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ladessa [460]
2 years ago
15

Write 11/4 as a mixed number. Give your answer in its simplest form.

Mathematics
2 answers:
AVprozaik [17]2 years ago
8 0

Answer:

2 3/4 (two and three fourths)

Step-by-step explanation:

sp2606 [1]2 years ago
3 0
I think the answer is 2 3/4
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What is the awnser to |2x-3|-4=3
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3

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X/3+5=10 solve for x only type the numerical answer below
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Answer:

x=15

Step-by-step explanation:

x/3=10-5

x/3=5

x=5*3

x=15

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If f(x) = 4x - 3, evaluate f(2).
hjlf

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yarga [219]

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Step-by-step explanation:

5 0
3 years ago
In a competition, a school awarded medals in different categiories.40 medals in sport 25 medals in danceand 212 medals in music,
aleksley [76]

Answer:

210

Step-by-step explanation:

Given:

Medals in sports = 40

Medals in dance = 25

Medals in music = 212

Total students that received medals = 55

Total students that received medals in all three categories = 6

Required:

How many students get medals in exactly two of these categories?

Take the following:

A = set of persons who got medals in sports.

B = set of persons who got medals in dance

C = set of persons who got medals in music.

Therefore,

n(A) = 40

n(B) = 25

n(C) = 212

n(A∪B∪C)= 55

n(A∩B∩C)= 6

To find how many students get medals in exactly two of these categories, we have:

n(A∩B) + n(B∩C) + n(A∩C) −3*n(A∩B∩C)

=n(A∩B) + n(B∩C) + n(A∩C) −3*6 ……............... (1)

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)

Thus, n(A∩B)+n(B∩C)+n(A∩C)=n(A)+n(B)+n(C)+n(A∩B∩C)−n(A∪B∪C)

Using equation 1:

=n(A)+n(B)+n(C)+n(A∩B∩C)−n(A∪B∪C)−18

Substitute values in the equation:

= 40 + 25 + 212 + 6 − 55 − 18

= 283 - 73

= 210

Number of students that get medals in exactly two of these categories are 210

6 0
4 years ago
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