Find the quotient: –1019÷(−57). a: 7095 B: 1419 c : 1419 d : 7095

The correct answer is:

She can ascertain that while there exists an association between the variables, a correlation does not exist, and causation cannot be established.

Explanation:

Due to the arch shape of her graph, it indicates that as one variable varies, the other also changes. This indicates an association exists between these variables.

Nonetheless, given that the graph does not follow a linear pattern, no correlation can be claimed. As there is no correlation, causation cannot be determined.

I believe the answer is A, since individuals employed at the largest corporation may hold different views compared to those not working in large companies. The benefits associated with one's employment can vary, and those at the major corporation may only reflect the interests of the upper class, leaving out perspectives from those who do not have high-paying corporate jobs. 

Answer:

a. A relative-frequency distribution relates to a variable just as a _____probability_____ distribution relates to a random variable. b. A relative-frequency histogram pertains to a variable similarly to how a _____probability_____ histogram pertains to a random variable.

Step-by-step explanation:

Probability, a mathematical concept, involves numerical assessments of how likely a certain event may occur, indicating the validity of its occurrence. The range for the probability of any event is typically between 0 and 1,

with 0 signifying impossibility and 1 indicating certainty of occurrence.

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'' =

Answer: 0.1289

Step-by-step explanation:

Given: The proportion of students absent on Mondays at a large university.:

Sample size:

Mean:

Standard deviation =

Let x represent a binomial variable.

Referencing the standard normal distribution table,

(1)

Z score for normal distribution:-

For x=4

For x=3

Thus, from (1)

Consequently, the likelihood of four students being absent =