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horrorfan [7]
2 years ago
7

Qual o resultado de 3842-269

Mathematics
1 answer:
marissa [1.9K]2 years ago
4 0

Answer:

es 3573

Step-by-step explanation:

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Help me with this answer I know there is no solution I just wanna know why
mylen [45]

Answer:

The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation. ... This is because there is truly no solution—there are no values for x that will make the equation true

4 0
3 years ago
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Find the quotient<br> 4/3÷1 1/12
nlexa [21]
The answer should be 1/9

8 0
2 years ago
Determine the remainder when (5x3 + 49x2 + 74x + 6) is divided by (x + 8).
blondinia [14]
The polynomial remainder theorem says that the remainder when dividing a polynomial P(x) by a linear divisor x-k is simply P(k).

If P(x)=5x^3+49x^2+74x+6, then the remainder upon dividing by x+8 is

P(-8)=-10

You could also verify this by actually computing the quotient and remainder.
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3 years ago
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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
What is the value of 5x+3 when x=4
mylen [45]

Answer:

23

Step-by-step explanation:

plug and chug: 5(4)+3=20+3=23

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