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3 years ago

Help me yall!!......

Mathematics

1 answer:

Ugo [173]3 years ago

8 0

This then allows us to see exactly how and where the subtended angle θ of a sector will fit into the sector formulas. Now we can replace the "once around" angle (that is, the 2π) for an entire circle with the measure of a sector's subtended angle θ, and this will give us the formulas for the area and arc length of that sector

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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

4 years ago

Subject differential equation<br>Day. Month.<br>Year. ..<br>(y-secx<br>) dx + tanxdy...o

Colt1911 [192]

$(y-\sec x)\,\mathrm dx+\tan x\,\mathrm dy=0$

Divide both side by $\mathrm dx$ and rearrange terms to get a linear ODE;

$\tan x\dfrac{\mathrm dy}{\mathrm dx}+y=\sec x$

Multiply both sides by $\cos x$ :

$\sin x\dfrac{\mathrm dy}{\mathrm dx}+\cos x\,y=1$

The left side can be condensed as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dx}(\sin x\,y)=1$

Integrate both sides, then solve for $y$ :

$\sin x\,y=x+C\implies\boxed{y(x)=x\csc x+C\csc x}$

8 0

3 years ago

Find the product by using box method. (x+2)(x-4) show work please

LenaWriter [7]

Attached is your solution: simply set up the box as shown, then find the product of each in individual square in the box. The solution is the sum of the products... Note that when multiplying two binomials, you'll have like terms on one diagonal.

Hope this helps.. let me know if you have questions.

3 0

3 years ago

I have picture of it

LiRa [457]

Answer:

basically multiply the prices by how much they filled up. That should give the answer.

Step-by-step explanation:

5 0

3 years ago

It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are giv

bonufazy [111]

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.9

p-value = 0.206

Since the p-value is greater than α therefore, we cannot reject the null hypothesis.

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

Step-by-step explanation:

Let σ₁² denotes the variance of night shift-workers

Let σ₂² denotes the variance of day shift-workers

State the null and alternative hypotheses:

The null hypothesis assumes that the variance of night shift-workers is equal to or less than day-shift workers.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance of night shift-workers is more than day-shift workers.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic or also called F-value is calculated using

Test statistic = Larger sample variance/Smaller sample variance

The larger sample variance is σ₁² = 38

The smaller sample variance is σ₂² = 20

Test statistic = σ₁²/σ₂²

Test statistic = 38/20

Test statistic = 1.9

p-value:

The degree of freedom corresponding to night shift workers is given by

df₁ = n - 1

df₁ = 9 - 1

df₁ = 8

The degree of freedom corresponding to day shift workers is given by

df₂ = n - 1

df₂ = 8 - 1

df₂ = 7

We can find out the p-value using F-table or by using Excel.

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.9, 8, 7)

p-value = 0.206

Conclusion:

p-value > α

0.206 > 0.05 ( α = 1 - 0.95 = 0.05)

Since the p-value is greater than α therefore, we cannot reject the null hypothesis corresponding to a confidence level of 95%

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

3 0

3 years ago

Help me yall!!......​

Help me yall!!......