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gladu [14]
3 years ago
7

Mark the statement either true (in all cases) or false (for at least one example). If false, construct a specific example to sho

w that the statement is not always true. Such an example is called a counterexample to the statement. If the statement is true, give a justification. If v1,…,v4 are in R4 and {v1,v2,v3} is linearly dependent then {v1,v2,v3,v4} is also linearly dependent.
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

(A) True. Because v3 = 2v1 + v2, v4 must be the zero vector. Thus, the set of vectors is linearly dependent.
(B) True. The vector v3 is a linear combination of v1 and v2, so at least one of the vectors in the set is a linear combination of the others and set is linearly dependent.
(C) True. If c1 =2, c2 = 1, c3 = 1, and c4 = 0, then c1v1 + ........... + c4v4 =0. The set of vectors is linearly dependent.
(D) False. If v1 = __, v2 =___, v3 =___, and v4 = [1 2 1 2], then v3 = 2v1 + v2 and {v1, v2, v3, v4} is linearly independent.
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0

Answer with Step-by-step explanation:

We are given that v_1,v_2,..,v_4 are in R^4 and v_1,v_2,v_3 is linearly dependent then {v_1,v_2,v_3,v_4}[/tex] is also linearly dependent.

We have to find that given statement is true or false.

Dependent vectors:Dependent vectors are those vectors in which atleast one vector is  a linear combination of other given vectors.

Or If we have vectors x_1,x_2,....x_n

Then their linear combination

a_1x_1+a_2x_2+.....+a_nx_n=0

There exist at least one scalar which is not zero.

If v_1,v_2,v_3 are dependent vectors then

a_1v_1+a_2v_2+a_3v_3=0 for scalars a_1,a_2,a_3

Then , by definition of dependent  vectors

There exist a vector which is not equal to zero

If vector v_3 is a linear combination of v_1\;and \;v_2, So at least one of vectors in the set is a linear combination of others and the set is linearly dependent.

Hence, by definition of dependent vectors

{v_1,v_2,v_3,v_4} is linearly dependent.

Option B is true.

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The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter
JulsSmile [24]

Answer:

1) n=114

2) n=59

3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by \alpha=1-0.80=0.2 and \alpha/2 =0.1. And the critical value would be given by:

z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28

Part 1

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.06 or 6% points, and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Since we don't have a prior estimate of \het p we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.06}{1.28})^2}=113.77  

And rounded up we have that n=114

Part 2

On this case we have a prior estimate for the population proportion and is \hat p =0.85 so replacing the values into equation (b) we got:

n=\frac{0.85(1-0.85)}{(\frac{0.06}{1.28})^2}=58.027

And rounded up we have that n=59

Part 3

On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

5 0
3 years ago
a volley ball player servers the ball. The ball follows a path given by the equation y=-0.01x^2+0.5x+3 where x and y are measure
dolphi86 [110]

Answer:

(a)

Distance from player should be 13.82 feet or 36.2 feet

(b)

The ball will go over the net

Step-by-step explanation:

we are given

The ball follows a path given by the equation

y=-0.01x^2+0.5x+3

where

x and y are measured in feet and the origin is on the court directly below where the player hits the ball

(a)

net height is 8 ft

so, we can set y=8

and then we can solve for x

8=-0.01x^2+0.5x+3

8\cdot \:100=-0.01x^2\cdot \:100+0.5x\cdot \:100+3\cdot \:100

800=-x^2+50x+300

-x^2+50x-500=0

x^2-50x+500=0

we can use quadratic formula

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}

x=5\left(5-\sqrt{5}\right),\:x=5\left(5+\sqrt{5}\right)

x=13.82,x=36.2

So, distance from player should be 13.82 feet or 36.2 feet

(b)

we can plug x=30 and check whether y=8 ft

y=-0.01(30)^2+0.5(30)+3

y=9ft

we know that

height of net is 8 ft

so, the ball will go over the net


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ladessa [460]

Answer:

25

Step-by-step explanation:

4×25 =100

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3 years ago
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Answer:

No solution

Step-by-step explanation:

There is no solution to the system of equations y=3x+2 and 6x-2y=8, meaning that they are parallel to each other on a graph, with no points ever intersecting on each other's lines.

6 0
2 years ago
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