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Gekata [30.6K]
3 years ago
6

A sporting goods store buys a basketball for $10.00 and then adds a 110% markup to the price. How much will the basketball sell

for in the store.​
Mathematics
2 answers:
wolverine [178]3 years ago
4 0

Answer:

The price will be $21 on the store.

Step-by-step explanation:

In order to solve this question we can use a rule of three in such a way that $10 represents 100% and "x" represents 110%, we then sum "x" to the value the ball was bought which is 10 and that will be it's price in the store. We have:

$10 -> 100%

$x -> 110%

10/x = 100/110

100*x = 110*10

100*x = 1100

x = 1100/100 = 11

store price = 10 + x = 10 + 11 = 21

The price will be $21 on the store.

sweet-ann [11.9K]3 years ago
3 0

Answer: The basketball will sell for $21 .

Step-by-step explanation:

Hi, to answer this question, first, we have to multiply the cost of the basketball (10) by the percent markup in decimal form (divided by 100).

Mathematically speaking:

10 x (110/100) = $11

Now, we have to add the markup amount (11) to the cost (10) to obtain the selling price:

10 +11 = $21

The basketball will sell for $21 .

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Answer:

1) χ² ≥ 11.07

2) Goodness of fit test, df: χ²_{3}

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The goodness of fit test has more degrees of freedom than the independence test.

3) e_{females.} = 80

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Step-by-step explanation:

Hello!

1)

The researcher took a sample of n=60 people and made them taste proof samples of six different brands of pizza and choose their favorite brand, their choose was recorded. So the study variable is the following:

X: favorite pizza brand, categorized in brand 1, brand 2, brand 3, brand 4, brand 5 and brand 6.

The Chi-square goodness of fit test is done with the following statistic:

χ²= ∑\frac{(O_i-E_i)^2}{E_i} ≈χ²_{k-1}

Where k represents the number of categories of the study variable. In this example k= 6.

Remember, the rejection region for the Chi-square tests of "goodnedd of fit", "independence", and "homogeneity" is allways one-tailed to the right. So you will only have one critical value.

χ²_{k-1; 1 - \alpha }

χ²_{6-1; 1 - 0.05 }

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This means thar the rejection region is χ² ≥ 11.07

If the Chi-Square statistic is equal or greather than 11.07, then you reject the null hypothesis.

2)

The statistic for the goodness of fit is:

χ²= ∑\frac{(O_i-E_i)^2}{E_i} ≈χ²_{k-1}

Degrees of freedom: χ²_{k-1}

In the example: k= 4 (the variable has 4 categories)

χ²_{4-1} = χ²_{3}

The statistic for the independence test is:

χ²= ∑∑\frac{(O_ij-E_ij)^2}{E_ij} ≈χ²_{(r-1)(c-1)} ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

If the information is in a contingency table

r= represents the total of rows

c= represents the total of columns

In the example: c= 2 and r= 2

Degrees of freedom: χ²_{(r-1)(c-1)}

χ²_{(2-1)(2-1)} = χ²_{1}

The goodness of fit test has more degrees of freedom than the independence test.

3)

To calculate the expected frecuencies for the independence test you have to use the following formula.

e_{ij} = n * P_i. * P_.j = n * \frac{o_i.}{n} * \frac{o_.j}{n}

Where o_i. represents the total observations of the i-row, o_.j represents the total of observations of the j-column and n is the sample size.

Now, this is for the expected frequencies in the body of the contingency table, this means the observed and expected frequencies for each crossing of categories is not the same.

On the other hand, you would have the totals of each category and population in the margins of the table (subtotals), this is the same when looking at the observed frequencies and the expected frequencies. Wich means that the expected frequency for the total of a population is the same as the observed frequency of said population. A quick method to check if your calculations of the expected frequencies for one category/population are correct is to add them, if the sum results in the subtotal of that category/population, it means that you have calculated the expected frequencies correctly.

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e_{females.} = 80

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There are two ways of writing down a null hypothesis for the independence test:

Way 1: using colloquial language

H₀: The variables X and Y are independent

Way 2: Symbolically

H₀: P_{ij}= P_{i.} * P_{.j} ∀ i= 1, 2, ..., r and j= 1, 2, ..., c

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5)

In this example, you have an independence test for two variables.

Variable 1, has 3 categories

Variable 2, has 4 categories

To follow the notation, let's say that variable 1 is in the rows and variable 2 is in the columns of the contingency table.

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χ²= ∑∑\frac{(O_ij-E_ij)^2}{E_ij} ≈χ²_{(r-1)(c-1)} ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

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χ²_{(3-1)(4-1)} = χ²_{6}

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