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KiRa [710]
2 years ago
13

Becky and Michele are both shopping for a new car at two different dealerships. Dealership A is offering $500 cashback on any pu

rchase, while dealership B
is offering $1,000 cash back. The tax rate is 5% at dealership A but 8% at Dealership B. Becky wants to buy a car that is $15,000, and Michele is planning to buy a car that costs $20,000. Use algebraic expressions to help you answer the following question.

1. At which dealership will Becky get the better deal? how much does she save?

2. At which dealership will Michele get the better deal? how much does she save?

3. What gerneralization can you make that would help any shopper know which dealership has the better deal? At which price point would the two deals be equal.
Mathematics
2 answers:
USPshnik [31]2 years ago
7 0
Dealership A : 500 cash back , 5% tax
Becky : 15,000 + 0.05(15000) - 500 = 15,000 + 750 - 500 = 15,250
Michele : 20,000 + 0.05(20,000) - 500 = 20,000 + 1000 - 500 = 20500

dealership B : 1000 cash back, 8% tax
Becky : 15,000 + 0.08(15,000) - 1000 = 15000 + 1200 - 1000 = 15,200
Michele : 20,000 + 0.08(20,000) - 1000 = 20,000 + 1600 - 1000 = 20600

1. Becky will get the better deal at dealership B.....she saves 50 bucks
2. Michele will get the better deal at dealership A...she saves 100 bucks
3. ?....sorry
motikmotik2 years ago
7 0
A. First write an algebraic expression to represent each dealership’s special. Dealership A p + 0.05p - 500 = 1.05p - 500 Dealership B p + 0.08p - 1000 = 1.08p - 1000 Now substitute the price that Becky plans to pay to see which deal is better for her. Dealership A: 1.05(15,000) - 500 = 15,750 - 500 = 15,250 Dealership B: 1.08(15,000) - 1000 = 16,200 - 1000 = 15,200 Dealership A > Dealership B So, Becky will get a better deal with Dealership B, since it is $50 cheaper. b. Substitute the price that Michele plans to pay into each expression. Dealership A: 1.05(20,000) - 500 = 21,000 - 500 = 20,500 Dealership B: 1.08(20,000) - 1000 = 21,600 - 1000 = 20,600 Dealership A < Dealership B So, Michele will get a better deal with Dealership A, since it is $100 cheaper. c. The two deals are equivalent when the additional 3% tax equals $500 (0.03p = 500), which occurs when the purchase price is $16,667. Students’ generalizations may suggest that any shopper planning to buy a car with a price less than $16,667 will get the better deal at Dealership B, while anyone planning to buy a car at a price higher than $16,667 will get a better deal at Dealership A.
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Every week a technician in a lab needs to test the scales in the lab to make sure that they are accurate. She uses a standard ma
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Answer:

a) \bar X=10.65

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Mode= 10.7

b) Range = 11.8-8.3=3.5

s= 0.948

c)IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

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Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got: \bar X=10.65

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

Median =\frac{10.7+10.7}{2}=10.7

The mode would be the most repeated value on this case:

Mode= 10.7

Part b

The range is defined as Range =Max-Min and if we replace we got:

Range = 11.8-8.3=3.5

We can calculate the standard deviation with the following formula:

s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s= 0.948

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and Q_1= \frac{10.6+10.7}{2}=10.55

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got Q_3 = \frac{10.9+11.2}{2}=11.05

Then we can find the IQR like this:

IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0
2 years ago
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