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Andrew [12]
4 years ago
8

In the flexed arm​ condition, what is the probability that a consumer has a choice score of

Mathematics
1 answer:
Flauer [41]4 years ago
6 0
<span>A study was conducted to determine if consumers are more likely to choose a vice product (e.g., a candy bar) when their arm is flexed (as when carrying a shopping basket) than when their arm is extended(as when pushing a shopping cart). The study measured choice scores (on a scale of 0 to 100, where higher scores indicate a greater preference for vice options) for consumer shopping under each of the two conditions. The average choice score for consumers with a flexed arm was 60, while the average for consumers with an extended arm was 43. For both conditions, assume that the standard deviation of the choice scores is 6. Also assume that both distributions are approximately normally distributed. Complete part a. a. In the flexed arm condition, what is the probability that a consumer has a choice score of 56 or greater? what is the probability that a consumer in the extended arm condition has a choice score of 56 or greater?</span>
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Answer:

I believe the answer is 8

Step-by-step explanation:

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For questions 9 and 11, solve for the remaining angles.
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Answer:

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*These angles are rounded to the nearest whole*

Step-by-step explanation:

Because you are only given sides you can find an individual angle measures with the inverse law of cosines.

Remember the side opposite of the angle corresponds to that angle.

a² = b² + c² - 2bc cos(A) →

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A group of children, adults, and senior citizens attended three different art exhibits that have different ticket prices for eac
jok3333 [9.3K]

Answer:

Option D) c = 100, a = 50, s = 20

Step-by-step explanation:

we have

6c + 7a + 2s = 990  -----> equation A

3c + 3a + 4s = 430  -----> equation B

2c + 4a + 4s = 480 ----> equation C

Isolate variable s in equation A

2s = 990 -6c-7a  

s=495-3c-3.5a -----> equation D

Substitute equation D in equation B and in equation C

3c + 3a + 4(495-3c-3.5a) = 430

3c + 3a +1,980-12c-14a= 430

-9c-11a=-1,550 -----> equation E

2c + 4a + 4(495-3c-3.5a) = 480

2c+4a+1,980-12c-14a=480

-10c-10a=-1,500 -----> 10c+10a=1,500 -----> equation F

Solve the system of equations E and F by graphing

using a graphing tool

Let

The variable c -----> the x-axis

The variable a ----> the y-axis

The solution is the point (50,100)

see the attached figure

so

c=50, a=100  

Find the value of s

s=495-3(50)-3.5(100)=-5 -----> is not make sense

therefore

Let

The variable c -----> the y-axis

The variable a ----> the x-axis

The solution is the point (50,100)

so

a=50, c=100  

Find the value of s

s=495-3(100)-3.5(50)=20

therefore

The solution is

c=100,a=50,s=20

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Answer:

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