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Gre4nikov [31]
2 years ago
14

Jeanne babysits for $7 per hour. She also works as a tutor for $9 per hour. She is only allowed to work 18 hours per week. She w

ants to make at least $65. Let x be the hours worked babysitting and let y be the hours worked tutoring. Model the scenario with a system of inequalities. Please show all work. Thanks <3
Mathematics
2 answers:
lutik1710 [3]2 years ago
8 0
7x + 9y > = 65
x + y < = 18
nata0808 [166]2 years ago
6 0

Answer: x+y\leq18 and 7x+9y\geq65


Step-by-step explanation:

Let x be the hours worked babysitting and let y be the hours worked tutoring.

As Jeanne babysits for $7 per hour. She also works as a tutor for $9 per hour.

Total hours she can work= 18 hours

Thus,  x+y\leq18

She wants to make at least $65.

It gives that 7x+9y\geq65

Thus, the system of inequalities are written as

x+y\leq18

7x+9y\geq65


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The given function is
f(x) = 4x - 3/2
where
f(x) = number of assignments completed
x =  number of weeks required to complete the assignments

We want to find f⁻¹ (30) as an estimate of the number of weeks required to complete 30 assignments.
The procedure is as follows:

1. Set y = f(x)
   y = 4x - 3/2

2. Exchange x and y
   x = 4y - 3/2

3. Solve for y
   4y = x + 3/2
   y = (x +3/2)/4

4. Set y equal to f⁻¹ (x)
  f⁻¹ (x) = (x + 3/2)/4

5. Find f⁻¹ (30)
  f⁻¹ (30) = (30 + 3/2)/4 = 63/8 = 8 (approxmately)

Answer:
Pedro needs about 8 weeks to complete 30 assignments.

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2 years ago
Show how to make a ten to solve 13-7. Write the number sentence.
VARVARA [1.3K]
-7 + 3 + 10 = -4 + 10 = 10 - 4 = 6. 

To turn 13 into 10, you need to break it up into 10 + 3, which does not change the value of 13. Then, you add 3 to -7, which results in -4. Next, you add -4 to 10 (or rather, subtract 4 from 10), which results in 6. 

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2 years ago
Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in
likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation  

n=500 represent the random sample taken    

X=7 represent the households with three or more large-screen TVs

\hat p=\frac{7}{500}=0.014 estimated proportion of households with three or more large-screen TVs

\alpha=0.05 represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

Part b

Yes is possible. We hav that np>10 and n(1-p)>10 so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370

0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

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Answer:

Kindly check explanation

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