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3 years ago

L (-1,-1)

2 answers:

8 0

The answer is B) (2,-1).If you don’t believe me punch in the coordinates into a graphing calculator and then punch in (2,-1). It will give you a perfect parallelogram.

zalisa [80]3 years ago

6 0

Answer: The correct option is (B) (2, -1).

Step-by-step explanation: Given that LMNO is a parallelogram where the co-ordinates of the vertices L,M and N are

L(-1,-1) , M(0,0) and N(3,0).

We are to find the co-ordinates of the vertex O.

Let, (a, b) be the co-ordinates of the vertex O.

Since LMNO is parallelogram, so the opposite sides will be parallel.

That is, LM is parallel to NO. So, we have

$\textup{slope of LM}=\textup{slope of NO}\\\\\Rightarrow \dfrac{0-(-1)}{0-(-1)}=\dfrac{b-0}{a-3}\\\\\\\Rightarrow \dfrac{1}{1}=\dfrac{b}{a-3}\\\\\Rightarrow b=a-3\\\\\Rightarrow a=b+3~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and MN is parallel to OL. So,

$\textup{slope of MN}=\textup{slope of OL}\\\\\\\Rightarrow \dfrac{0-0}{3-0}=\dfrac{-1-b}{-1-a}\\\\\\\Rightarrow 0=\dfrac{b+1}{a+1}\\\\\\\Rightarrow b+1=0\\\\\Rightarrow b=-1.$

Therefore, from equation (i), we get

$a=-1+3=2.$

Thus, the co-ordinates of vertex O are (2, -1).

Option (B) is correct.

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1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

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3 years ago

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alexdok [17]

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Rashid [163]

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