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3 years ago

A point is represented by a single dot and

1 answer:

5 0

The most possible answer is B- an uppercase letter because a point is pointed towards the highest elevation and back to the lowest elevation making the highest elevation an uppercase and the lowest elevation an uppercase. For example, lowest elevation would be B whilst highest elevation is H. That would make it (A,H). Unless you were doing a negative point but nobody uses those these days.

Hope this helps. :)

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10 2⁄3 yards = how many feet

ollegr [7]

Answer:

10 2/3 yards = 32 feet

Step-by-step explanation:

there is 3 feet in one yard

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2 years ago

!!!!!!!!!!!!!!!!!!!!!1

Aleksandr [31]

Answer:

3x36

Step-by-step explanation:

Because he is getting 36 3 times so 3x36

By a Fourth grader

8 0

1 year ago

Read 2 more answers

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

belka [17]

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%.

4 0

2 years ago

The 5th and 6th characters of "x" in code o40.1xx0 are called __________.

Maksim231197 [3]

<span>Representation of any diagnosis, procedures and medical services in alphanumeric codes that can be used universally is called as Medical Coding. In the concepts of medical coding the 5th and 6th characters of "x" in code o40.1xx0 are called as placeholders and the characters 4, 5, 6 are called as subcategory characters.</span>

8 0

2 years ago

Any help on either of these questions would be great. Thank you

riadik2000 [5.3K]

I think problem 4 is 5

6 0

3 years ago