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konstantin123 [22]
2 years ago
9

Based on data from a statistical abstract, only about 16% of senior citizens (65 years old or older) get the flu each year. Howe

ver, about 30% of the people under 65 years old get the flu each year. In the general population, there are 14% senior citizens (65 years old or older). (Round your answers to three decimal places.)
What is the probability that a person selected at random from the general population is senior citizen who will get the flu this season?
Mathematics
2 answers:
yarga [219]2 years ago
4 0

Answer:

Probability that a person selected at random from the general population is senior citizen who will get the flu this season = 0.0224 .

Step-by-step explanation:

We are given that in the general population, there are 14% senior citizens (65 years old or older).

Let Probability of senior citizens (65 years old or older) in the population, P(S) = 0.14

Probability of citizens under 65 years old in the population, P(NS) = 1 - 0.14 = 0.86

Now, let event F = citizens getting flu each year

Probability of senior citizens (65 years old or older) getting flu each year, P(F/S) = 0.16

Probability of citizens under 65 years getting flu each year, P(F/NS) = 0.30

So, probability that a person selected at random from the general population is senior citizen who will get the flu this season = Probability that person is senior citizen(65 years old or older) * Probability of senior citizens (65 years old or older) getting flu each year

= P(S) * P(F/S) = 0.14 * 0.16 = 0.0224.

SSSSS [86.1K]2 years ago
3 0

Answer: The required probability is 2.24%.

Step-by-step explanation:

Since we have given that

Percentage of people are senior citizens ( 65 years old or older ) = 14%

Percentage of people are under 65 years old = 100-14 = 86%

Probability that senior citizens get the flu each year = 16%

Probability that under 65 years old get the flu each year = 30%

So, Probability that a person selected at random from the general population is senior citizen who get the flu this season is given by

P(S\cap F)={P(F|S)\times P(S)\\\\P(S\cap F)=0.16\times 0.14\\\\P(S\cap F)=0.0224\\\\P(S\cap F)=2.24\%Hence, the required probability is 2.24%.

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Answer:

Except k=7, any real number for k would cause the system of equations to have no solution.

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In general a system of equations can be represented as ax+by=c and dx+ey=f. In order this system of equations to have NO SOLUTIONS a/d=b/a≠c/f. In our example a=6, b=4, c=14, d=3, e=2 and f=k. To apply the formula above, 6/3=4/2≠14/k. Hence k≠7. It can be concluded that except k=7, any real number for k would cause the system of equations to have no solutions.

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(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho
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Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

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