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2 years ago

The dimensions of one of the sails on a sailboat are 9 feet, 12 feet, and 15 feet. Is the sail a right triangle?

Mathematics

2 answers:

kipiarov [429]2 years ago

7 0

Answer:

yes

Step-by-step explanation:

Using the converse of Pythagoras' identity

If the square of the longest side is equal to the sum of the squares of the other 2 sides then the triangle is right.

longest side = 15 , then 15² = 225

9² + 12² = 81 + 144 = 225

Thus the sail is a right triangle.

Eduardwww [97]2 years ago

5 0

In order to test if it’s a right triangle, we can try to apply Pythagorean theorem, which is x^2+y^2=z^2
Plug in the values: 9^2+12^2=15^2 to get 81+144=225
By far we can tell that this formula works, meaning that the sail is a right triangle.

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3 years ago

What is the average rate of change of the functionf(x)=20(14)x from x = 1 to x = 2?

atroni [7]

<h2>Answer-Average rate of change(A(x)) of f(x) over a interval [a,b] is given by:</h2><h2 /><h2>A(x) = \frac{f(b)-f(a)}{b-a}A(x)= </h2><h2>b−a</h2><h2>f(b)−f(a)</h2><h2> </h2><h2> </h2><h2 /><h2>Given the function:</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^xf(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>x</h2><h2> </h2><h2 /><h2>We have to find the average rate of change from x = 1 to x= 2</h2><h2 /><h2>At x = 1</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^1 = 5f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>1</h2><h2> =5</h2><h2 /><h2>At x = 2</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^2=20 \cdot \frac{1}{16} = 1.25f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>2</h2><h2> =20⋅ </h2><h2>16</h2><h2>1</h2><h2> </h2><h2> =1.25</h2><h2 /><h2>Substitute these in above formula we have;</h2><h2 /><h2>A(x) = \frac{f(2)-f(1)}{2-1}A(x)= </h2><h2>2−1</h2><h2>f(2)−f(1)</h2><h2> </h2><h2> </h2><h2 /><h2>⇒A(x) = \frac{1.25-5}{1}=-3.75A(x)= </h2><h2>1</h2><h2>1.25−5</h2><h2> </h2><h2> =−3.75</h2><h2 /><h2>therefore, average rate of change of the function f(x) from x = 1 to x = 2 is, -3.75</h2>

<h2>Please Mark me as brainlist. </h2>

8 0

2 years ago

A researcher used a sample of n = 60 individuals to determine whether there are any preferences among six brands of pizza. Each

Blizzard [7]

Answer:

1) χ² ≥ 11.07

2) Goodness of fit test, df: χ² $_{3}$

Independence test, df: χ² $_{1}$

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

3) $e_{females.} = 80$

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

5) χ² $_{6}$

Step-by-step explanation:

Hello!

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 $_{6-1; 1 - 0.05 }$

χ² $_{5; 0.95 }$ = 11.070

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.

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.

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.

The expected frequency for the total of females is 80

Using the formula:

(If the females are in a row) $e_{females.} = 100 * \frac{80}{100} * \frac{0}{100}$

$e_{females.} = 80$

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

This type of hypothesis follows from the definition of independent events, where if we have events A and B independent of each other, the probability of A and B is equal to the product of the probability of A and the probability of B, symbolically: P(A∩B) = P(A) * P(B)

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.

The statistic for this test is:

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

In the example: c= 3 and r= 4

Degrees of freedom: χ² $_{(r-1)(c-1)}$

χ² $_{(3-1)(4-1)}$ = χ² $_{6}$

I hope you have a SUPER day!

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