Use the following vectors to answer parts (a) and (b). v1equals=[start 3 by 1 matrix 1st row 1st column 1 2nd row 1st column negative 4 3rd row 1st column 2 endmatrix ]1 −4 2 , v2equals=[start 3 by 1 matrix 1st row 1st column negative 4 2nd row 1st column 16 3rd row 1st column negative 8 endmatrix ]−4 16 −8 , v3equals=[start 3 by 1 matrix 1st row 1st column 5 2nd row 1st column 7 3rd row 1st column h endmatrix ]5 7 h (a) for what values of h is v3 in span{v1, v2}? (b) for what values of h is {v1, v2, v3} linearly dependent?

(1) No matter what's the value of , is never in the span of and .

(2) The three vectors , , and are always linearly dependent for all real .

Step-by-step explanation:

If is in the span of and , there need to exist real and such that

.

Assume that such and do exist.

In other words,

.

.

.

Rewrite as an augmented matrix and row-reduce:

.

(Add four times row one to row two and times row one to row three.)

.

Note that in row two,

In other words, this system is inconsistent. There's no real and that would satisfy the condition

.

Hence .

There's no real that allows , to be part of the span of and .

If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.

Note that

. In other words, can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of is zero, neither the exact value of nor the value of will make a difference. Therefore, for all , the three vectors , , and are linearly dependent.

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step-by-step explanation:

the probability is   .24

the first player had the highest probability of getting a silver marble

the game could be made fair by adding back the marble that was drawn after each draw

d

step-by-step explanation: