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Marta_Voda [28]
3 years ago
5

A teacher finishes explaining a new concept to the class and wants to check that all the students have grasped the concept. The

teacher asks those who do not understand to raise their hands. This process leads to which of the following bias(es)?
a) Response bias
b) Measurement bias
c) Both response bias and measurement bias
d) neither response bias nor measurement bias
Mathematics
1 answer:
faust18 [17]3 years ago
3 0

Answer:

response bias

Step-by-step explanation:

because whan you are asked a question you expect response in return

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Answer:

<u>The correct answer is 130.9 kilograms.</u>

Step-by-step explanation:

We want to calculate what is the result of 7.7 kilograms multiplied by 17 and the answer should be in kilograms

Result = 7.7 kilograms * 17

Result = 7.7 * 17

Result = 130.9

Result = 130.9 kilograms

<u>The result of the multiplication of 7.7 kilograms by 17 is exactly 130.9 kilograms</u>

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All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0
3 years ago
The probabilty that a student owns a car is 0.65 the porbability that a student owns a compuer is 0.82 the probability that a st
DanielleElmas [232]

Question:  The probability that s student owns a car is 0.65, and the probability that a student owns a computer is 0.82.

a. If the probability that a student owns both is 0.55, what is the probability that a randomly selected student owns a car or computer?

b. What is the probability that a randomly selected student does not own a car or computer?

Answer:

(a) 0.92

(b) 0.08

Step-by-step explanation:

(a)

Applying

Pr(A or B) = Pr(A) + Pr(B) – Pr(A and B)................. Equation 1

Where A represent Car, B represent Computer.

From the question,

Pr(A) = 0.65, Pr(B) = 0.82, Pr(A and B) = 0.55

Substitute these values into equation 1

Pr(A or B) = 0.65+0.82-0.55

Pr(A or B) = 1.47-0.55

Pr(A or B) = 0.92.

Hence the probability that a student selected randomly owns a house or a car is 0.92

(b)

Applying

Pr(A or B) = 1 – Pr(not-A and not-B)

Pr(not-A and not-B) = 1-Pr(A or B) ..................... Equation 2

Given: Pr(A or B)  = 0.92

Substitute these value into equation 2

Pr(not-A and not-B) = 1-0.92

Pr(not-A and not-B) = 0.08

Hence the probability that a student selected randomly does not own a car or a computer is 0.08

8 0
2 years ago
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