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4 years ago

20 points !!!

Mathematics

2 answers:

nekit [7.7K]4 years ago

6 0

Answer:

Since it is an isoceles triangle...both the base angles will be equal to 40 degrees

by angle sum property

Third angle + 40+40=180

The third angle will be equal to 100 degrees

Step-by-step explanation:

hope it helps..

77julia77 [94]4 years ago

5 0

Answer:

100

Step-by-step explanation:

We will denote the missing angle as x.

As the base angles of an isosceles triangles are equal, so it would look like this:

x + 40 + 40 = 180

x + 80 = 180

x = 180 - 80

x = 100

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Based on aâ poll, among adults who regret gettingâ tattoos, 18â% say that they were too young when they got their tattoos. Assum

Sergeeva-Olga [200]

Answer:

a) 20.44% probability that none of the selected adults say that they were too young to get tattoos.

b) 35.90% probability that exactly one of the selected adults says that he or she was too young to get tattoos.

c) 56.34% probability that the number of selected adults saying they were too young is 0 or 1.

d) No

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they say they were too young when they got their tattoos, or they don't say that. Each adult is independent of each other. So we use the binomial probability distribution 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.

18% say that they were too young when they got their tattoos.

This means that $p = 0.18$

Eight adults who regret getting tattoos are randomly selected

This means that $n = 8$

a. Find the probability that none of the selected adults say that they were too young to get tattoos.

This is P(X = 0).

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

$P(X = 0) = C_{8,0}.(0.18)^{0}.(0.82)^{8} = 0.2044$

20.44% probability that none of the selected adults say that they were too young to get tattoos.

b. Find the probability that exactly one of the selected adults says that he or she was too young to get tattoos.

This is P(X = 1).

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

$P(X = 1) = C_{8,1}.(0.18)^{1}.(0.82)^{7} = 0.3590$

35.90% probability that exactly one of the selected adults says that he or she was too young to get tattoos.

c. Find the probability that the number of selected adults saying they were too young is 0 or 1.

Either a. or b.

20.44 + 35.90 = 56.34

56.34% probability that the number of selected adults saying they were too young is 0 or 1.

d. It we randomly select 9 adults. Is 1 a significantly low number who day that they were too young to get tattoos?

Now $n = 9$

It is significantly low if it is more than 2.5 standard deviations below the mean.

The mean is $E(X) = np = 9*0.18 = 1.62$

The standard deviation is $\sqrt{V(X)} = \sqrt{n*p*(1-p)} = \sqrt{9*0.18*0.82} = 1.15$

1 > (1.62 - 2.5*1.15)

So the answer is no.

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