7. A company manufactures shaving sets for $5 each and sells them for $7 each. How many shaving sets must be sold for the

Let u,v,wu,v,w be three linearly independent vectors in R7R7. Determine a value of kk, k=k= , so that the set S={u−3v,v−2w,w−ku}

11Alexandr11 [23.1K]

Answer:

1/6

Step-by-step explanation:

For a set of vectors a, b and c to be linearly dependent, the linear combination of the set of vectors must be zero.

C1a + C2b + C3c = 0

From the question, we are given the set S={u−3v,v−2w,w−ku}, the corresponding vectors are a(0, -3, 1), b(-2,1,0) and c(1,0,-k)

The values in parenthesis are the ccoefficients of w, v and u respectively.

On writing this vector as a linear combination, we will have;

C1(0, -3, 1) + C2(-2,1,0) + C3(1,0,-k) = (0,0,0)

0-2C2+C3 = 0........ 1

-3C1+C2 = 0 ........... 2

C1-kC3 = 0 ….......... 3

From equation 2, 3C1 = C2

Substituting into 1, -2(3C1)+C3 = 0

-6C1+C3 = 0

-6C1 = -C3

6C1 = C3.…..4

Substitute 4 into 3 to have

C1-k(6C1) = 0

C1 = k6C1

6k = 1

k = 1/6

Hence the value of k for the set of vectors to be linearly dependent is 1/6

6 0

3 years ago

Can anyone figure this out???

Anestetic [448]

It’s a right triangle because it has a 90 degree angle

5 0

3 years ago

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Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

3 years ago

Help please?? I’m being timed!!

weqwewe [10]

Answer:

answer is is option 3 all the best for examination

6 0

3 years ago

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Identify the volume of the figure in cubic units.

Alekssandra [29.7K]

Answer:

8 cubic units

Step-by-step explanation:

Count it. It's all visible for this question.

6 0

3 years ago