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3 years ago

Evaluate the expression

Mathematics

1 answer:

beks73 [17]3 years ago

8 0

Answer:

Solution given:

$\frac{1 + {4}^{2} \times 3}{7}$

$\frac{1 + 16 \times 3}{7}$

$\frac{1 + 48}{7}$

$\frac{49}{7}$

7 is a required expression.

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The hypotenuse of a right triangle is 95 inches long. One leg is 4 inch(es) longer than the other. Find the lengths of the legs

diamong [38]

Answer:

65.1 and 69.1

Step-by-step explanation:

a^2+b^2=c^2

c=95

b=a+4

Solve for a^2+(a+4)^2=95^2

a=65.1

b=a+4=69.1

6 0

3 years ago

Read 2 more answers

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.

(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

4 years ago

Jonathan works at a garden center. He earns $550 per month, plus an addtional $3 for each plant that he sells. Jonathan hopes to

Wewaii [24]

Answer:

550 + 3x > 1000, where x > 150

The quantities true for x are: x must be greater than 150

Step-by-step explanation:

The question claims he wants to earn OVER 1,000 that month. He already makes $550 and $3 PER PLANT he sells. Subtract $550 from $1000 and you would get 450. Then divide by 3. That makes 150 plants. So now we know the x needs to be GREATER THAN 150 plants to make over 1,000. The x would also go next to the 3 because the 3 represents how much per plant sold, and the 150 represents how many plants. The equation would be $550 + 3x > 1000, where x > 150.

4 0

3 years ago

-2x + 3 = -3x + 2 what solution is this?

shusha [124]

Answer:

x=-1

Step-by-step explanation:

-2x+3=-3x+2

-2x+3x=2-3

x=-1

6 0

3 years ago

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when making a deposit into a finanical insitution such ad a bank this best reflects which funiction of money

ozzi

Store of value is the answer

7 0

3 years ago