P= RE^2 (R+b)^2 make R subject of the Formular

Help ASAP I need a answer please

lesya [120]

Answer:

The common difference is 10, so the function is f(n + 1) = f(n) + 10 where f(1) = 14.

Step-by-step explanation:

14 , 24 , 34 , 44 , 54

The common difference(d) is 24 - 14 = 10 , 34 - 24 = 10 , 44 - 34 = 10 , 54 - 44 = 10

f(n + 1) = f(n) + 10 where f(1) = 14.

6 0

3 years ago

Please help, really need it. ill give points

Sedaia [141]

Answer:

A=1/5

B=177147

C=1/1024

D= - 1/7

6 0

3 years ago

Read 2 more answers

What is the degree of (2x + 45)

lesya692 [45]

Answer:180 degrees

Step-by-step explanation:Let "x" be the measure of the angle the problem is asking for.

"when twice of a angle is added to 45" is the angle (2x+45) degrees.

The problem says that the angle (2x+45) is the supplement to the given angle.

It means that

x + (2x + 45) = 180 degrees. <<<---=== This equation is the exact meaning of the condition.

Simplify and solve this equation for x:

3x + 45 = 180 ====> 3x = 180 - 45 = 135

====> x = 135/3 = 45 degrees.

5 0

3 years ago

Carolyn and Stefanie want to build a fans for their front yard. They want to have a length of 4x-5 yards and a width of 5x-4. Wh

Inessa [10]

As it is perimeter you would have to multiple by 2 for both as l and w are the same on both sides 2(5x-4) + 2(4x-5)

7 0

2 years ago

A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with

Vikentia [17]

Answer:

(a) The PMF of X is: $P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let X = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable X follows a Geometric distribution.

The probability mass function of X is:

$P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....$

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of X is:

$P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

= 1 - P (X = 1) - P (X = 2)

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64$

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

$E(X)=\frac{1}{p}$

Compute the expected number of opponents contested in a game as follows:

$E(X)=\frac{1}{p}=\frac{1}{0.20}=5$

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

= 1 - P (X = 1) - P (X = 2) - P (X = 3)

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512$

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

$E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2$

Thus, the expected number of game plays until a player contests four or more opponents is 2.

4 0

3 years ago