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3 years ago

Solve for z : 4z-6/4=3z+1

Mathematics

1 answer:

ruslelena [56]3 years ago

7 0

Note the equal sign, what you do to one side, you do to the other side. Isolate the variable z.

First, subtract 3z and add 6/4 to both sides

4z (-3z) - 6/4 (+6/4) = 3z (-3z) + 1 (+6/4)

4z - 3z = 1 + 6/4

Simplify. Note that 1 = 4/4.

z = 4/4 + 6/4

Simplify. Combine like terms

z = 10/4

Simplify. Change the improper fraction into mixed fraction and simplify.

z = 10/4 = 2 2/4 = 2 1/2

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igomit [66]

Answer:

x = 18√3

Step-by-step explanation:

From the picture attached,

In a right triangle,

Measure of one angle = 30°

Measure of adjacent side of the angle = 9 units

We have to find the measure of Hypotenuse from the given properties.

By cosine rule,

cos(30°) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\frac{\sqrt{3}}{2} =\frac{9}{x}$

$x=18\sqrt{3}$

Therefore, measure of Hypotenuse is 18√3 units.

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

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

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3 years ago

Brandon is shopping at Old Navy during a sale. All shorts are $16 each and all t-shirts are $10 each.He has $100 to spend and wo

timama [110]

Old Navy sucks btw and Brendon is poor

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4 years ago

en una fiesta se cómo aparte un pastel y al final solo quedan 2/5 del mismo ¿si Andrés se come 1/4 de lo que quedó que fracción

Morgarella [4.7K]

Answer: 2/125

Step-by-step explanation: No esto 100% seguro pero hice esto:

2/5 % 4 = 2/125

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3 years ago

Help pls lol it’s due today:)))

Andrei [34K]

Answer:

4. supplementary

5. complementary

Step-by-step explanation:

4 is supplementary because the sum of the two angles is 180°.

5 is complementary because the sum of the two angles is 90°.

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3 years ago

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