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2 years ago

Problem 2 (4 points) The Better Business Bureau conducts a survey of

2 answers:

8 0

The probability that all three would recommend their attorney to a friend is 18.19%, and the probability that the first would not, but the second and third would recommend their attorney to a friend is 8.56%.

<h3><u>Probabilities</u></h3>

Since the Better Business Bureau conducts a survey of 30 people who recently hired an attorney, and 17 of them say that they would recommend their attorney to a friend, while 8 say that they would not recommend their attorney to a friend, and the remaining 5 say that they are not sure, and then three people are selected from this group of 30 people at random, to determine what is the probability that A) all three would recommend their attorney to a friend, and B) the first would not, but the second and third would recommend their attorney to a friend, the following calculation must be performed:

17/30 x 17/30 x 17/30 = X
0.5666 x 0.5666 x 0.5666 = X
0.1819 = X

8/30 x 17/30 x 17/30 = X
0.2666 x 0.5666 x 0.5666 = X
0.0856 = X

Therefore, the probability that all three would recommend their attorney to a friend is 18.19%, and the probability that the first would not, but the second and third would recommend their attorney to a friend is 8.56%.

Learn more about probabilities in brainly.com/question/24217562

8_murik_8 [283]2 years ago

5 0

Probabilities are used to determine the chances of events

The probability that all would recommend their attorney to a friend is 0.1675
The probability the first would not, but the second and third would recommend their attorney to a friend is 0.0893

The given parameters are:

n = 30 -- the sample size

Recommend = 17

Not recommend = 8

Not sure = 5

The probability that all would recommend their attorney to a friend is calculated as:

$p = \frac{17}{30} * \frac{16}{29} * \frac{15}{28}$

Evaluate the product

$p = 0.1675$

The probability the first would not, but the second and third would recommend their attorney to a friend is calculated as:

$p = \frac{8}{30} * \frac{17}{29} * \frac{16}{28}$

$p = 0.0893$

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3 years ago

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Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

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Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

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2 years ago

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son4ous [18]

Answer:

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Step-by-step explanation:

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2 years ago

Vincent and Aidan buy the same number of pens. Vincent buys packs of 6 pens, and Aidan buys boxes of 20 pens. What is the least

Shtirlitz [24]

The least number of pens each boy can have brought is 60

<h3>How to determine the least number of pen?</h3>

We have:

Vincent = 6 packs

Aidan = 20 boxes

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20 = 1 * 2 * 2 * 5

Multiply all factors

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Evaluate the product

LCM = 60

Hence, the least number of pens each boy can have brought is 60