The degree of the polynomial function f(x) is 3.

Find the equation of a line that has a slope of 1/2 and passes through the origin.

omeli [17]

Step-by-step explanation:

The equation of a line with slope m and y-intercept b is y = m x + b.

This line goes through the point ( 0 ,0 ), so the y-intercept is zero.

We are given the slope as − 2 / 3.

The equation of the line is y= – 2 / 3 x + 0 = – 2 / 3 x

y = – 2 / 3 x

8 0

2 years ago

We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent

Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(c) P(X = −2) = 1/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

(c) P(Y = 2) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

( X = -2 ): { X_2 = 1 , X_1 = 3 }

P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }

P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 3 / 9 ) = ( 1 / 3 )

( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

P ( X = 1 ): P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 2 ): { X_2 = 1 , X_1 = 3 }

P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

- The distribution Y = X_2,

P(Y=0) = 0

P(Y=1) = 1/3

P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

7 0

3 years ago

I need help plz do each problem

Aleks [24]

1. 4n + 28
2. 30-10x
3. 3m+3n
4. 8+2j

3 0

2 years ago

Find the value of x. The angle measures of a right triangle are <br>30°<br>2x°<br>3x°

Radda [10]

Answer:

30

Step-by-step explanation:

Exterior angle equals sum of opposite interior angles

3x = 2x + 30

3x - 2x = 30

x = 30

3 0

2 years ago

Read 2 more answers

A uniform bar of mass $m$ and length $l$ is suspended on a frictionless hinge. A horizontally launched blob of clay of mass $m$

Irina-Kira [14]

Answer:

conservation of angular momentum ; conservation of energy

Step-by-step explanation:

The complete Question is given as follows:

" A uniform bar of mass (m) and length (L) is suspended on a frictionless hinge. A horizontally launched blob of clay of mass (m) strikes the bottom end of the bar and sticks to it. After that, the bar swings upward. What is the minimum initial speed (v) of the blob of clay that would enable the rod to swing a full circle? Which concepts/laws would be most helpful in solving this problem? Select the best answer from the options below. "

CHOICES

kinematics of rotational motion; conservation of energy

conservation of momentum ; conservation of energy

conservation of angular momentum ; conservation of momentum

conservation of angular momentum ; conservation of energy

conservation of energy ; Newton's laws

Newton's laws ; conservation of angular momentum

conservation of angular momentum ; kinematics of rotational motion

Newton's laws, kinematics of rotational motion

Solution:

- We will apply the conservation of angular momentum M. Note the linear momentum does not remains conserved as the rod stores some energy as the clay sticks to the rod:

M_i = M_f

- Initially the rod was at rest and clay had velocity of v, then M_i can be written as:

M_i = m*v*L

- The final momentum is the combined effect of clay and rod:

M_f = ( m*L^2 + I_rod )*w

- Where w is the angular speed of the rod after impact. And I_rod is the moment of inertia of rod.

I_rod = mL^2 / 3

M_f = ( m*L^2 + m*L^2 /3 )*w = (4*m*L^2 / 3)*w

- Formulate w in terms of initial velocity v:

m*v*L = (4*m*L^2 / 3)*w

0.75*v / L = w

- The minimum amount of velocity required would be enough to complete half of a circle.

- Apply conservation of Energy principle:

T_i + V_i = T_f + V_f

Where, T is the kinetic energy soon after impact and at top most position.

Assuming, T_f = 0 , for minimum velocity required to complete on circle.

T_i = 0.5*I_combined*w^2

Where, I_combined = I_clay + I_rod = 4*m*L^2 / 3

And w = 0.75*v / L:

T_i = 0.5*[4*m*L^2 / 3]*[0.75*v / L]^2

T_i = 0.5*[m]*[v^2]

Also, V is the potential energy of the clay plus rod system soon after impact and at top most point.

V_i = 0 ( Datum )

V_f = V_rod + V_clay

V_f = m*g*L + m*g*2L = 3*m*g*L

- Plug in the expressions in the energy balance and we get:

0.5*[m]*[v^2] + 0 = 0 + 3*m*g*L

v_min = sqrt ( 6*g*L)

- So the choices used were:

conservation of angular momentum ; conservation of energy

6 0

3 years ago