Dan invests £1200 into his bank account.

Answer:

Step-by-step explanation:

Assuming a fair die is rolled.

• The sample space comprises 1, 2, 3, 4, 5, 6, with all results being equally probable.

Let X represent the collection of all outcomes. Let A represent a specific outcome.

<pThus, the probability of event A occurring is:

                                                       

Considering that the set of all possible outcomes for a singular die roll is:

Notably,

Here,  

  since 8 is not included in the sample space. Therefore, rolling an 8 is impossible within the defined outcomes.

<pThis leads to the conclusion that the probability is zero.

In other terms,

                       

                         

<pAs a result,

                   

Response:

b

Detailed explanation:

<span>For two vectors u and v, the cross product can be expressed as follows: (u2v3 - u3v2) i + (u3v1 - u1v3) j + (u1v2 - u2v1) k. Consequently, we have m x n: ( (2 *-4) - (-5 * -2) ) i + ( (-5*6) - (3 *-4) )j + ( ( 3*-2) - (2*6))k, which simplifies to ( -8 - 10) i + ( -30 + 12) j + ( -6 - 12) k. Thus, the product of vectors m and n yields -18 i - 18j -18k</span>.

Answer:

link:

Step-by-step explanation:

Answer:

Step-by-step explanation:

The total count of laptops, N = 20

Defective laptops, n = 3

The school is purchasing 2 laptops

Case 1:

Both purchased laptops are non-defective

Probability of selecting 2 non-defective laptops

The count of non-defective laptops equals 20 - 3 = 17

Overall laptop count = 20

Probability, P = 17/20 = 0.85

Case 2:

One laptop is found defective

The probability of choosing 1 defective and 1 non-defective laptop

P' = 3/20 x 17/19 = 0.134

Case 3:

Both laptops are defective

Choosing 2 laptops from 3 = 3 C 2 = 3

Selecting 2 from 20 = 20 C 2 = 190

The probability of selecting two defective laptops equals 3 / 190 = 0.0158

P'' =