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Sindrei [870]
3 years ago
6

A 3ed degree binomial with a constant term of 8

Mathematics
1 answer:
Svet_ta [14]3 years ago
7 0

Answer:

4m³+8 is a 3rd degree binomial with constant term of 8.

Step-by-step explanation:

  • A binomial expression consists of two terms.

For example,

m+n is a binomial, where m and n are two terms.

  • As we know that the degree of the polynomial is said to be the highest exponent of any of the terms.

For example, in expression 7m + 3, the exponent of m is 1. So it is 1st degree polynomial.

Similarly, in the expression 10m²+9m, variable m has the highest exponent which is 2. So it is 2nd degree polynomial.

Now, let us write a 3rd degree binomial with a constant term of 8

4m³+8

As it has two terms which are 4m³ and 8, Therefore, it would be called binomial.

Also the highest exponent with variable m is 3. Therefore it would be a 3rd degree binomial with constant term of 8.

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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
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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

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(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.

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you want to put 2000 in a simple intrest account. it has 2.5 annual intrest rate. how long will it take you to earn 500 in intre
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