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KengaRu [80]
2 years ago
6

Which of the following is a valid reason for stating that <1 <2

Mathematics
1 answer:
Slav-nsk [51]2 years ago
7 0
Because 1<2 thus anything <1 must be <2
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elena-s [515]

Answer:

h ≤ (approx) 3.7

Step-by-step explanation:

To start, we can take our knowledge that we have to spend 20$, at minimum. Therefore, we can only allocate (150-20)=130 to our payment per hour. As we pay 35$ per hour, the amount of money relative to hours would be 35*h. Therefore, as we can allocate 130 dollars or less for our hourly payments, and our money is 35*h, we can day that 35*h(the money we spend on hourly rates) ≤ 130. Then, as 35*h ≤ 130, we can divide 35 from both sides to get h ≤ 130/35, or h ≤ (approx) 3.7

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If 596 people voting in the election how many were over 65 years old.
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596*25%=149
149 people voted over age 65
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What is estimate of 1/8+1/4=
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Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoti
Archy [21]

This question is incomplete, the complete question is;

Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen.

a) What is E[G] or range of G

b) Give the distribution over the random variable G

Answer:

a) E[G] is [ 0, 1 , 2 ]

b)

the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

Step-by-step explanation:

Given the data in the question;

Number to be chosen is 2

from 7 girls and 3 boys

a) range of G

number of girls can be; [ 0, 1 , 2 ]

Therefore, E[G] is [ 0, 1 , 2 ]

b) he distribution over the random variable G.

Distribution of the random variable G is Hypergeometric;

so

with P( G=g ) = P( getting g girls from 7 and 2-g boys from 3)

= ((^7C_g) × (^3C_{2-g)) / ^{10}C_2

now since, the range of G is [ 0, 1 , 2 ]

P( G=0 ) = ((^7C_0) × (^3C_{2)) / ^{10}C_2

= [(7!/(0!(7-0)!)) × (3!/(2!(3-2)!))] / (10!/(2!(10-2)!))

= [1 × 3] / 45 = 3/45 = 0.0667

P( G=1 ) = ((^7C_1) × (^3C_{1)) / ^{10}C_2

= [(7!/(1!(7-1)!)) × (3!/(1!(3-1)!))] / (10!/(2!(10-2)!))

= [ 7 × 3 ] / 45 = 21/45 = 0.4667

P( G=2 ) = ((^7C_2) × (^3C_{0)) / ^{10}C_2

= [(7!/(2!(7-2)!)) × (3!/(0!(3-0)!))] / (10!/(2!(10-2)!))

= [ 21 × 1 ] / 45 = 21/45 = 0.4667

Therefore, the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

3 0
3 years ago
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