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san4es73 [151]
2 years ago
5

1367324+3469345-75369x3429x0=??

Mathematics
1 answer:
Mnenie [13.5K]2 years ago
5 0

Answer:

it's ans is 4761300

Step-by-step explanation:

by using BODMAS

75369×3429×0

=0

0+1367324+3469345

=4836669

4836669-75369

=4761300

plz mark me as brainliest plz

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A student skipped a step when she tried to convert 18 hours into seconds and she got the following result:
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Multiply 18hours by 60 seconds, the answer is 2880 seconds.

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Is the relationship a linear function?(0,-1), (1,5),(2,11),(3,17)
cricket20 [7]

Answer: Yes

Step-by-step explanation: x goes up by 1 each time and y goes up by 5 each time, making the function linear.

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The ordered pair (1,81), (2,100), (3,121), 5,169).. represent a function. What is a rule that represent this function?
zloy xaker [14]

<span>A rule that represents this function is y = (x + 8)².</span>

 

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Now square both sides

<span>y = (x + 8)²</span>

y = x² + 16x + 64

 

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3 years ago
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
2 years ago
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