If 4 people enter a bus that has 9 available seats, how many different ways can they be seated?

Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay p

garik1379 [7]

Answer:

m = $\frac{1}{20}$
μ = 20
σ = 20

The probability that a person is willing to commute more than 25 miles is 0.2865.

Step-by-step explanation:

Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.

A random variable X follows an exponential distribution with parameter m.

The decay parameter is, m.

The probability distribution function of an Exponential distribution is:

$f(x)=me^{-mx}\ ;\ m>0, x>0$

Given: The decay parameter is, $\frac{1}{20}$

X is defined as the distance people are willing to commute in miles.

The decay parameter is m = $\frac{1}{20}$ .
The mean of the distribution is: $\mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20$ .
The standard deviation is: $\sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20$

Compute the probability that a person is willing to commute more than 25 miles as follows:

$P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865$

Thus, the probability that a person is willing to commute more than 25 miles is 0.2865.

7 0

4 years ago

The sum of two numbers is 41. The larger number is one less than twice the smaller number. Find the numbers.

Arte-miy333 [17]

I hope this helps you

small a

larger 2a-1

a+2a-1=41

3a=42

a=14

2a-1=2.14-1=27

6 0

3 years ago

-3x=50 what is the answer I just need some help lol

____ [38]

Answer:

assuming that you have solve x, the the answer is x=-16.6 repeating

Step-by-step explanation:

4 0

3 years ago

Can someone smart please help me?

ollegr [7]

Hello!

For a:

How I would do this, is I would first say if all 46 animals (heads) were chickens, how many legs would there be? Each chicken has 2 legs, so 46 * 2 = 92. The total amount of legs is 96 as stated in the question, so if all of the animals were chickens, the farmer would be 4 legs short.

Now to add rabbits into the equation. Rabbits have 4 legs, and chickens have 2. You want to find the difference between the two, because as you add rabbits to the animals the farmer has, then you have to take away chickens at the same time. 4-2 = 2, so for each rabbit you replace, you add 2 legs.

Since the farmer is 4 legs short with all chickens, then you just divide that 4 by the 2 legs you add by replacing a chicken with a rabbit.

4 / 2 = 2 rabbits

So that means there are 2 rabbits. Since there are 46 heads in total, if 2 are rabbits, that means there are 44 chickens.

So there are 44 chickens and 2 rabbits.

b)

You can follow the same steps: I'm assuming all are child tickets for now:

3.05 * 100 = $305

And now you find how much money short you are.

498.6 - 305 = 193.6

Next, you find the difference in the ticket costs.

5.25 - 3.05 = 2.20

And you divide to find the number of adult tickets.

193.6 / 2.2 = 88

Since 100 tickets were sold, and 88 adult tickets were sold, that means 12 child tickets were sold.

7 0

3 years ago

The manager of a department store uses the given function to model the monthly opening inventory of down comforters at the store

tangare [24]

Answer: $d(x)=-750x^2-3750x-10500$

Step-by-step explanation:

Given: The manager of a department store uses the given function to model the monthly opening inventory of down comforters at the store, in dollars, where x represents the number of months from September.

$d(x)=-750(x^2-5x-14)$

To rewrite this hat reveals the store's opening inventory of down comforters in the month of September, multiply -750 on the bracket, we get

$d(x)=-750x^2-3750x-10500$

6 0

3 years ago

Read 2 more answers