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3 years ago

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Lia must work at least 5 hours per week in her family's restaurant for $8 per hour. She also does yard work for $12 per hour. Li

a's parents allow her to work a maximum of 15 hours per week overall. Lia's goal is to earn at least $120 per week.

1 answer:

agasfer [191]3 years ago

6 0

Answer:

Step-by-step explanation:

let r= number of hours worked at the restaurant

y=number of work hours in the yard

Maximum number of work per week=r+y ≤ 15

Lia must work at least 5 hours in the restaurant

x ≥ 5

Lia wants to earn at least $120 $8/hr in the restaurant and $12/hr in the yard:

8r + 12y ≥ 120

What is the maximum of hours Lia can work in the restaurant and still make at least 120 hours?

Assume Lia worked all 15 hours in the restaurant

she would earn $8*15hours = $120.

The maximum number of hours she can work in the restaurant is therefore 15 hours

What is the maximum amount of money Lia can earn in a week?

Lia has to work a minimum of 5 hours in the restaurant. She earns more from yard work, so she should devote the rest of her available work hours to yard work.

That means that, given her 15 hour work limit, she will maximize her income by working 5 hours in the restaurant and 10 hours in the yard.

5r+10y

5*8 + 10*12

= 40 + 120

= $160

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Answer:

See explanation

Step-by-step explanation:

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Answer:

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Zielflug [23.3K]

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From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

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Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

Answer:

a

$\= x = 13.4$ . $E = 2.3$

b

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

=> $\= x = 13.4$

Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

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Answer: 0.714286

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