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4 years ago

The yearbook club had a meeting. The meeting had

1 answer:

5 0

If 30 people is three fifths of the club, then you can divide 30 by 3 to find out how many people make up one fifth of the club. So, 10 people make up one fifth of the club.

30 + 20 (which is two more fifths of the club) = 50 people (the entire club)

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(x-4)^2+12(x-4)+20=0

Ratling [72]

Answer:

x=2 or x=−6

Step-by-step explanation:

4 0

3 years ago

Eddie the elf made 13 toys one day in the workshop. some were dolls and the rest were action figures. the number of action figur

Lunna [17]

To solve this problem we first call x = number of action figures, y = number of dolles. A system of two equations with two unknowns must be made to describe the problem. The system is the following:
(x + 1) + y = 13
1/2 * x = y.
Then solving the system we have that x = 8 and y = 4.
Since we know that the number of action figures is twice as many dolls plus one, then x = 8 + 1 = 9.
Thus,
dollos = 4
action figures = 9

3 0

3 years ago

Please answer !!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!

zvonat [6]

i would probably use an iv or a blood simple

3 0

3 years ago

I only have 2 hours to do this can someone plz help

wolverine [178]

Figure it out tonta tonta

8 0

3 years ago

At a high school, students can choose between three art electives, four history electives, and five computer electives. Each stu

V125BC [204]

Answer:

$\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Step-by-step explanation:

Given,

Art electives = 3,

History electives = 4,

Computer electives = 5,

Total number of electives = 3 + 4 + 5 = 12,

Since, if a student chooses an art elective and a history elective,

So, the total combination of choosing an art elective and a history elective = $^3C_1\times ^4C_1$

Also, the total combination of choosing any 2 subjects out of 12 subjects = $^{12}C_2$

Hence, the probability that a student chooses an art elective and a history elective = $\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}$

$=\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Which is the required expression.

3 0

3 years ago

Read 2 more answers