years ago

Which quadratic function only has one x-intercept?

jasenka [17]

Answer:

The answer is a m8

Step-by-step explanation:

Idk why the other kid deleted it

4 0

4 years ago

Find the value of x if a, b, and c are collinear points and b is between a and c. ab=x,bc=x 2,ac=14

Aleks04 [339]

The value of x is 7 for the given collinear points a, b, and c as b is the mid-point of ac.

According to the given question.

a, b, and c are collinear points. Which means all points a, b, and c lie in a same plane.

b is lies between a and c.

Also,

ab = x, bc = x and ac = 24.

Since, b lies in between a and c, and it is given that ab = x and bc = x.

Which means b is the mid point of ac.

Therefore,

ab + bc = ac

Substitute the values of ab, bc and ac in the above equation.

⇒ x + x = 14

⇒ 2x = 14

⇒ c = 14/2

⇒ c = 7

Therefore, the value of x is 7.

Find out more information about mid-point and collinear points here:

brainly.com/question/1593959

#SPJ4

8 0

2 years ago

BRIA is a rectangle. AN=5. NR=x-3. please solve for x, NR, and BI

Lena [83]

Answer:

x=8

NR=5 units

BI=10 units

Step-by-step explanation:

In a rectangle BRIA

AN=5 units

NR=x-3

We have to solve for x , NR and BI.

We know that

Diagonals of rectangle bisect to each other.

BI and AR are the diagonals of rectangle BRIA and intersect at point N.

AN=NR

$5=x-3$

$x=5+3=8$

$x=8$

Substitute the value of x

NR=8-3=5

By property of rectangle

BI=AR=AN+NR=5+5=10 unit

BI=10 units

7 0

3 years ago

Let X and Y have the joint density f(x, y) = e −y , for 0 ≤ x ≤ y. (a) Find Cov(X, Y ) and the correlation of X and Y . (b) Find

adoni [48]

a. I assume the following definitions for covariance and correlation:

$\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

$\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}$

Recall that

$E[g(X,Y)]=\displaystyle\iint_{\Bbb R^2}g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

where $f_{X,Y}$ is the joint density, which allows us to easily compute the necessary expectations (a.k.a. first moments):

$E[XY]=\displaystyle\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3$

$E[X]=\displaystyle\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1$

$E[Y]=\displaystyle\int_0^\infty\int_0^yye^{-y}\,\mathrm dx=2$

Also, recall that the variance of a random variable $X$ is defined by

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

We use the previous fact to find the second moments:

$E[X^2]=\displaystyle\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2$

$E[Y^2]=\displaystyle\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6$

Then the variances are

$\mathrm{Var}[X]=2-1^2=1$

$\mathrm{Var}[Y]=6-2^2=2$

Putting everything together, we find the covariance to be

$\mathrm{Cov}[X,Y]=3-1\cdot2\implies\boxed{\mathrm{Cov}[X,Y]=1}$

and the correlation to be

$\mathrm{Corr}[X,Y]=\dfrac1{\sqrt{1\cdot2}}\implies\boxed{\mathrm{Corr}[X,Y]=\dfrac1{\sqrt2}}$

b. To find the conditional expectations, first find the conditional densities. Recall that

$f_{X,Y}=f_{X\mid Y}(x\mid y)f_Y(y)=f_{Y\mid X}(y\mid x)f_X(x)$

where $f_{X\mid Y}$ is the conditional density of $X$ given $Y$ , and $f_X$ is the marginal density of $X$ .

The law of total probability gives us a way to obtain the marginal densities:

$f_X(x)=\displaystyle\int_x^\infty e^{-y}\,\mathrm dy=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^ye^{-y}\,\mathrm dx=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$

Then it follows that the conditional densities are

$f_{X\mid Y}(x\mid y)=\begin{cases}\frac1y&\text{for }0\le x$

$f_{Y\mid X}(y\mid x)=\begin{cases}e^{x-y}&\text{for }0\le x$

Then the conditional expectations are

$E[X\mid Y=y]=\displaystyle\int_0^y\frac xy\,\mathrm dy\implies\boxed{E[X\mid Y=y]=\frac y2}$

$E[Y\mid X=x]=\displaystyle\int_x^\infty ye^{x-y}\,\mathrm dy\implies\boxed{E[Y\mid X=x]=x+1}$

c. I don't know which theorems are mentioned here, but it's probably safe to assume they are the laws of total expectation (LTE) and variance (LTV), which say

$E[X]=E[E[X\mid Y]]$

$\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]$

We've found that $E[X\mid Y]=\frac Y2$ and $E[Y\mid X]=X+1$ , so that by the LTE,

$E[X]=E[E[X\mid Y]]=E\left[\dfrac Y2\right]\implies E[Y]=2E[X]$

$E[Y]=E[E[Y\mid X]]=E[X+1]\implies E[Y]=E[X]+1$

$\implies2E[X]=E[X]+1\implies\boxed{E[X]=1}$

Next, we have

$\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\dfrac{Y^2}3-\left(\dfrac Y2\right)^2\implies\mathrm{Var}[X\mid Y]=\dfrac{Y^2}{12}$

where the second moment is computed via

$E[X^2\mid Y=y]=\displaystyle\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3$

In turn, this gives

$E\left[\dfrac{Y^2}{12}\right]=\displaystyle\int_0^\infty\int_0^y\frac{y^2e^{-y}}{12}\,\mathrm dx\,\mathrm dy\implies E[\mathrm{Var}[X\mid Y]]=\frac12$

$\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\dfrac Y2\right]=\dfrac{\mathrm{Var}[Y]}4\implies\mathrm{Var}[E[X\mid Y]]=\dfrac12$

$\implies\mathrm{Var}[X]=\dfrac12+\dfrac12\implies\boxed{\mathrm{Var}[X]=1}$

5 0

3 years ago