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Studentka2010 [4]
3 years ago
9

One hundred football players were asked how many broken bones they have suffered. The probability distribution shows the probabi

lity of the number of bones broken by a football player.
What is the probability that a football player has suffered more than 3 broken bones?

Enter your answer, as a decimal, in the box.

Mathematics
1 answer:
777dan777 [17]3 years ago
6 0
The probability for a football player to suffer more than three roken bones will be the the sum of probabilities for 4 broken bones and five broken bones. Looking at the chart provided we can see that the probablity for 4 four broken bones is P(4)=0.10, and for 5 broken bones is P(5)=0.15. Thus the probaility,
P(x>3)=0.10+0.15=0.25.
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I have a Algebra 2 question <br><br> f(x) = 5x +4 and g(x) = x^2 - 5 <br><br> find g(f(4))
victus00 [196]

Answer: 571

To get this answer, we first need to compute the value of f(4). This happens when we plug x = 4 into f(x)

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f(4) = 5*4+4 ... each x replaced with 4

f(4) = 20+4

f(4) = 24

Now replace every 'x' in g(x) with f(4) like so

g(x) = x^2 - 5

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g(f(4)) = ( f(4) )^2 - 5 .... every x replaced with f(4)

g(f(4)) = ( 24 )^2 - 5 ... see note below

g(f(4)) = 576 - 5

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About 1111​% of the population of a large country is math phobicmath phobic. If two people are randomly​ selected, what is the p
yKpoI14uk [10]

Answer:

a) P(X=2) = (2C2) (0.11)^2 (1-0.11)^{2-2}= 0.0121

b) P(X=0) = (2C0) (0.11)^0 (1-0.11)^{2-0}= 0.7921

P(X \geq 1)= 1-P(X  

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem  

Part a

Let X the random variable "number of people math phobicmath phobic" , on this case we now that the distribution of the random variable is:  

X \sim Binom(n=2, p=0.11)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}

And we want this probability:

P(X=2)

If we use the probability mass function we got:

P(X=2) = (2C2) (0.11)^2 (1-0.11)^{2-2}= 0.0121

Part b

For this case we want this probability:

P(X \geq 1)= 1-P(X  

We can find first this probability

P(X=0) = (2C0) (0.11)^0 (1-0.11)^{2-0}= 0.7921

P(X \geq 1)= 1-P(X  

5 0
3 years ago
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