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Katyanochek1 [597]
3 years ago
15

Which function is a quadratic function?

Mathematics
2 answers:
Irina18 [472]3 years ago
8 0
F(x) is the only quadratic function here. Any term with a degree of 2 starting the equation is a quadratic. 
VLD [36.1K]3 years ago
3 0
Quadratic= 2nd degrree, means highest exponent of placeholder is 2

f(x)=2x^2 is 2 exposnnet, that is quadratic
g(x)=5x^3, 3rd degree, cubic
k(x)=-x, 1st degree, linear
b(x)=-9x^4, 4th degree, quartic


answer if f(x)
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Which bear describes the relationship between the successive terms in the sequence shown
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Idk what awser is just tryns get points
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How do u covert 372in.=yd 1ft what's the answer
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3 0
3 years ago
A law firm has five senior and six junior partners. A committee of three partners is selected at random to represent the firm at
satela [25.4K]

Answer:

the probability that at least one of the junior partners is on the committee is 0.9393

Step-by-step explanation:

The probability that a event happens is calculate as the quotient between the numbers of possibilities in which that event happens and the numbers of total possibilities.

Then for calculate the number of possibilities is necessary to introduce the concept of combination. The number of combinations are the ways that we can form groups with size k from a bigger group of size n and it is calculate as:

nCk=n!/(k!(n-k)! )

In other words nCk give as the number of ways that we can form groups of k objects from a group of n objects.

Taking into account the last explanation, the number of groups of 3 that we can create with the 11 partners (5 junior and 6 senior) is 165 and is calculate as:

11C3= 11!/(3!(11-3)! )

11C3=165

Then, for have at least 1 junior partner, we have 3 options:

1. A group formed by 1 junior partner and 2 senior partner

2. A group formed by 2 junior partner and 1 senior partner

3. A group formed by 3 junior partner and 0 senior partner

For the first option: we have to find the number of ways to choose one junior partner from 6 general options and multiply to the number of ways to choose two senior partner from 5 general options. This is calculate as:

Possibilities 1 = 6C1*5C2 = 6*10 = 60

That mean that there are 60 ways to choose group of 3 formed by one junior partner and 2 senior partners.

For the second option: we have to find the number of ways to choose two junior partner from 6 general options and multiply to the number of ways to choose one senior partner from 5 general options. This is calculate as:

Possibilities 2 = 6C2*5C1 = 15*5 = 75

That mean that there are 75 ways to choose group of 3 formed by 2 junior partner and 1 senior partners.

For the third option: we have to find the number of ways to choose three junior partner from 6 general options and multiply to the number of ways to choose zero partner from 5 general options. This is calculate as:

Possibilities 3 = 6C3*5C0 = 20*1 = 20

That mean that there are 20 ways to choose group of 3 formed just by junior partners.

So, the number of possibilities to have at least one junior partner is the sum of the three last options (75 + 60 + 10 = 155).

Finally the probability that at least one of the junior partner is on the committee:

P=155/165  

P= 0.9393

So, the probability that at least one of the committee be a junior partner is 0.9393

6 0
4 years ago
How do I solve the equation m+15=-22
MariettaO [177]
m+15=-22 

15m=-22-15
 

m=-37
7 0
3 years ago
Read 2 more answers
Ricardo has 2 cases of video games Edith the same number of games in each case. He gives 4 games to his brother. Ricardo has 10
alisha [4.7K]

Answer:

7 games

Step-by-step explanation:

Ricardo has 2 cases with x number of games in each. This means 2x is the amount of video games in both. He gives away 4. He now has 2x-4 and has 10 left. This is 2x-4=10. Solve for x.

2x - 4 = 10

2x = 14

x=7

7 0
3 years ago
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