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makvit [3.9K]
3 years ago
13

let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

endent? prove your answere?
Mathematics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer with Step-by-step explanation:

We are given that

x_1,x_2 and x_3 are linearly independent.

By definition of linear independent there exits three scalar a_1,a_2 and a_3 such that

a_1x_1+a_2x_2+a_3x_3=0

Where a_1=a_2=a_3=0

y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1

We have to prove that y_1,y_2 and y_3 are linearly independent.

Let b_1,b_2 and b_3 such that

b_1y_1+b_2y_2+b_3y_3=0

b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0

b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0

(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0

b_1+b_3=0

b_1=-b_3...(1)

b_1+b_2=0

b_1=-b_2..(2)

b_2+b_3=0

b_2=-b_3..(3)

Because x_1,x_2 and x_3 are linearly independent.

From equation (1) and (3)

b_1=b_2...(4)

Adding equation (2) and (4)

2b_1==0

b_1=0

From equation (1) and (2)

b_3=0,b_2=0,b_3=0

Hence, y_1,y_2 and y_3 area linearly independent.

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<h2>Explanation:</h2><h2></h2>

In this problem, we know that Harry saved $100 each week for 8 weeks. In other words, he saved a total amount of money:

\text{Amount of money saved}=8\times 100=\$800

We know that he earned $48 on his savings of $800, so for every $100 the interest (I) he earns is:

I=\frac{48}{8}=\$6

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4 years ago
A Family purchased tickets to the movies in spent a total of $40.75. The family purchase five tickets there was a $.90 processi
muminat

Answer:

$7.25

Step-by-step explanation:

Total amount spent on the ticket   = $40.75

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Processing fee per ticket  = $0.90

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Cost of each ticket = ?

Solution:

To solve this problem, we need to find the processing fee for the 5 tickets purchase;

      Processing fee = number ticket x processing fee per ticket

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Now, the real cost a ticket = $40.75  - $4.5  = $36.25

So, cost per ticket is;

        Cost  = \frac{36.25}{5}   = $7.25

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One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo
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Answer:

(1) P(\bar X < 26 inches) = 0.0436

(2) P(27.5 inches < \bar X < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let \bar X = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

            Z = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean vertical  leap = 28 inches

            \sigma = standard deviation = 7 inches

            n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P(\bar X < 26 inches)

   P(\bar X < 26) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{26-28}{\frac{7}{\sqrt{36} } } ) = P(Z < -1.71) = 1 - P(Z \leq 1.71)

                                                 = 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < \bar X < 28.5 inches) = P(\bar X < 28.5 inches) - P(\bar X \leq 27.5 inches)

    P(\bar X < 28.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{28.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z < 0.36) = 0.64058 {using z table}                      

    P(\bar X \leq 27.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{27.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z \leq -0.36) = 1 - P(Z < 0.36)

                                                        = 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < \bar X < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

6 0
3 years ago
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