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Serggg [28]
3 years ago
6

The cost of renting a fishing boat is $25 per hour for the first 7 hours, $10 per hour for the next 10 hours, and $5 per hour fo

r any additional hours. Identify the cost of renting the boat for 30 hours.
Mathematics
2 answers:
yKpoI14uk [10]3 years ago
5 0

Answer: $340

Step-by-step explanation:

To identify the cost of renting the boat for 30 hours, breakdown the 30hours into specific renting hours

i.e 30 hours = (7 hours + 10 hours + 13 hours)

Then calculate each of the renting cost based on the hour breakdown given above

- Since the Cost of renting a fishing boat is $25 per hour for the first 7 hours, then (25 x 7 hours = $175)

- For the $10 per hour for the next 10 hours, then (10 x 10 hours = $100)

- and for $5 per hour for any additional hours. (5 x 13 additional hours = $65)

Now, sum up all of the cost of the hourly breakdown

i.e 30 hours = (7 hours + 10 hours + 13 hours)

= $175 + $100 + $65

= $340

Thus, the cost of renting the boat for 30 hours is $340

Tom [10]3 years ago
4 0

Answer:

Step-by-step explanation:

The cost of renting a fishing boat is $25 per hour for the first 7 hours. This means that the cost of renting it for the first 7 hours is

25 × 7 = $175

It costs $10 per hour for the next 10. hours. This means that the cost of renting it for the next 10 hours is

10 × 10 = $100

Since the total number of hours for which the boat will be rented is 30, then the number of additional hours is

30 - (7 + 10) = 13 hours

It costs $5 per hour for any additional hours. The cost of renting it for 13 additional hours is

13 × 5 = $65

The total cost of renting the boat for 30 hours is

175 + 100 + 65 = $340

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A random variable X with a probability density function () = {^-x > 0
Sliva [168]

The solutions to the questions are

  • The probability that X is between 2 and 4 is 0.314
  • The probability that X exceeds 3 is 0.199
  • The expected value of X is 2
  • The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

P(x) = \int\limits^a_b {f(x) \, dx

So, we have:

P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx

Using an integral calculator, we have:

P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2

Expand the expression

P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx

Using an integral calculator, we have:

P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3

Expand the expression

P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

E(x) = \int\limits^a_b {x * f(x) \, dx

So, we have:

E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx

This gives

E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx

Using an integral calculator, we have:

E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx

This gives

E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx

Using an integral calculator, we have:

E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

brainly.com/question/15318348

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<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

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d1i1m1o1n [39]

Answer:

a. OM is congruent to ON.

Step-by-step explanation:

To use the HL Theorem, you must have a congruent hypotenuse and a congruent leg. In this case, you have two congruent hypotenuses. You just need to find two congruent legs.

a. OM is congruent to ON. This says that two legs are congruent, so this is your answer.

b. LM is congruent to ML. This does not help as it is saying a segment is the same as the same segment.

c. and d. Both of these use angle measurements, which does not help with the HL Theorem.

Hope this helps!

7 0
3 years ago
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