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3241004551 [841]
1 year ago
9

Paloma has 3 jackets, 6 scarves, and 4 hats. What is the number of different outfits consisting of a jacket, a scarf, and a hat

that Paloma can wear?
Answer:
Submit Answer
Mathematics
1 answer:
Charra [1.4K]1 year ago
5 0

Answer:

72

Step-by-step explanation:

We need to multiply all of them because there are 3 choices for jackets, 6 for scarves and so on.

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Your question
Alex777 [14]
2x+4.7=77.8

double the unknown number then add 4.7
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an isosceles triangle contains two equal angles. Each of these angles is 25 degrees larger than twice the smallest angle. What a
Helga [31]
Well, the sum of all internal angles in a triangle, isosceles or whatever else, is always 180°, so the sum of all those three folks, will be 180 then

so.. hmmm let's say the smallest angle is "a" units

twice that is, 2*a, or 2a
25 larger than that? is well 2a + 25

so... the other two twin angles, are 2a+25 each

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3 years ago
Please simplify these expressions
Oksana_A [137]

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A survey was conducted of 42 students who take math, Eng lish, or history. In the survey, 15 students said they take math and hi
strojnjashka [21]

Answer:

16 students take only English or only Math.

Step-by-step explanation:

We can solve this problem by treating these values as sets, and building the Venn Diagram.

I am going to say that:

A is the number of students who take Math.

B is the number of students who take English.

C is the number of students who take History.

We have that:

A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)

In which a is the number of students that only take Math, A \cap B is the number of students who take both Math and English, A \cap C is the number of students that take both Math and History, and A \cap B \cap C is the number of students that take all these classes.

By the same logic, we have:

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

This diagram has the following subsets:

a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)

There were 42 students suveyed. This means that:

a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 42

We start finding the values from the intersection of three sets.

12 students said they take math, English, and history. This means that:

A \cap B \cap C = 12

18 students said they take English and history. This also takes into account those who take math, english and history. So:

(B \cap C) + (A \cap B \cap C) = 18

B \cap C = 6

17 students said they take math and English.

(A \cap B) + (A \cap B \cap C) = 17

A \cap B = 5

15 students said they take math and history

(A \cap C) + (A \cap B \cap C) = 15

A \cap C = 3

2 students said they only take history.

c = 2

How many students take only English or only math?

This is a + b, that we can find by the following formula:

a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 42

a + b + 2 + 5 + 3 + 6 + 12 = 42

a + b = 16

16 students take only English or only Math.

4 0
3 years ago
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