Determine which of the following statements is true. A: If V is a 6-dimensional vector space, then any set of exactly 6 elements in V is automatically a basis for V. B: If there exists a set that spans V, then dim V = 3. C: If H is a subspace of a finite-dimensional vector space V, then dim H ≤ dim V

A. This statement A is false.

B. This statement A is false.

C. This statement is true .

Step-by-step explanation:

Determine which of the following statements is true.

From the statements we are being given , we are to determine if the statements are valid to be true or invalid to be false.

SO;

A: If V is a 6-dimensional vector space, then any set of exactly 6 elements in V is automatically a basis for V

This statement A is false.

This is because any set of exactly 6 elements in V is linearly independent vectors of V . Hence, it can't be automatically a basis for V

B. If there exists a set that spans V, then dim V = 3

The statement B is false.

If there exists a set , let say , then any set of n vector (i.e number of elements forms the basis of V)  spans V. ∴ dim V < 3

C. If H is a subspace of a finite-dimensional vector space V  then dim H ≤ dim V is a correct option.

This statement is true .

We all know that in a given vector space there is always a basis, it is equally important to understand that there is a cardinality for every basis that exist ,hence the dimension of a vector space is uniquely defined.

SO,

If  H is a subspace of a finite-dimensional vector space V  then dim H ≤ dim V is a correct option.

i think the answer is fried chicken ; >

step-by-step explanation:

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

happy to

pls mark as brainliest