Help please..................thank you

Are These two aright?

Over [174]

Answer:

◑__◐

Step-by-step explanation:

3 0

3 years ago

in new england 47% of the houses have a garage and 36% of the houses have a garage and a backyard what is the probability that a

miv72 [106K]

Answer:

The probability that a house has a backyard given that has a garage is 76.60%.

Step-by-step explanation:

Given that in New England 47% of the houses have a garage and 36% of the houses have a garage and a backyard, to determine what is the probability that a house has a backyard given that has a garage the following crossed multiplication must be performed:

47 = 100

36 = X

(36 x 100) / 47 = X

3,600 / 47 = X

76.5957 = X

Therefore, the probability that a house has a backyard given that has a garage is 76.60%.

7 0

3 years ago

The pipe fitting industry had 546.5 thousand jobs in 2015 and is expected to decline at an average rate of 3 thousand jobs per y

lesya692 [45]

Answer:

The amount of jobs from fitting industry shall decline in 5.5 percent from 2015 to 2025.

Step-by-step explanation:

Due to the assumption of a yearly average rate, a linear function model shall be used. The expected amount of jobs ( $n$ ) after a certain amount of years (t) is given by the following formula:

$n = n_{o} + \frac{\Delta n}{\Delta t}\cdot t$

Where:

$n_{o}$ - Initial amount of jobs in pipe fitting industry, measured in thousands.

$\frac{\Delta n}{\Delta t}$ - Average yearly rate, measured in thousands per year. (A decline is indicated by a negative sign)

If $n_{o} = 546.5$ , $t = 2025-2015 = 10\,years$ and $\frac{\Delta n}{\Delta t} = -3\,\frac{1}{years}$ , then:

$n = 546.5+\left(-3\,\frac{1}{year}\right)\cdot (10\,years)$

$n = 516.5$

The percent change in jobs from pipe fitting industry is calculated as follows:

$\%n = \left(1-\frac{n}{n_{o}}\right)\times 100\,\%$

$\% n = \left(1-\frac{516.5}{546.5}\right)\times 100\,\%$

$\%n = 5.5\,\%$

The amount of jobs from fitting industry shall decline in 5.5 percent from 2015 to 2025.

4 0

3 years ago

Please help i have to write and i dont understand

Lostsunrise [7]

Answer:

A. This is a vertical angle set. As you can see, they are cut by the transversal and have the exact same interior angle, so it is a vertical angle.

B. Since these too are vertical angles, all you have to do is just put each expression on the sides of an equation like so:

5y-29=3y+19

Once you do this, you just need to simplify the equation by putting like terms together, and then you will find the solution for y, which is 24.

C. We know that the solution for y is 24, so we need to plug it into the equation to find, the angle, so if we do 3*24-19, we get 53, so we need to subtract that from 180 degrees, which is the total angle of the straight line, then we can get 127 degrees, which is what angles <7 and <8 are.

Step-by-step explanation:

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!

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

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.

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.

The expected frequency for the total of females is 80

Using the formula:

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

$e_{females.} = 80$

4)

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)

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.

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