In the set of all past due accounts, let the event A mean the account is between 31 and 60 days past due and the event B mean th

The likelihood that John scores on the first attempt and misses the second is:

Solution:

[(2x² + 5x) + (4x² – 4x)] + 5x³ =

(2x² + 5x) + [(4x² – 4x) + 5x³]

Step-by-step breakdown:

To determine the variance, we need to first calculate the second moment as follows: The variance can be determined using this equation: The standard deviation is simply the square root of the variance, which gives us the result.

¡Hola! Bienvenido a!

Vamos a sumar cuántas canicas tenemos en total.

12+11+17+5=45

Queremos hallar la probabilidad de elegir una canica que no sea azul. Observemos cuántas canicas no son azules.

12+11+5=28

Tendremos esta probabilidad sobre 48.

28/48

Al simplificar, obtenemos 7/12 o alrededor de 58.33%.

¡Espero que esto ayude!

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'' =