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zimovet [89]
3 years ago
14

A study published in 2010 showed that city dwellers have a higher risk of developing anxiety disorders and a higher risk of deve

loping mood disorders than those who live in the country. A follow-up study published in 2011 used brain scans of city dwellers and country dwellers as they took a difficult math test.1 To increase the stress of the participants, those conducting the study tried to humiliate the participants by telling them how poorly they were doing on the test. The brain scans showed very different levels of activity in stress centers of the brain, with the urban dwellers having greater brain activity than rural dwellers in areas that react to stress.
Required:

a. Is the 2010 study an experiment or an observational study?
i. Experiment
ii. Observational study

b. Can we conclude from the 2010 study that living in a city increases a person's likelihood of developing an anxiety disorder or mood disorder?
i. Yes
ii. No

c. Is the 2011 study an experiment or an observational study?
i. Experiment
ii. Observational study

d. Can we conclude from the 2011 study that living in a city increases activity in stress centers of the brain when a person is under stress?
i. Yes
ii. No
Mathematics
1 answer:
user100 [1]3 years ago
4 0

Step-by-step explanation:

a.It is a controlled experiment because it is clearly seen that an established hypothesis or study is being verified through an experiment, now in this case the hypothesis is tested through an experiment where excessive stress is placed on the Participants during their math exam to effectively verify that people in urban or city areas live with more stress than people in rural areas.

2..As now that the previously established hypothesis has been verified, it is concluded that people who live in the city live with more anxiety and stress than people who live in rural areas.

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Only 12 people can ride a trolley at one time. This is 20% of the people waiting to ride. How many people are waiting to ride?​
Thepotemich [5.8K]

Answer:

2 people are waiting if you want a whole number

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2 years ago
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The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars ar
Korolek [52]

Complete Question:

The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars are selected at random to be inspected. Do not simplify your answers. Leave in combinatorics form. What is the probability that:

a. Every car selected is in poor shape

b. At least two cars selected are in good shape.

c. Exactly three cars selected are in great shape.

d. Two cars selected are in great shape and two are in good shape.

e. One car selected is in good shape but the other 3 selected are in poor shape.

Answer:

a

   P_A = \frac{^4 C_4}{^{30}C_{4}}

b

  P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}

c

  P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}

d

   P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}

e

   P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}

Step-by-step explanation:

From the question we are told that

 The number of car in the parking lot is  n =  30  

  The number of cars in great shape is  k =  10

   The number of cars in good shape is r = 16

    The number of cars in poor shape is q = 4

The number of cars that were selected at random is N= 4

Considering question a

    Generally the number of way of selecting four cars that are in a poor shape from number of cars that are in poor shape is

            ^{q} C_{N}

=>        ^{4} C_{4}

Here C stands for combination.

 Generally the number of way of selecting four cars that are in a poor shape from total number of cars  in the parking lot is

          ^{n} C_{N}

=>      ^{30} C_{4}

Generally the probability that every car selected is in poor shape  is mathematically represented as

       P_A = \frac{^4 C_4}{^{30}C_{4}}

Considering question b

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{2}

=>        ^{16} C_{2}

Here C stands for combination.

 Generally the number of way of selecting the remaining 2 cars  from the remaining number of cars  in the parking lot is

          ^{n-r} C_{2}

=>      ^{30-16} C_{2}

=>      ^{14} C_{2}

Generally the number of way of selecting 3 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{3}

=>        ^{16} C_{3}

 Generally the number of way of selecting the remaining 1 cars  from the remaining number of cars  in the parking lot is

          ^{n-r} C_{1}

=>      ^{30-16} C_{1}

=>      ^{14} C_{1}

Generally the number of way of selecting 4 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{4}

=>        ^{16} C_{4}

Generally the probability that at least two cars selected are in good shape

       P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}

Considering question c

Generally the number of way of selecting 3 cars that are in great shape from number of cars that are in great shape is      

            ^{k} C_{3}

=>        ^{10} C_{3}

 Generally the number of way of selecting the remaining 1 cars  from the remaining number of cars  in the parking lot is

          ^{n-k} C_{1}

=>      ^{30-10} C_{1}

=>      ^{20} C_{1}

Generally the probability of selecting exactly three cars selected are in great shape is

        P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}

Considering question d

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{2}

=>        ^{16} C_{2}

Generally the number of way of selecting 2 cars that are in great shape from number of cars that are in great shape is      

            ^{k} C_{2}

=>        ^{10} C_{2}

Generally the probability that two cars selected are in great shape and two are in good shape.

              P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}

Considering question e

Generally the number of way of selecting 1 cars that is in good shape from number of cars that are in good shape is

            ^{r} C_{1}

=>        ^{16} C_{1}

    Generally the number of way of selecting 3 cars that are in a poor shape from number of cars that are in poor shape is

            ^{q} C_{3}

=>        ^{4} C_{3}

Generally the probability that one car selected is in good shape but the other 3 selected are in poor shape is

         P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}

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Step-by-step explanation:

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Help please! the question is in the picture below for reference when answering! thank you so much!
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Answer:

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the way from A to C goes through B.

so, the total way of A to C is the sum of the way from A to B and then from B to C.

=>

(5x + 10) + (2x + 25) = 28

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