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stellarik [79]
2 years ago
13

Consider the function represented by the equation 6q = 3s - 9. Write the equation in function notation, where q is the independe

nt variable.
Mathematics
2 answers:
Firlakuza [10]2 years ago
8 0
Basically they want you to solve for q
6q=3s-9
divide both sides by 6
q=1/2s-3/2
bearhunter [10]2 years ago
4 0

Answer:

Given the equation:

6q = 3s -9

Add 9 to both sides we have;

6q+9= 3s -9+9

Simplify:

6q+9= 3s

Divide by 3 to both sides we have;

\frac{1}{3}(6q+9)=s

Using the distributive property, a \cdot (b+c) = a\cdot b+ a\cdot c

2q+3 = s

or

s = 2q+3 where,  q is independent variable and s is the dependent variable.

Use notation: s = f(q)

Therefore, the equation in function notation:

f(q) = 2q+3, where q is the independent variable

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jose is choosing a 3-letter password from the letters a,b,c,d. the password cannot have the same letter repeated in it. how many
kvasek [131]
Alright, so we are dealing with permutations. Permutations are the number of combinations in a specific order possible for the set. 

There's 4 letters, each one can be paired with 3 other letters, but those letters could be in different orders. To figure out how many variations of each combination there are (aka the number of permutations) use this formula:

_{n}P_{r}= \frac{n!}{(n-r)!}

r= number of elements in the subset

n= number of elements in the set

P= permutations of the set

There are only 3 elements in the subset because there is 1 that will not be repeated in each set, and there are 4 elements in the set.

Here's the math:

_{n}P_{r}= \frac{n!}{(n-r)!}

_{4}P_{3}= \frac{4!}{(4-3)!}

_{4}P_{3}= \frac{4!}{(1)!}

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There are 24 permutations. I can prove this by showing you the model:

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2 years ago
The arena wants to estimate the previous two month’s ticket-sale income. In total, 71,115 people attended different events at th
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Hey there,
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7 0
2 years ago
among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play
kondaur [170]

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

3 0
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