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OverLord2011 [107]
3 years ago
11

A line passes through the points (p, a) and (p, –a) where p and a are real numbers and p ≠ 0. Describe each of the following. Ex

plain your reasoning. slope of the line equation of the line y-intercept slope of a line perpendicular to the given line
Mathematics
1 answer:
irinina [24]3 years ago
6 0
The slope of the line that passes through (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)
given
(p,a) and (p,-a)
slope=(-a-a)/(p-p)=-2a/0=undefined
means
the slope is undefined
x=something
(x,y)
(p,a)
x=p is the equation
the equation of a perpendicular line is y=k where k is any real number
the y intercept of y=k would be y=k or the pont (0,k)
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What is the solution to this system of linear equations? x − 3y = −2 x + 3y = 16
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Answer:

so x=5 1/3 and

y= -2 2/3

Step-by-step explanation:

x-3y= -2x+3y=16. First we need to move all variables to one side of the equation and whole numbers to the other side of the equation. I see

x-3y+2x-3y=16. -3y-3y equals to -6y. 2x+x=3x. so 3x-6y=16. Lets take out the -6y so our equation would be 3x=16. x would equal 5 1/3. Now lets put back -6y into our equation. Let's now substitute x as 0. 3 times 0 equals 0 so our equation would now be -6y=16 which equals to -2 2/3.

so x=5 1/3 and

y= -2 2/3

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3 years ago
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Lee purchased groceries for $33.29. He paid with two $20 bills. How much change should Lee receive?
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Answer: $6.71

Step-by-step explanation:

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As part of a community clean-up project, several artists are painting murals on cylindrical trash cans. Each trash can has a hei
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3 years ago
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.

5 0
3 years ago
A highway engineer knows that his crew can lay 5 miles of highway on a clear day, 2 miles on a rainy day, and only 1 mile on a s
lesya692 [45]

Answer:

Mean = 3.7

Variance = 2.61

Step-by-step explanation:

From the data given; we can represent our table into table format for easier solution and better understanding.

Given that:

A highway engineer knows that his crew can lay 5 miles of highway on a clear day, 2 miles on a rainy day, and only 1 mile on a snowy day

Let X represent the crew;

P(X) represent their respective probabilities

                   clear  day           rainy day            snowy day

X                  5                          2                           1

P(X)              0.6                       0.3                       0.1

From Above; we can determine our X*P(X) and X²P(X)

Let have the two additional columns to table ; we have

X                      P(X)                      X*P(X)                       X²P(X)

5                        0.6                          3                               15

2                        0.3                        0.6                             1.2

1                         0.1                          0.1                            0.1

Total                  1.0                        3.7                             16.3

The mean \mu can be calculated by using the formula:

\sum \limits ^n _{i=1}X_i P(X_i)

Therefore ; mean \mu = 3.7

Variance \sigma^2 = \sum \limits ^n _{i=1}X^2_i P(X_i)- \mu^2

Variance = 16.3 -3.7²

Variance = 16.3 - 13.69

Variance = 2.61

4 0
4 years ago
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