# A conference consists of 5 sessions: A, B, C, D, and E. Here are the costs of the sessions.Session A: $50Session B:$50Session C: $100Session A conference consists of 5 sessions: A, B, C, D, and E. Here are the costs of the sessions. Session A:$50
Session B: $50 Session C:$100
Session D: $150 Session E:$200

A participant plans to attend 3 sessions. Here is a list of all possible samples of size 3 sessions from this population of 5 sessions: ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, and CDE. If sessions A, B, and C are selected, what is the sample minimum session cost?

$50$66.67
$100$200

## This Post Has 9 Comments

1. Megcuttie101 says:

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2. robert7248 says:

$\begin\\\begin{tabular}{|l|l|l|}No&Sample&Median\\01&58,42,37 &42\\02&58,42,34&42\\03&58,42,23& 42\\04&37,34,23 & 34\\05&58,37,34 & 37\\06&58,37,23 & 37\\07&58,34,23 &34\\08&42,37,34 &37\\09&42,37,23 &37\\10&42,34,23 &34\end$

Step-by-step explanation:

From the question we are told that

5 samples(23 34 37 42 58)

To select 3 random sample

a) Generally to find all 10 possible outcomes and there median we make a table

Therefore

L1

58,42,37  Median=42

L2

58,42,34  Median=42

L3

58,42,23  Median=42

L4

37,34,23  Median=34

L5

58,37,34  Median=37

L6

58,37,23  Median=37

L7

58,34,23  Median=34

L8

42,37,34  Median=37

L9

42,37,23  Median=37

L10

42,34,23  Median=34

$\begin\\\begin{tabular}{|l|l|l|}No&Sample&Median\\01&58,42,37 &42\\02&58,42,34&42\\03&58,42,23& 42\\04&37,34,23 & 34\\05&58,37,34 & 37\\06&58,37,23 & 37\\07&58,34,23 &34\\08&42,37,34 &37\\09&42,37,23 &37\\10&42,34,23 &34\end$

3. firenation18 says:

$200 ABC 50+50+100=200 4. masonorourke says: D Step-by-step explanation: A C B 5. kydog00 says: Porch typ cu lu po me der al les una 6. Expert says: How much are you paying ? 7. emmanolan112 says: All possible are: (G,L,S) (G,L,R) (G,L,P) (G,S,R) (G,S,P) (G,R,P) (L,S,R) (L,S,P) (S,R,P) {L,R,P) Probability of 1st/2nd/10th sample = 1/10 Step-by-step explanation: All the possible combinations of the 3 size samples from a 5 size population have been listed without repetition. Total Numbers of Samples = 10 To find the probability of finding the first sample from random sampling procedure, Probability = Number of desired outcomes/ Total number of outcomes Where Number of desired outcome is 1 and total number of outcomes is 10. Probability = 1/10 Similarly, to find 2nd sample or 10th sample, the number of desired outcomes is same i.e 1, hecne the probability remains the same i.e 1/10 8. Expert says: she would want to sell 10 full-year subscriptions and at least 9 half-year subscriptions to get a total of at least$300.

i hope this answer you! if you have any further questions or concerns, feel free to ask! : )

step-by-step explanation:

mark brainliest!

$Kelly sells magazine subscriptions after work. the half-year subscriptions cost 12, and the full-ye$

9. mariap3504 says:

There are 9 possible samples for selection of two with replacement.

They are: (56, 56) (56, 46) (56, 60) (46, 46) (46, 56) (46, 60) (60, 60) (60, 46) (60, 56)

Remember that when making solution sets with replacement you have the possibility of choosing the same number (or data point) twice. So in this case, you will have the number of possible answers squared for total possible answers.