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stepan [7]
2 years ago
7

Given the following sets:

Mathematics
1 answer:
hjlf2 years ago
8 0
<h3>Answer:  10</h3>

===========================================================

Explanation:

Even though your teacher doesn't want you to list the items of the set, it helps to do so.

We'll be working with these two sets

A = {b, d, f, h, j, I, n, p, r, t}

C =  {d, h, I, p, t}

When we union them together, we combine the two sets together. Think of it like throwing all the letters in one bin rather than two bins.

A u C = {b, d, f, h, j, I, n, p, r, t,   d, h, I, p, t  }

The stuff that isn't bolded is set A, while the stuff that is bolded is set C

After we toss out the duplicates, we end up with this

A u C = {b, d, f, h, j, I, n, p, r, t}

But wait, that's just set A. Notice how everything in set C can be found in set A. This indicates set C is a subset of set A.

That's why all of the stuff in bold was tossed out (because they were duplicates of stuff already mentioned).

Once we determine what set A u C looks like, we count out the number of items in that set to determine the final answer.

There are 10 items in {b, d, f, h, j, I, n, p, r, t} which means 10 is the final answer.

----------------------------

An alternative method is to use the formula below

n(A u C) = n(A) + n(C) - n(A and C)

n(A u C) = 10 + 5 - 5

n(A u C) = 10

The notation n(A and C) counts how many items are found in both sets A and C at the same time. But as mentioned earlier, this is identical to just counting how many items are in set C. So we'll have n(C) cancel out with itself.

In short, n(A u C) = n(A) = 10

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The average of a list of 4 numbers is 90.0. A new list of 4 numbers has the same first 3 numbers as the original list, but the f
Thepotemich [5.8K]

Answer:

the average of this new list of numbers is 94

Step-by-step explanation:

Hello!

To answer this question we will assign a letter to each number for the first list and the second list of numbers, remembering that the last number of the first list is 80 and the last number of the second list is 96

for the first list

\frac{a+b+c+80}{4} =90

for the new list

\frac{a+b+c+96}{4} =X

To solve this problem consider the following

1.X is the average value of the second list

2. We will assign a Y value to the sum of the numbers a, b, c.

a + b + c = Y to create two new equations

for the first list

\frac{y+80}{4} =90

solving  for Y

Y=(90)(4)-80=280

Y=280=a+b+c

for the second list

\frac{y+96}{4} =X\\

\frac{280+96}{4} =X\\x=94

the average of this new list of numbers is 94

4 0
3 years ago
Read 2 more answers
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
Select the assumption necessary to start an indirect proof of the statement. If 3a+7 &lt; 28, then a&lt;7.
pashok25 [27]
When applying indirect proofs, we assume the negation of the conclusion is true, and show that this assumption would lead to nonsense, or contradiction.

In our case we assume a is not smaller than 7, that is we assume a≥7.

a≥7 then, multiplying both sides by 3:
3a≥21, then, adding both sides 7:

3a+7≥28,

which is a contradiction because 3a+7 is smaller than 28.

So our assumption is wrong, which means the opposite of it is correct.


Answer: assume a≥7
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In any case.  if  the x values (-6) are the same for both points the equation of line would be x = -6
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Which of the diagrams below represents the statement if it is a rectangle, then it is a square
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All the rectangle are square if length becomes equal to breath !
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