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Brut [27]
3 years ago
9

Dawn raises money for her school in a jog-a-thon. She will get three dollars for every lap she completes. If it takes 4 laps to

jog 1 mile, and Dawn jogs a total of 11 miles, how much money will Dawn raise for her school?
Mathematics
2 answers:
AlexFokin [52]3 years ago
7 0
$132 dollars cause she must run 44 laps in order to reach 11 miles.
sesenic [268]3 years ago
5 0

Answer:

Dawn raises $132 for her school.

Step-by-step explanation:

Dawn rises money for her school in a jog-a-thon.

She will get $3 for each lap she completes.

It takes 4 laps to jog = 1 mile

For 11 miles it takes laps = 11 × 4 = 44 laps

∵ 3 Dollars for 1 lap

∴ for 44 laps = 44 × 3 = $132

Dawn raises $132 for her school.

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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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