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Nikolay [14]
3 years ago
10

At a competition with 5 runners, medals are awarded for first, second, and

Mathematics
2 answers:
dlinn [17]3 years ago
7 0

Answer:

B: Permutation;  Number of ways are 60

Step-by-step explanation:

At a competition with 5 runners, medals are awarded for first, second, and third places.

Each of the 3 medals is different.

This is a case of Permutation and the Number of ways are 60.

We have to award 3 medals among 5 runners.

This can be done in 5P3 ways.

= \frac{5!}{(5-3)!}

= \frac{5\times4\times3\times2\times1}{2\times1}

= 5\times4\times3=60

Therefore, the answer is option A.

Tpy6a [65]3 years ago
4 0

Answer:

a

Step-by-step explanation:

thanks

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34566+986994689999999
Jlenok [28]

Answer:

9.8699469e+14 and/or 986994690000000

Step-by-step explanation:


4 0
3 years ago
Unit activity: exponential and logarithmic functions
nirvana33 [79]

We will conclude that:

  • The domain of the exponential function is equal to the range of the logarithmic function.
  • The domain of the logarithmic function is equal to the range of the exponential function.

<h3>Comparing the domains and ranges.</h3>

Let's study the two functions.

The exponential function is given by:

f(x) = A*e^x

You can input any value of x in that function, so the domain is the set of all real numbers. And the value of x can't change the sign of the function, so, for example, if A is positive, the range will be:

y > 0.

For the logarithmic function we have:

g(x) = A*ln(x).

As you may know, only positive values can be used as arguments for the logarithmic function, while we know that:

\lim_{x \to \infty} ln(x) = \infty \\\\ \lim_{x \to0} ln(x) = -\infty

So the range of the logarithmic function is the set of all real numbers.

<h3>So what we can conclude?</h3>
  • The domain of the exponential function is equal to the range of the logarithmic function.
  • The domain of the logarithmic function is equal to the range of the exponential function.

If you want to learn more about domains and ranges, you can read:

brainly.com/question/10197594

3 0
1 year ago
Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose
Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

In which

x is the number of sucesses

&#10;e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Poisson mean:

\mu = 0.2

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so p = 0.181

We want to find P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671

0.671 = 67.1% probability that neither contains a missing pulse

8 0
3 years ago
PLEASE ANSWER WILL GIVE BRAINLIEST
torisob [31]

Answer:

volume of box = l*b*h

= 8/9*5/12*3/10

= 1/9 cubic feet

Step-by-step explanation:

3 0
2 years ago
What is the coefficient of 8x
IrinaK [193]

' 8 ' is the numerical coefficient.

' x ' is the literal coefficient.
4 0
2 years ago
Read 2 more answers
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