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3 years ago

Is my answer correct

Mathematics

1 answer:

Katen [24]3 years ago

5 0

Answer:

Yes your answer is correct

Step-by-step explanation:

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Let Y be a random variable with a density function given by

Neporo4naja [7]

From the given density function we find the distribution function,

$F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y$

(a)

$F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)$

$\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1$

$\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}$

(b)

$F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y$

$\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2$

$\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}$

(c)

$F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})$

$\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3$

$\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}$

5 0

3 years ago

Find each measurement indicated. Round your answers to the nearest tenth.

Law Incorporation [45]

Answer:

52.08°

Step-by-step explanation:

Law of sine: $\frac{93}{25}$ = $\frac{x}{14}$

93*14=25x

1302=25x

52.08=x

5 0

3 years ago

The percentage difference from 239,900 to 251,895

Katarina [22]

Answer:

4.87805%

Step-by-step explanation:

yuh yw

4 0

2 years ago

Read 2 more answers

Futhe Mathematics <img src="https://tex.z-dn.net/?f=%28Cos%20%7B%7D%5E%7B4%7Dt%20-Sin%20%7B%7D%5E%7B4%7Dt%20%29%20%20%5Cdiv%2

Nastasia [14]

Answer:

cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

$\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t }$

We can write the numerator as ,

$\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t }$

Recall the identity ,

$\longrightarrow (a-b)(a+b)=a^2-b^2$

Using this we have ,

$\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}$

Again , as we know that ,

$\longrightarrow sin^2\phi + cos^2\phi = 1$

Therefore we can rewrite it as ,

$\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}$

Again using the first identity mentioned above ,

$\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}$

Or else we can also write it using ,

$\longrightarrow cos2\phi = cos^2\phi - sin^2\phi$

Therefore ,

$\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}$

And we are done !

$\rule{200}{4}$

Additional info :-

Derivation of cos²x - sin²x = cos2x :-

We can rewrite cos 2x as ,

$\longrightarrow cos(x + x )$

As we know that ,

$\longrightarrow cos(y + z )= cosy.cosz - siny.sinz$

So that ,

$\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x)$

On simplifying,

$\longrightarrow cos(x+x) = cos^2x - sin^2x$

Hence,

$\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }}$

$\rule{200}{4}$

7 0

2 years ago

A Crane Lifts a load of 4000N to a height of 20m in 5 seconds. Calculate the power of the Crane in horse power

Digiron [165]

Answer:

2.54379284 hp.

Step-by-step explanation: Torque is defined as the engines rotational force. First we find the distance = m in equation- then we use the formula Power = work/time = 16.7/ 5 = 3.34 and Horsepower formula is given by. Horsepower = (Torque × Speed) / 5252. HP = (3.34 × 4000) / 5252. HP = 13360/5252. HP = 2.54379284 hp.

Given : W=4000 m=? h=20 m

Time of operation t =5 s

Work done in lifting a box W=mgh= ? ×12×20=4000 = m = 4000/240=16.6666667 = 16.7KG

Power required P= W/t = w/p = 120000/120 =1000 watts We divide 60/5 = 120 then divide by 10 = 12 (being g)

3 0

3 years ago