answer.
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
Dmitry [639]
2 years ago
9

The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me

an 8 minutes.
(a) Find the probability that the time interval between two successive barges is less than 5 minutes.
(b) Find a time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t.
Mathematics
1 answer:
lisov135 [29]2 years ago
6 0

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

You might be interested in
Show that if the vector field F = Pi + Qj + Rk is conservative and P, Q, R have continuous first-order partial derivatives, then
nekit [7.7K]

Answer: The field F has a continuous partial derivative on R.

Step-by-step explanation:

For the field F has a continuous partial derivative on R, fxy must be equal to fyx and since our field F is ∇f,

∇f = fxi + fyj + fzk.

Comparing the field F to ∇f since they at equal, P = fx, Q = fy and R = fz

Since P is fx therefore;

∂P ∂y = ∂ ∂y( ∂f ∂x) = ∂2f ∂y∂x

Similarly,

Since Q is fy therefore;

∂Q ∂x = ∂ ∂x( ∂f ∂y) = ∂2f ∂x∂y

Which shows that ∂P ∂y = ∂Q ∂x

The same is also true for the remaining conditions given

4 0
2 years ago
A rocket leaves the surface of Earth at time t=0 and travels straight up from the surface. The height, in feet, of the rocket ab
erastovalidia [21]

Answer:

Remember that:

Speed = distance/time.

Then we can calculate the average speed in any segment,

Let's make a model where the average speed at t = t0 can be calculated as:

AS(t0) = (y(b) - y(a))/(b - a)

Where b is the next value of t0, and a is the previous value of t0. This is because t0 is the middle point in this segment.

Then:

if t0 = 100s

AS(100s) = (400ft - 0ft)/(200s - 0s) =   2ft/s

if t0 = 200s

AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s

if t0 = 300s

AS(300s) = (3200ft - 400ft)/(400s - 200s) =  14ft/s

if t0 = 400s

AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s

So for the given options, t = 400s is the one where the velocity seems to be the biggest.

And this has a lot of sense, because while the distance between the values of time is constant (is always 100 seconds) we can see that the difference between consecutive values of y(t) is increasing.

Then we can conclude that the rocket is accelerating upwards, then as larger is the value of t, bigger will be the average velocity at that point.

3 0
2 years ago
What is the slope intercept form of (3,3) (7,-1)
kow [346]
Slope intercept form follows this general equation...
y=mx+b
With this information you have posted, all you can do is find slope, your m. To calculate for slope, you need to use this following equation...
m= \frac{y_2-y_1}{x_2-x_1}
Then, you pick one of your coordinate pairs to be 1 and the other to be 2. I will choose the first coordinate pair as 1, so...
m=\frac{-1-3}{7-3} =-1
You plug in -1 into your slope formula...
y=-1x+b
Now, to find your y-intercept, or b, you must plug in one of your coordinate pairs. In that equation, you may plug in the values (3,3), so you'll have...
3=-1(3)+b
Just solve for b and you will be done
5 0
2 years ago
A shower has a flow rate of 2.3 gallons per minute. If a person takes an average of 6 showers per week and the average length of
shusha [124]

Answer:

161,460 gallons are used in a year.

Step-by-step explanation:

1 shower- 2.3×15=34.5

Number of minutes spent in the shower in a week- 15×6=90

Gallons used per week- 34.5×90=3,105

There is 52 weeks in a year.

52×3,105=<u>161,460</u>

8 0
2 years ago
Read 2 more answers
in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is  

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298

The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0
2 years ago
Other questions:
  • The Jenkins family's monthly budget is shown in the circle graph. The family has a monthly income of $4,800. How much money do t
    10·1 answer
  • Ms.graves gave her class 12 minutes to read. Carrie read 5 1/2 pages in that time. At what rate, in oages per hour, did Carrie r
    6·1 answer
  • The average age three people running for election is 42. A fourth person joins the race and the average drops to 40. What is the
    15·1 answer
  • The current dividend yield on CJ's common stock is 1.89 percent. The company just paid an annual dividend of $1.56 and announced
    15·1 answer
  • Alexis, Becca, and Cindy do volunteer work at Adopt-a-Pet Animal Shelter. They worked a total of 285 hours at the shelter last s
    7·1 answer
  • A group of students is given a 10 by 10 grid to cut into individual unit squares. The challenge is to create two squares using a
    13·1 answer
  • Bartolo is running a race. He determines that if he runs at an average speed of 15 feet/second, he can finish the race in 6 seco
    11·1 answer
  • A person invests 9500 dollars in a bank. The bank pays 5.75% interest compounded quarterly. To the nearest tenth of a year, how
    5·1 answer
  • Two hot-air balloons, one red and one blue, took off at the same time from different platforms. Each began ascending at a consta
    6·1 answer
  • There are between 60 and 100 children at the sunny side holiday camp. When the children are put into pairs, there is 1 child lef
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!