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2 years ago

Need help ASAP!!!( Algebra 1)

Mathematics

1 answer:

Sphinxa [80]2 years ago

5 0

Answer:

[negitive]56 plus positive 56+15=-15

Step-by-step explanation:

i explained it up there

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At which value will the graph of y=tanx have a zero? Please explain the correct answer!

Rina8888 [55]

Answer: None of the above are correct.

Step-by-step explanation:

Given function: $y=\tan x$

We know that the value of tan x =0 at x=0.

Also, $At\ x=\frac{\pi}{3},\ \tan\frac{\pi}{3}=\sqrt{3}$

$At\ x=\frac{\pi}{2},\ \tan\frac{\pi}{2}=\infty$

$At\ x=\frac{\pi}{3},\ \tan\frac{\pi}{4}=1$

Hence, the only option is correct is "None of the above are correct".

4 0

3 years ago

Id : 688 737 6744 pas : c8SNRa

Travka [436]

Answer:

ill joinn

Step-by-step explanation:

thank you

4 0

2 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

What does a represent? A.foci B. Minor axis C. Major axis S. Vertices.

bulgar [2K]

C) Major axis (because its facing your upper right)

8 0

2 years ago

A bacterial culture grows exponentially according to the function n(t)=n(e^n), where n(t) is the quantity after t hours and n is

mart [117]

Answer:

$n(t=2) = 2 e^{1.386294361(2)}=32 grams$

So then after 2 hours we will have 32 grams.

Step-by-step explanation:

For this case we have the followin exponential model:

$n(t) = n e^{rt}$

n(t) is the quantity after t hours, n is the original quantityand t represent the hours and r the rate constant.

For this case we know that n(0) = 2 grams and n(3) = 128 grams and we want to find n(2)=?

From the initial condition we know that n = 2, and we have the model like this:

$n(t) = 2 e^{rt}$

Now if we apply the other conditionn(3) = 128 we got:

$128 = 2 e^{3r}$

If we divide both sides by 2 we got:

$64= e^{3r}$

If we apply natural log for both sides we got:

$ln(64) = 3r$

$r = \frac{ln(64)}{3}=1.386294361$

And our model is this one:

$n(t) = 2 e^{1.386294361t}$

And if we replace t = 2 hours we got:

$n(t=2) = 2 e^{1.386294361(2)}=32 grams$

So then after 2 hours we will have 32 grams.

5 0

2 years ago