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3 years ago

The perimeter of a triangle is 170 feet and the sides are in the ratio of 25:14:12. Find the area of the triangle.

1 answer:

6 0

Let
x,y, and z the long sides of the triangle

we know that

x+y+z=170 ft------> equation 1
x/y=25/14----> y=14x/25------> equation 2
y/z=14/12-----> equation 3
x/z=25/12----> z=12x/25------> equation 4

substitute equation 2 and equation 4 in equation 1
x+[14x/25]+[12x/25]=170------> multiply by 25 both sides
25x+14x+12x=4250
51x=4250
x=4250/51
x=250/3
y=14x/25------> y=(250/3)*(14/25)----> y=140/3
z=12y/14-----> (140/3)*12/14----> z=40

Using Heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
where s is the semiperimeter
s=170/2-----> s=85 ft
Area=√85*[85-250/3]*[85-140/3]*[85-40]
Area=9.22*[1.67]*[38.33]*[45]------> Area=26558.21 ft²

the answer is
26558.21 ft²

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According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se

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Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability 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.

50% of adult smokers started smoking before 21 years old.

This means that $p = 0.5$

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that $n = 5$ .

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

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

In which

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

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

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875$

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

$P(X \leq 4) = 1 - P(X = 5)$

In which

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

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875$

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

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

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

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