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andreev551 [17]
3 years ago
6

Is 7/10 or 11/20 greater than less than or equal too

Mathematics
2 answers:
m_a_m_a [10]3 years ago
7 0

Answer:

Step-by-step explanation:

11/20 < 7/10

dem82 [27]3 years ago
5 0

Answer:

11/20 < 7/10

Step-by-step explanation:

11/20 is less than 7/10 because 11/20 as a decimal is <em>0</em><em>.</em><em>5</em><em>5</em><em> </em>while 7/10 as a decimal is <em>0</em><em>.</em><em>7</em>

<em>PLEASE DO</em><em> </em><em>MARK ME</em><em> </em><em>AS BRAINLIEST</em><em> </em><em>IF MY</em><em> </em><em>ANSWER IS</em><em> </em><em>HELPFUL</em><em> </em><em>:</em><em>)</em><em> </em>

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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
How many solutions does the equation have? show your work.
Inessa05 [86]

All real numbers

7w-(2+w) = 2(3w-1)

Expand

7w-(2+w)= 6w-2

7w-2-w=6w-2

Group like terms

6w-2=6w-2

Add 2 to both sides

6w=6w

subtract 6w from both sides

0=0

True for all w

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Anne has 51. Freddy has 48.
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liubo4ka [24]

Answer:

85

Step-by-step explanation:

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