What is the weight of a 48kg girl on earth round the answer to the nearest whole number

Suppose that the time to failure (in hours) of fans in a personal computer can be modeled by an exponential distribution with λ

Reptile [31]

Answer:

1. 0.0498

2. 0.8775

Explanation:

Let X represent the time of failures (in hours) of PC fans

From the question, X is exponentially distributed with the parameter λ = 0.0003

Also, it's to be noted that X is a continuous random variable

The probability distribution function of X is f(x) = λe^-(λx) for x >= 0

To solve this, we need to get the CDF (cumulative distribution function) of the above PDF

CDF

The CDF is simply the integration (integral calculus) of the PDF

Let F(x) represent the CDF

1. What proportion of the fans will last at least 10,000 hours

F(x) = 1 - e^-(λx) for x>=0

For x = at least 10000 hours and λ = 0.0003 -------- The keyword here is at least

Because P(True) + P(False) = 1 ----------- (Probability of true + Probability of false = 1),

We have

F(x>= 10000) = 1 - F(x<= 10000) -------- Notice the change in the inequality sign

This then equates to

F(x>=10000) = 1 - (1 - e^-(λx)) ---------- Open the bracket

F(x>=10000) = 1 - 1 + e^-(λx)

F(x>=10000) = e^-(λx)

F(x>=10000) = e^-(0.0003 * 10000)

F(x>=10000) = e^-(3)

F(x>=10000) = 0.049787068367863

F(x>=10000) = 0.0498 ---------- Approximated

2. What is the proportion of the fans will last at most 7000 hours

X = at most 7000 and λ = 0.0003

The keyword here is at most, so we make use of

F(x) = 1 - e^-(λx) for x>=0

F(x<=7000) = 1 - e^-(λx)

F(x<=7000) = 1 - e^-(0.0003 * 7000)

F(x<=7000) = 1 - e^-(2.1)

F(x<=7000) = 1 - 0.122456428252981

F(x<=7000) = 0.877543571747018

F(x<=7000) = 0.8775 ---------- Approximated

5 0

4 years ago

Wire coils spinning between two magnets will produce …?

klemol [59]

Answer:

B

Explanation:

I don't know but its b

6 0

2 years ago

Read 2 more answers

HELP ASAP due in 5m

Papessa [141]

Answer:

I belive it is the one at the bottom of the table visable in the picture with 27.4 radioactive age

Explanation:

8 0

3 years ago

A 65 Kg roller-blade is accelerating at 5 m/s/s across the side walk. What force would be necessary for this acceleration to occ

Neporo4naja [7]

Newton's second law states that the force applied to an object is equal to the product between the mass m of the object and its acceleration a:
$F=ma$
Using $m=65 kg$ and $a=5 m/s^2$ , we can find the value of the force applied to the roller-blade to obtain this acceleration:
$F=(65 kg)(5 m/s^2)=325 N$