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3 years ago

Solve the system of equations

Mathematics

2 answers:

Elina [12.6K]3 years ago

7 0

The correct answer is letter C, (3,5)

When you substitute this ordered pair on the equations, you would found out that it is the correct one

Helen [10]3 years ago

6 0

The correct answer is c

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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

Question 4 of 40

UkoKoshka [18]

Answer:

A. The ability to do something well without wasted time or effort.

3 0

2 years ago

I need help! I forgot how to do this!! Please help!

Dovator [93]

First change the mixed fraction into an improper fraction

7 1/2 = 15/2

Next, change multiply 2 to both numerator and denominator

15 x 2 = 30

2 x 2 = 4

30/4

Each piece needs 3/4 of an inch. Divide 30 with 3

30/3 = 10

c) 10 pieces is your answer

hope this helps

8 0

2 years ago

Read 2 more answers

Let U = {a, b, c, e, d, e, f, g, h, i, j, k}; A = {a, b, e, f, h, j}; B = {b, c, d, f, j}. Draw a Venn Diagram with a rectangle

marishachu [46]

<h3>Answer: Venn Diagram is below</h3>

The rectangle represents the universal set (marked as "set U"). Everything is contained in this rectangle.

Circle A represents set A = {a, b, e, f, h, j}

Circle B represents set B = {b, c, d, f, j}

The common elements between the two sets are {b, f, j}, so we'll write these three values in the overlapping portion of the two circles.

The items {a,e,h} will go in circle A, but not in the overlapping portion since they are found only in set A.

Similarly we'll have {c,d} go in circle B but not in the overlapping portion since these items are only found in set B.

The remaining terms {g, i, k} go outside both circles, but inside the rectangle, because neither of these elements are in set A or set B.

6 0

3 years ago

What kind of word problem can be solved with a two-variable equation?

Annette [7]

Answer: Here’s one I made up! Combined, Megan and Kelly worked 60 hours, Kelley worked twice as many hours as Megan. How many hours did they each work? Hope this helps!

5 0

3 years ago