1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vitek1552 [10]
3 years ago
9

6.81 The length of time between breakdowns of an es- sential piece of equipment is important in the decision of the use of auxil

iary equipment. An engineer thinks that the best model for time between breakdowns of a generator is the exponential distribution with a mean of 15 days. (a) If the generator has just broken down, what is the probability that it will break down in the next 21 days? (b) What is the probability that the generator will op- erate for 30 days without a breakdown?
Mathematics
1 answer:
Reptile [31]3 years ago
7 0

Answer:

a) 75.34%

b) 13.53%

Step-by-step explanation:

First, let us consider the cumulative exponential distribution

F(t) = P[T\leq t] = 1 - e^{-rt}

Where

r=\frac{1}{m} and <em>m</em> denotes the mean

Therefore, our cumulative distribution for this exercise is as it follows:  

P[T\leq t] = 1-e^{-\frac{1}{15}t }

a) According to the distribution, the first question can be calculated as it follows:

P[T\leq 21]=1-e^{-\frac{1}{15}*21 } =1-e^{-\frac{7}{5} }

P[T\leq 21]=0.7534

Hence, the probability that the piece will break down in the next 21 days is 75.34%

b) Now, take into account that we now need the probability complement for applying the cumulative distribution like this:

P[T\geq 30]=1-P[T

P[T\geq 30]=1-(1-e^{-\frac{1}{15}*30 })=e^{-2}

P[T\geq 30]=0.1353

Therefore, we have that the probability that the generator will operate for 30 days without a breakdown is 13.53%

You might be interested in
Which type of interest rate is not adjusted for different factors, including inflation?
Murrr4er [49]

Answer:

C: Nominial interest rate

Step-by-step explanation:

C

8 0
2 years ago
Which ordered pair is a solution of the equation y equals x minus 3
Rom4ik [11]
Y = x-3

Basically you just need to plug in an x value and solve for y
y = x - 3
y = (4) - 3
y = 1

In this case, the ordered pair of this equation would be (4, 1)

You can do this given any x or y coordinate

Hope this helps 
6 0
3 years ago
Consider this function rule: multiply the input by two and then subtract one to get the output. Write an equation that gives thi
rusak2 [61]

Answer:

f(x)=2x-1

Step-by-step explanation:

The output is f(x). The input is x.

The prompt says to multiply the input by 2, so 2x.

Then subtract 1 to get f(x).

Hope this helps!

7 0
3 years ago
REALLY NEED HELP
artcher [175]

Answer:

Y=-5/3x

Step-by-step explanation:

Use the slope formula and slope-intercept form y=mx+b to find the equation.

3 0
3 years ago
Can someone help me answer this pls.
Natali5045456 [20]
Y=-12

explanation multiply x by -3 to get y
4 0
3 years ago
Read 2 more answers
Other questions:
  • If four bells toll at intervals of 8,9,12 and 15 minutes respectively. If they toll at 3 pm when will they toll together
    7·1 answer
  • Whats 2/5 + 1/3 please answer quickly
    15·2 answers
  • Bob Ross has a charge card. Ross's previous monthly balance if $167.42. He made a payment of $159.15. There is no finance charge
    12·1 answer
  • Four rectangles are shown in the diagram. For which pair of rectangles is the ratio of the side lengths 1:3?
    9·1 answer
  • A stray dog ate 40 of your muffins. That was 5/6 of all them! How many are left?
    15·2 answers
  • What are 2 challenging equations that equal 100
    7·2 answers
  • i don't know if this makes sense but I have to make 8X to the 12 power a number in parentheses. I have to fill in the boxes with
    13·1 answer
  • HELP!!!
    9·1 answer
  • If 3/p = 6 and 3/q = 15 then p - q =?
    11·1 answer
  • find the length of the missing of the missing triangle. Round your answer to the nearest tenth. Show your work​
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!