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11Alexandr11 [23.1K]
2 years ago
8

A junior college is going to choose a Homecoming Queen, and the candidates are the Freshman Queen and the Sophomore Queen. If th

ere are 600 freshman and 400 sophomore voters, which would be a FAIR way of conducting the vote? [Fair, in this case, means that the Freshman Queen does not have an automatic advantage.]
A) Count the votes, then deduct 100 from the freshman total.
B) Count the votes and the Queen with the highest total becomes Homecoming Queen.
C) Count the votes, counting each sophomore ballot as 1.5 votes and each freshman ballot as 1 vote.
D) Count the votes, counting each freshman ballot as 1.5 votes and each sophomore ballot as 1 vote.
Mathematics
1 answer:
KonstantinChe [14]2 years ago
5 0
Count the votes, counting each sophomore ballot as 1.5 votes and each freshmen ballot as 1 vote.
ur doing this because there is 200 more freshmen then sophomores...and if u count each sophomore vote as 1.5, it would make up for the 200 more freshmen
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6) Supplementary Exercise 5.51
tresset_1 [31]

Answer:

P(X \le 4) = 0.7373

P(x \le 15) = 0.0173

P(x > 20) = 0.4207

P(20\ge x \le 24)= 0.6129

P(x = 24) = 0.0236

P(x = 15) = 1.18\%

Step-by-step explanation:

Given

p = 80\% = 0.8

The question illustrates binomial distribution and will be solved using:

P(X = x) = ^nC_xp^x(1 - p)^{n-x}

Solving (a):

Given

n =5

Required

P(X\ge 4)

This is calculated using

P(X \le 4) = P(x = 4) +P(x=5)

This gives:

P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^{5-4} + ^5C_5*0.8^5*(1 - 0.8)^{5-5}

P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0

P(X \le 4) = 0.4096 + 0.32768

P(X \le 4) = 0.73728

P(X \le 4) = 0.7373 --- approximated

Solving (b):

Given

n =25

i)

Required

P(X\le 15)

This is calculated as:

P(X\le 15) = 1 - P(x>15) --- Complement rule

P(x>15) = P(x=16) + P(x=17) + P(x =18) + P(x = 19) + P(x = 20) + P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)

P(x > 15) = {25}^C_{16} * p^{16}*(1-p)^{25-16} +{25}^C_{17} * p^{17}*(1-p)^{25-17} +{25}^C_{18} * p^{18}*(1-p)^{25-18} +{25}^C_{19} * p^{19}*(1-p)^{25-19} +{25}^C_{20} * p^{20}*(1-p)^{25-20} +{25}^C_{21} * p^{21}*(1-p)^{25-21} +{25}^C_{22} * p^{22}*(1-p)^{25-22} +{25}^C_{23} * p^{23}*(1-p)^{25-23} +{25}^C_{24} * p^{24}*(1-p)^{25-24} +{25}^C_{25} * p^{25}*(1-p)^{25-25}

P(x > 15) = 2042975 * 0.8^{16}*0.2^9 +1081575* 0.8^{17}*0.2^8 +480700 * 0.8^{18}*0.2^7 +177100 * 0.8^{19}*0.2^6 +53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0  

P(x > 15) = 0.98266813045

So:

P(X\le 15) = 1 - P(x>15)

P(x \le 15) = 1 - 0.98266813045

P(x \le 15) = 0.01733186955

P(x \le 15) = 0.0173

ii)

P(x>20)

This is calculated as:

P(x>20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)

P(x > 20) = 12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0

P(x > 20) = 0.42067430925

P(x > 20) = 0.4207

iii)

P(20\ge x \le 24)

This is calculated as:

P(20\ge x \le 24) = P(x = 20) + P(x = 21) + P(x = 22) + P(x =23) + P(x = 24)

P(20\ge x \le 24)= 53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1

P(20\ge x \le 24)= 0.61291151859

P(20\ge x \le 24)= 0.6129

iv)

P(x = 24)

This is calculated as:

P(x = 24) = 25* 0.8^{24}*0.2^1

P(x = 24) = 0.0236

Solving (c):

P(x = 15)

This is calculated as:

P(x = 15) = {25}^C_{15} * 0.8^{15} * 0.2^{10}

P(x = 15) = 3268760 * 0.8^{15} * 0.2^{10}

P(x = 15) = 0.01177694905

P(x = 15) = 0.0118

Express as percentage

P(x = 15) = 1.18\%

The calculated probability (1.18%) is way less than the advocate's claim.

Hence, we do not believe the claim.

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How to work -1x-2y=-13 x-2y=11 equation by substitution
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Well if you're wanting to use substitution, you first have to end up with one term on either side of the equation. Use the second one as it's easiest. 

So: x-2y=11, so find -2y as there is a -2y in the first equation. 
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x=12. then chuck the x into any of the equations to find what y equals. 

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