Richard is buying a subscription for video game rentals. The plan he has chosen has aninitial fee of $20 plus $2 per video game rented. This plan can be represented by the
function f(x) = 2x + 20. How much money will Richard pay this month if he rents 5 video
games?

Answers

Answer 1

Answer:

Answer:

Richard will pay $30.

Step-by-step explanation:

Because "x" is equivalent to the amount of video games he rents, you would replace "x" with 5. Do the math, and you would get 10+20=30! Hope this helps!

Answer 2

Answer: he will pay $30 this month

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The slope of the line below is -2. Use the coordinates of the labeled point tofind a point-slope equation of the line.

5. What is the value of x if the quadrilateral is a kite? B X+2 C С 13 A Xth12 D

Answers

Answer: just had this problem! X = 11

Which of the following expressions are equivalent to - -a/b

Answers

Answer:

B) $-(a)/(-b)$

Step-by-step explanation:

1. First, we have to know that two negatives equal positive.

2. Given the information above, $-(-a)/(b)$ can simplify to $(a)/(b)$ .

3. Let's go through each answer choice and see which one also simplifies to $(a)/(b)$ .

$(a)/(-b)$
$-(a)/(b)$

$-(a)/(-b)$
$(a)/(b)$

Therefore, the answer is B) $-(a)/(-b)$ .

A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is no net force on the system, what is the velocity of the pitcher? −0.1 m/s −0.2 m/s −0.7 m/s −1.4 m/s

Answers

Answer:

The velocity of the pitcher is −0.1 m/s

Step-by-step explanation:

Given : Mass of pitcher = 50 kg

Mass of Baseball= 0.15kg

Velocity of Ball = 35m/s

To Find : velocity of the pitcher

Solution :

The total momentum of the system is conserved when no external force acts on a system .The total initial momentum of the system is equal to the total final momentum of the system.

Since , the ball and the pitcher are initially at rest, therefore, the total initial momentum of the system is zero.

Since we are given that no external forces act on the system , the total final momentum of the system is also equal to zero.

Let us suppose the mass of the pitcher is $m_(p)$

Speed of pitcher = $v_(p)$

The mass of the ball is $m_(b)$

Speed of ball = $v_(b)$

So, the final momentum of the system of pitcher and the ball is given by:

$momentum =m_(p) v_(p) +m_(b) v_(b) =0$

⇒ $50* v_(p) +0.15*35 =0$

⇒ $50* v_(p) +5.25 =0$

⇒ $v_(p) = (-5.25)/(50)$

⇒ $v_(p) = -0.105$

Thus , The velocity of the pitcher is -0.105m/s≈−0.1 m/s

Negative sign shows the opposite direction.

Hence The velocity of the pitcher is −0.1 m/s

the answer is -0.1 m/s if you're looking for the Edgnuity answer. I just took the test. Hope this helps!

A fair die is rolled 8 times. What is the probability that the die comes up 6 exactly twice? What is the probability that the die comes up an odd number exactly five times? Find the mean number of times a 6 comes up. Find the mean number of times an odd number comes up. Find the standard deviation of the number of times a 6 comes up. Find the standard deviation of the number of times an odd number comes up.

Answers

Answer:

0.2605, 0.2188, 1.33, 4, 1.0540, 1.4142

Step-by-step explanation:

A fair die is rolled 8 times.

a. What is the probability that the die comes up 6 exactly twice?

b. What is the probability that the die comes up an odd number exactly five times?

c. Find the mean number of times a 6 comes up.

d. Find the mean number of times an odd number comes up.

e. Find the standard deviation of the number of times a 6 comes up.

f. Find the standard deviation of the number of times an odd number comes up.

a. A die is rolled 8 times. If A represent the number of times a 6 comes up. For a fair die the probability that the die comes up 6 is 1/6 - Thus A ~ Bin(8, 1/6)

The probability mass function of the random variable A is

$p(A) = \left \{ {(8!)/(x!(8 - x)!)*((1)/(6) )^(A)*((5)/(6) )^(8-A) } \right. for A=0,1, ...8$

hence, p(6 twice) implies P(A=2)

that is P(2) substitute A = 2

$p(2) = \left \{ {(8!)/(2!(8 - 2)!)*((1)/(6) )^(2)*((5)/(6) )^(8-2) } \right. for A=0,1, ...8$

$p(2)=(8!)/(2!6!) *((1)/(6) )^(2) *((5)/(6) )^(6)$

p(2) = 0.2605

b. If B represent the number of times an odd number comes up. For the fair die the probability that an odd number comes up is 0.5.

Thus B ~ Bin(8, 1/2 )

The probability mass function of the random variable B is given by

$p(B) = \left \{ {(8!)/(B!(8 - B)!)*((1)/(2) )^(B\n)*((1)/(2) )^(8-B) } \right. for B=0,1, ...8$

hence p(odd comes up 5 times) is

$p(x=5) = p(2)=(8!)/(5!3!) *((1)/(2) )^(5) *((1)/(2) )^(3)$

p(5) = 0.2188

c. let the mean no of times a 6 comes up be μₐ

and let the total number of outcomes be n

using the formula μₐ = nρₐ

μₐ = 8 * 1/6

= 1.33

d. let the mean nos of times an odd nos comes up beμₓ

let the total outcomes be n = 8

let the probability odd be pb = 1/2

μₓ = npb

= 8 * (1/2)

= 4

e. the standard deviation of a random variable A is given as follows

σₐ $= √(np(1-p))$

where p = 1/6 (prob 6 outcome)

n = total outcomes = 8

$= \sqrt{8*(1)/(6)*(5)/(6) }$

= 1.0540

f. the standard dev of the binomial random variable Y is given by

σ $= √(np(1-p))$

where p = 1/2 and n = 8

= $\sqrt{8*(1)/(2) *(1)/(2) }$

= 1.4142

Equations
Solve T = C(9+ AB) for B

Answers

Answer:

Simplifying

T = C(9 + AB) * forB

Reorder the terms for easier multiplication:

T = C * forB(9 + AB)

Multiply C * forB

T = forBC(9 + AB)

T = (9 * forBC + AB * forBC)

Reorder the terms:

T = (forAB2C + 9forBC)

Solving

T = forAB2C + 9forBC

Solving for variable 'T'.

Move all terms containing T to the left, all other terms to the right.

Simplifying

T = forAB2C + 9forBC

Step-by-step explanation:

Simplifying

T = C(9 + AB) * forB

Reorder the terms for easier multiplication:

T = C * forB(9 + AB)

Multiply C * forB

T = forBC(9 + AB)

T = (9 * forBC + AB * forBC)

Reorder the terms:

A small regional carrier accepted reservations for a particular flight with 17 seats. 14 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 52% chance.A. Find the probability that overbooking occurs. B. Find the probability that the flight has empty seats.