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Monica [59]
3 years ago
12

Earnings per hour and greatest hourly rate

Mathematics
2 answers:
egoroff_w [7]3 years ago
8 0
Money is earned by the hour.
Therefore all you have to do is divide the second column by the first column.

Caleb: 36.25/5; or $7.25 an hour.
Jeremy: 65.25/7.5; or $8.70 an hour.
Maria: 34.00/4.25; or $8.00 an hour.
Rosa: 54.00/8; or $6.70 an hour
horrorfan [7]3 years ago
5 0

This can be solved by using a simple algebraic equation.

For example, in the first column, Caleb worked 5 hours. We do not know how much he makes per hour, so let that be x. He earned a total of $36.25, so the equation can be written as 5x=36.25.

Divide by 5 on both sides, and you get x= 7.25

He earns $7.25 per hour of work.

Repeat this for all the workers:

Jeremy: 7.5x=65.25

x= $8.70

Maria: 4.25x=34.00

x= $8

Rosa: 8x=54.00

x= $6.75


Jeremy has the greatest hourly rate


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4. Is rap music more popular among young Indians than among young Asians? A sample survey compared 634 randomly chosen Indians a
bulgar [2K]

Answer:

H0 is accepted

there is no difference between the proportions of Indian and Asian young people who listen to rap music every day.

Step-by-step explanation:

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p_{1} = \frac{368}{634} =0.580

It found that 368 of the Indians and 130 Asians listened to rap music every day.

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Null hypothesis H0: there is no difference between the proportions of Indian and Asian young people who listen to rap every day.

p_{1} = p_{2}

Alternative hypothesis:- p_{1} \neq  p_{2}

Level of significance α = 0.05

The test of statistic

z = \frac{p_{1} -  p_{2}}{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} })  } } }

where p = \frac{n_{1}p_{1} + n_{1}p_{2}}{n_{1}+n_{2}} = \frac{634(0.580)+567(0.229)}{634+567}

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and q = 1-p = 1- 0.414 =0.586

Z = \frac{0.580-0.229}{\sqrt{0.414(0.586)(\frac{1}{634} +\frac{1}{567} } } }

on calculation , we get

z = 0.300 ><1.96 at 95 % level of significance

H0 is accepted

there is no difference between the proportions of Indian and Asian young people who listen to rap every day.

3 0
3 years ago
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