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3 years ago

8x=63-x what is the solution?

Mathematics

1 answer:

Gre4nikov [31]3 years ago

3 0

Answer:

X = 7. You can confirm the answer yourself by plugging in this number. I hope I helped :)

Step-by-step explanation:

<h3>Please mark as brainliest.</h3>

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Calculate the difference between the two times.<br> 8: 15 am. or pm <br><br> 3: 30 am or pm.

Vlad1618 [11]

Answer:

5:15 pm is the answer for the question

8 0

3 years ago

At the farmers market you buy 2 pounds of blueberries for $5.00. What is the cost per pound?

Tju [1.3M]

Answer:

$2.50

Step-by-step explanation:

Divide 5 by 2.................

3 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Alborosie

Answer: $tan(V)=2.92$

Step-by-step explanation:

For this exercise you need to remember the following Trigonometric Identity:

$tan\alpha =\frac{opposite}{adjacent}$

You must observe the figure given in the exercise.

You can notice that the given triangle UVW is a Right triangle (because it has an angle that measures 90 degrees).

So, you can identify in the figure that:

$\alpha =V\\\\opposite=UW=35\\\\adjacent=VW=12$

Knowing these values, you can substitute them into $tan\alpha =\frac{opposite}{adjacent}$ :

$tan(V)=\frac{35}{12}$

Now you must evaluate:

$tan(V)=2.916$

Finally, rounding to the nearest hundreth, you get:

$tan(V)=2.92$

7 0

3 years ago

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago

Answer 1. And 2.for Brainliest

Alekssandra [29.7K]

Answer:

Step-by-step explanation:

10% discount means you pay 90% of the original cost x.

90% of x = $40.50

0.90x = $40.50

x = $40.50/0.90 = $45

:::::

total cost = 1.065·$40.50 ≅ $43.13

4 0

3 years ago