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zepelin [54]
3 years ago
14

What is the value of f(3) in the function below?

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
5 0

Answer:

A. 8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Functions
  • Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = 2ˣ

<u>Step 2: Evaluate</u>

  1. Substitute in <em>x</em> [Function f(x)]:                                                                           f(3) = 2³
  2. Evaluate:                                                                                                           f(3) = 8
tamaranim1 [39]3 years ago
5 0

Answer:

it's 8

Reason:

substitute x=3 into the function....it'd be 2³, which is 8

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On a circle of radius 5 feet, give the degree measure of the angle that would subtend an arc of length 1 feet. Round your answer
zmey [24]

Answer:

11.46°

Step-by-step explanation:

Let x be the angle that yields an arc length of 1 feet, if r= 5 feet, applying the circumference length equation and assuming that a full circumference has 360 degrees:

1=\frac{x}{360}*2\pi r \\1=\frac{x}{360}*2\pi 5\\x=\frac{360}{2 \pi 5}=11.46^o

Rounding to the nearest hundredth, the angle should be 11.46°

4 0
3 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 <em>X</em> 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 <em>X</em> = number of games played.

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

(a)

The random variable <em>X</em> follows a Geometric distribution.

The probability mass function of <em>X</em> 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 <em>X</em> 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
Which of the following is an equivalent expression to the given expression using the commutative property of addition?
olga nikolaevna [1]
I think it’s the second equation (2x)y+5x
7 0
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Read 2 more answers
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Arada [10]

Answer:

oof tough

Step-by-step explanation:

3 0
3 years ago
On day 1, i'm going to run 12 laps at the fitrec. then, for each of the next six days, i'm going to roll a six-sided die, and th
sveticcg [70]
Short answer: 36.
This is a combination/permutation problem.
To put it simple, a die has 6 sides, there are 7 days but 1 is already determined (12 laps).
So if we multiple the options (sides on a die) × the number of days (6) we get:
6 × 6 = 36 possible outcomes.

To show this we can get
12, 13, 14, 15, 16, 17, 18
12, 13, 14, 15, 16, 17, 19
12, 13, 14, 15, 16, 17, 20
... all the way to
12, 18, 24, 30, 36, 42, 48
5 0
4 years ago
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