Researchers wanted to determine if having a refrigerator in the bedroom is associated with obesity. The researchers administered a questionnaire to 353 twelve-year-old adolescents. After analyzing the results, the researchers determined that the body mass index of the adolescents who had a refrigerator in their bedroom was significantly higher than that of the adolescents who did not have a refrigerator in their bedroom. Complete parts (a) through (e) below. Why is this an observational study?

(A) This is an observational study because the researchers observe the behavior of the individuals in the study without trying to influence an explanatory variable of the study.
(B) This is an observational study because the researchers try to influence the outcome of the study.
(C) It is not an observational study

Answers

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Final answer:

The question's study is an observational study because researchers observed the behavior of individuals without influencing any aspect of the study, such as the presence of a refrigerator or the adolescent's BMI.

Explanation:

This study is indeed considered an observational study. The answer is (A) This is an observational study because the researchers observe the behavior of the individuals in the study without trying to influence an explanatory variable of the study. In the study described, researchers merely observed and recorded the behavior (having a refrigerator in the bedroom) and the outcome (body mass index or BMI). There was no attempt by the researchers to manipulate or control either the presence of the refrigerator or the adolescent's BMI, which is characteristic of an observational study.

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The number of milligrams of a certain medicine a veterinarian gives to a dog varies directly with the weight of the dog.If the veterinarian gives a 30-pound dog 5 milligram of the medicine, which equation relates the weight, w, and the
dosage, o?

Answers

Question: The number of milligrams of a certain medicine a veterinarian gives to a dog varies directly with the weight of the dog. If the veterinarian gives a 30-pound dog 3/5 milligram of the medicine, which equation relates the weight,w, and the dosage, d?

Answer: d= 1/50w

Explanation: I took the test in Edgenuity.

Hope this helps!

2/3 of the pizza was left over last night. Then, Marcy ate another 2/9 of it. How much is left over now?

Answers

Answer:

4/9 is left over.

Step-by-step explanation:

First make the denominators the same. 2/3=6/9. 6/9-2/9=4/9.

You have a plate of 50 cookies. Ten have chocolate chips and 14 have pecans. On the cookies mentioned in the preceding sentence, 6 have both chocolate chips and pecans. You select a cookie at random. What is the probability that your cookie has chocolate chips or pecans

Answers

Answer:

The probability is 0.36

Step-by-step explanation:

For calculated the probability that a cookie has chocolate chips or pecans, we need to divide the number of cookies that has chocolate chips or pecans between the total number of cookies.

So, the number of cookies that has chocolate chips or pecans is calculate as: A + B - C

Where A are the number of cookies that have chocolate chips, B are the number of cookies that have Pecans and C are the number of cookies that have both chocolate chips and pecans. Then, the number of cookies that has chocolate chips or pecans is:

10 + 14 - 6 = 18

Finally, the total number of cookies is 50 so the probability is:

$(18)/(50) =0.36$

a group of students is donating blood during a blood drive. a student has 9/20 probability of having type 'o' blood and a 2/5 probability of having type 'A'.what is the probability that a student has type 'o' or type 'a' blood and why?

Answers

For the answer to the question above,
the answer is "The student can have only one blood type, so the actual events are mutually exclusive. "

The probabilities are not mutually exclusive. Based on the group

P(Type O) = 9/20 = 45% or 0.45
P(Type A) = 2/5 = 40% or 0.40
P(Other) = 3/20 = 15% or 0.15
I hope my answer helped you. Feel free to ask more questions. Have a nice day!

Hello there.

A group of students is donating blood during a blood drive. a student has 9/20 probability of having type 'o' blood and a 2/5 probability of having type 'A'.what is the probability that a student has type 'o' or type 'a' blood and why?

The student can have only one blood type, so the actual events are mutually exclusive.

Consider the following functions. f(x) = x − 3, g(x) = x2 Find (f + g)(x). Find the domain of (f + g)(x). (Enter your answer using interval notation.) Find (f − g)(x). Find the domain of (f − g)(x). (Enter your answer using interval notation.) Find (fg)(x). Find the domain of (fg)(x). (Enter your answer using interval notation.) Find f g (x). Find the domain of f g (x). (Enter your answer using interval notation.)

Answers

Answer:

$(f+g)(x)=x-3+x^2$ ; Domain = (-∞, ∞)

$(f-g)(x)=x-3-x^2$ ; Domain = (-∞, ∞)

$(fg)(x)=x^3-3x^2$ ; Domain = (-∞, ∞)

$((f)/(g))(x)=(x-3)/(x^2)$ ; Domain = (-∞,0)∪(0, ∞)

Step-by-step explanation:

The given functions are

$f(x)=x-3$

$g(x)=x^2$

$(f+g)(x)=f(x)+g(x)$

Substitute the values of the given functions.

$(f+g)(x)=(x-3)+x^2$

$(f+g)(x)=x-3+x^2$

The function $(f+g)(x)=x-3+x^2$ is a polynomial which is defined for all real values x.

Domain of (f+g)(x) = (-∞, ∞)

$(f-g)(x)=f(x)-g(x)$

Substitute the values of the given functions.

$(f-g)(x)=(x-3)-x^2$

$(f-g)(x)=x-3-x^2$

The function $(f-g)(x)=x-3-x^2$ is a polynomial which is defined for all real values x.

Domain of (f-g)(x) = (-∞, ∞)

$(fg)(x)=f(x)g(x)$

Substitute the values of the given functions.

$(fg)(x)=(x-3)x^2$

$(fg)(x)=x^3-3x^2$

The function $(fg)(x)=x^3-3x^2$ is a polynomial which is defined for all real values x.

Domain of (fg)(x) = (-∞, ∞)

$((f)/(g))(x)=(f(x))/(g(x))$

Substitute the values of the given functions.

$((f)/(g))(x)=(x-3)/(x^2)$

The function $((f)/(g))(x)=(x-3)/(x^2)$ is a rational function which is defined for all real values x except 0.

Domain of (f/g)(x) = (-∞,0)∪(0, ∞)

$(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

$(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

$(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

$f(g(x)) = x^2 - 3$ , domain: all real numbers.

To find (f + g)(x), we need to add the functions f(x) and g(x).

The function f(x) = x - 3 and the function $g(x) = x^2.$

So, $(f + g)(x) = f(x) + g(x) = (x - 3) + (x^2).$

Expanding this equation, we get $(f + g)(x) = x^2 + x - 3.$

To find the domain of (f + g)(x), we need to consider the domain of the individual functions f(x) and g(x).

Since both f(x) = x - 3 and $g(x) = x^2$ are defined for all real numbers, the domain of (f + g)(x) is also all real numbers.

To find (f - g)(x), we need to subtract the function g(x) from f(x).

So, $(f - g)(x) = f(x) - g(x) = (x - 3) - (x^2).$

Expanding this equation, we get $(f - g)(x) = -x^2 + x - 3.$

The domain of (f - g)(x) is also all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find (fg)(x), we need to multiply the functions f(x) and g(x).

So, $(fg)(x) = f(x) * g(x) = (x - 3) * (x^2).$

Expanding this equation, we get $(fg)(x) = x^3 - 3x^2.$

The domain of (fg)(x) is all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find f(g(x)), we need to substitute g(x) into the function f(x).

So, $f(g(x)) = f(x^2) = x^2 - 3.$

The domain of f(g(x)) is also all real numbers, as $g(x) = x^2$ is defined for all real numbers, and f(x) = x - 3 is defined for all real numbers.

In summary:

- $(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

- $(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

- $(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

- $f(g(x)) = x^2 - 3$ , domain: all real numbers.

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The means and mean absolute deviations of the amount of rain that fell each day in a local city, last week and this week, are shown below. Means and Mean Absolute Deviations of Rainfall Last Week and This Week Last Week This Week Mean 19.7 cm 19.3 cm Mean Absolute Deviation 4.6 cm 5.2 cm The difference in the mean rainfall is approximately what percent of the mean absolute deviation of the data sets? 8% 23% 27% 67%

Answers

Answer:

Step-by-step explanation:

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