.

The South Coast Air Basin (Los Angeles, Orange, and San Bernardino counties) does not meet the air quality standards for carbon

Archy [21]

Answer:

0.2225 gr/mile

Step-by-step explanation:

Let's work out first the amount of CO sent out in 1990.

The population we estimated in 12.5 million people with a total of driven miles per year of about

12.5 million*8,700 = 108,750 million miles.

With an CO emission factor of 0.9 g per mile, we would have a total of CO emitted rounding 0.9*180,750 = 97,875 million grams

Now, we must estimate the population for 2020.

Since we are assuming an exponential growth, the population in year t is given by a function

$\bf P(t)= Ce^{kt}$

where C and k are constants to be determined.

We can take 1980 as year 0. This way calculations are lighter. 1990 is year 10 and 2020 is year 20.

So P(0) = 10.3 and C=10.3

So far we have

$\bf P(t)= 10.3e^{kt}$

Given that P(10)=12.5

$\bf 10.3e^{k*10}=12.5\rightarrow e^{10k}=\frac{12.5}{10.3}=1.21359\rightarrow \\10k=log(1.21359)\rightarrow k=0.01936$

And the function that models the population growth is

$\bf P(t)= 10.3e^{0.01936t}$

We need P(20)

$\bf P(20)= 10.3e^{0.01936*20}=10.3e^{0.38717}=15.17\;million$

If the miles driven per person per year remains constant at 8700 mi/yr.person, then we have a total miles driven of

15.17*8,700=131,979 million miles, so the CO emitted would be 0.9*131,979=118,781.1 million grams.

The 30% of the CO sent out in 1990 is 0.3*97,875=29,362.5 million grams.

We must reduce 118,781.1 down to 29,362.5

Hence the new CO emission factor would be

29,362.5/131,979 = 0.2225 gr/mile

6 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!

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

Use each set of line segments to sketch a triangle. If a triangle cannot be drawn, explain why.

melamori03 [73]

I don't think it can becauze of u count the squares and try 2 plan out the triangle u cant or i should say me i cant see a triangle being formed

8 0

3 years ago

Matt Kaminsky bought a Volvo that is 3 3/4 times as expensive as the car his parents bought. If his parents paid $8,000 for thei

Anna35 [415]

What we know... Parents car is $8,000 --- Volvo costs $8,000 x 3 3/4

3 3/4 is the same as 3.75

$8,000 x 3.75 = $30,000

Matt Kaminsky's Volvo costs $30,000

5 0

3 years ago

Find the exact value of tan(theta) for an angle (theta) with sec(theta) = 3 and with its terminal side in Quadrant I.

stiks02 [169]

Answer:

c

Step-by-step explanation:

1 + tan²theta = sec²theta

tan²theta = 3² - 1

tan²theta = 8

tan theta = sqrt(8)

Positive because Quadrant 1

sqrt(8) = sqrt(4×2) = sqrt(4)×sqrt(2)

= 2×sqrt(2)

3 0

3 years ago