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3 years ago

A triangular playground has angles with measures in the ratio 8 : 6 : 4. What is the measure of the smallest angle?

Mathematics

2 answers:

Airida [17]3 years ago

8 0

Answer:

20°

Step-by-step explanation:

One way of doing this is to find the constant of proportionality, k:

8k + 6k + 4k = 90° Then 18k = 90°, and k turns out to be 90/18, or 5.

Then the angles are 8(5), 6(5) and 4(5). The smallest of these angles is thus 20°

SVETLANKA909090 [29]3 years ago

5 0

let's recall that the sum of all interior angles in a triangle is 180°.

we know the angles are in a 8:6:4 ratio, so we simply divide 180 by (8+6+4) and then distribute accordingly.

$\bf 8:6:4\qquad \qquad \left( 8\cdot \cfrac{180}{8+6+4} \right) : \left( 6\cdot \cfrac{180}{8+6+4} \right) : \left( 4\cdot \cfrac{180}{8+6+4} \right) \\\\\\ (8\cdot 10):(6\cdot 10):(4\cdot 10)\implies 80~:~60~:~\stackrel{\textit{smallest}}{40}$

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marta [7]

Answer:

no solution

Step-by-step explanation:

no solution x=0

4 0

3 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!

4 0

3 years ago

A package of 3 pairs of insulated gloves costs $20.37. What is the unit price of the pairs of gloves?

Alinara [238K]

Answer: 6.79

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

How do you convert 1 1/3 as a decimal

Nesterboy [21]

1/3 would be 0.333...

1+0.333...=1.333

3 0

4 years ago

Read 2 more answers

For each of the following studies, say whether you would use a t test for dependent means or a t test for independent means.

Shtirlitz [24]

Answer:

1. T test for independent means

2. T test for dependent means

3. T test for dependent means

Step-by-step explanation:

In number 1, the two groups are unrelated. The first group has 25 subjects and they're all unemployed. The second group has 24 subjects and their employment status is not stated and might not be the same all through. Also, the first group is receiving a new type of job skills program while the second group is taking the standard job skills program.

- The groups in the experiment are unrelated

- The tests in the research are unrelated

- The purpose of the research is unreasonable - the researcher seeks to measure how well all 49 subjects perform on 'a' job skills test! No comparison between the scores or mean scores of the two groups.

In number 2, the researcher uses the same subjects and also measures the same variable but twice. This is a good example of a study where the t test for dependent means can be taken. Same applies in case 3.

5 0

3 years ago