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2 years ago

A manager wants to select one group of 4 people from his 28 assistants.

1 answer:

8 0

There are, 20475 different groups are possible if the manager wants to select one group of 4 people from his 28 assistants.

<h3>What is permutation and combination?</h3>

A permutation is the number of different ways a set can be organized; order matters in permutations, but not in combinations.

We have:

A manager wants to select one group of 4 people from his 28 assistants.

The total number of groups possible = C(28, 4)

$= \rm \dfrac{28!}{4!(28-4)!}$

After calculating:

= 20475

Thus, there are, 20475 different groups are possible if the manager wants to select one group of 4 people from his 28 assistants.

Learn more about permutation and combination here:

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Line segment addition !!!!!! SOLVE FOR X !!!!!

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Answer:

x = 11

Step-by-step explanation:

JM = JK + KL + LM

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23 = 9 + 2x - 18 + x - 1
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Each tire of a bicycle has a diameter of 26 inches. If you start pedaling and the tires make 10 complete rotations, how far has

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New legislation passed in 2017 by the U.S. Congress changed tax laws that affect how many people file their taxes in 2018 and be

stiks02 [169]

Answer:

a. Yes

b. 0.35

c. 0.5

E(x)=34.25

V(x)=73.1875

SD(x)=8.555

Step-by-step explanation:

x 20 25 30 35 40 45 50

f(x) 0.05 0.2 0.25 0.15 0.15 0.1 0.1

This will be valid probability distribution if

1. All probabilities are between 0 and 1 inclusive.

2. Sum of probabilities must be 1.

Condition 1 is satisfied in this distribution. Now, we will check second condition

sum(f(x))=0.05+0.2+0.25+0.15+0.15+0.1+0.1=1.

As the both conditions are satisfied, thus the given distribution is a valid probability distribution.

P(x≥40)=P(X=40)+P(X=45)+P(X=50)

P(x≥40)=0.15+0.1+0.1

P(x≥40)=0.35

So, the probability that Backens and Hayes LLC will obtain 40 or more new clients is 0.35.

P(x<35)=P(X=20)+P(X=25)+P(X=30)

P(x<35)=0.05+0.2+0.25

P(x<35)=0.5

So, the probability that Backens and Hayes LLC will obtain fewer than 35 new clients is 0.5.

Expected value of x=E(x)= sum[x*f(x)]

E(x)= 20*0.05+25*0.2+30*0.25+35*0.15+40*0.15+45*0.1+50*0.1

E(x)=1+5+7.5+5.25+6+4.5+5

E(x)=34.25

Variance of x=V(x)=sum[x²*f(x)]-(sum[x*f(x)])²

sum[x²*f(x)]= 20²*0.05 +25²*0.2 +30²*0.25 +35²*0.15 +40²*0.15 +45²*0.1 +50²*0.1

sum[x²*f(x)]=20+1255+225+183.75+240+202.5+50

sum[x²*f(x)]=1246.25

V(x)=1246.25-(34.25)²

V(x)=73.1875

Standard deviation of x=SD(x)=√V(x)

SD(x)=8.555

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3 years ago