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BigorU [14]
1 year ago
6

Anna wanted to buy a camera. The first discount store sold her favorite camera for $95. The second store sold the same camera fo

r $115, but it was on sale for 20% off. The third store carried the camera for $105 but offered it at 10% off with a coupon. which store had the better buy?
Mathematics
1 answer:
lions [1.4K]1 year ago
5 0

The second store offered the better buy at the price of $92.00.

What is better buy?

Better buy refers to the lowest price out of the prices offered by the three stores.

In order to determine the lowest price, the no discount price of the first store needs to be compared to the prices of  two other stores, bearing that after-discount price is the pre-discount price multiplied by 1 minus the discount rate.

First store price=$95.00

Second store after-discount price=pre-tax discount price*(1-discount rate)

pre-discount price=$115

discount rate=20%

Second store after-discount price=$115*(1-20%)

Second store after-discount price=$92.00

Third store after-discount price=pre-tax discount price*(1-discount rate)

pre-discount price=$105

discount rate=10%

Third store after-discount price=$105*(1-10%)

Third store after-discount price=$94.50

Find out more about discounted price on:brainly.com/question/1153322

#SPJ1

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Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the
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Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million)     Number of companies

25 up to 35                                                    4

35 up to 45                                                    19

45 up to 55                                                    27

55 up to 65                                                    16

65 up to 75                                                     9

TOTAL                                                            75

Let's find the probabilities first:

P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}

For 35 up to 45

P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}

For 45 up to 55

P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}

For 55 up to 65

P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}

For 65 up to 75

P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}

Chi-Square Table can be computed as follows:

Expense   No of   Probabilities(P)  Expe                (O-E)^2   \dfrac{(O-E)^2}{E}

             compa                                 cted E (n*p)

             nies (O)  

25-35           4      0.0612    75*0.0612 = 4.59        0.3481       0.0758

35-45           19     0.2218   75*0.2218 = 16.635     5.5932       0.3362

45-55           27     0.3568   75*0.3568 = 26.76     0.0576      0.021

55-65           16      0.2552   75*0.2552 = 19.14      9.8596      0.5151

65-75           9      0.0839     75*0.0839 = 6.2925   7.331         1.1650

                                                                                           \sum \dfrac{(O-E)^2}{E}= 2.0492                                                                                                      

Using the Chi-square formula:

X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square  \ X^2 = 2.0942

Null hypothesis:

H_o: \text{The population of advertising expenses follows a normal distribution}

Alternative hypothesis:  

H_a: \text{The population of advertising expenses does not follows a normal distribution}

Assume that:

\alpha = 0.02

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from X^2 = 11.667

Decision rule: To reject H_o  \  if \  X^2  test statistics is greater than X^2 tabulated.

Conclusion: Since X^2 = 2.0942 is less than critical value 11.667. Then we fail to reject H_o

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