All categories

3 years ago

Use the quadratic formula to find the solutions to the equation. 3x2 - 10x + 5 = 0

Mathematics

2 answers:

krok68 [10]3 years ago

4 0

Answer:

The answer is A 10 +/ $\sqrt{40}$ /6

Step-by-step explanation:

I just took the test

Nitella [24]3 years ago

3 0

Answer: A

Step-by-step explanation:

3x² - 10x + 5 = 0

a = 3 b = -10 c = 5

Δ = b²- 4ac = 100 - 4.3.5 = 40 > 0 => have 2 solutions

=> x = $\frac{10+\sqrt{40} }{6}$ or x = $\frac{10-\sqrt{40} }{6}$

You might be interested in

Pls answer i will give brainliest pls

Gelneren [198K]

Answer:

Step-by-step explanation:

In the slope intercept equation of the line the number that not multiplies the x or the y indicates where the line intercept the y axis

so in this cas the line intercept the y axis in the point (0,2)

so the answer is B

5 0

3 years ago

Please help with the equation

Anastaziya [24]

Answer:

The distance is dependent

Step-by-step explanation:

Time is always the dependent variable because time will always stay the same when other things change, such as distance.

3 0

3 years ago

Please help <br>I will give brainlist

beks73 [17]

-3 it’s the rise over run I believe & because it’s going down it’s negative

6 0

2 years ago

Read 2 more answers

What is a linear system?

Darya [45]

Answer:

Linear systems are systems of equations in which the variables are never multiplied with each other but only with constants and then summed up

8 0

3 years ago

Read 2 more answers

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