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Misha Larkins [42]
3 years ago
15

The photography club decided to go on a field trip to the Andy Warhol museum. They had to break into smaller groups once they go

t there. The first group had 25 students and 2 teachers. Their cost was $97.50. The second group had 32 students and 3 teachers. Their cost was $127. What is the cost of each student and each teacher?
Part A: Define your variables

Part B: Write a system of equations to represent this situation

Part C: Solve the system of equations you wrote in Part B.

Part D: Interpret what your answer means in context to this situation.
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0
A)
Let x represent the cost of 1 student, and y the cost of 1 teacher.

B)
In the first group, there's 25 students and 2 teachers. Their total cost is $97.50
So 25x + 2y = 97.50
In the second group, there's 32 students and 3 teachers. Their total cost is $127
So 32x + 3y = 127

We get the following system of equations:
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)

C)
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)

In equation (1)
25x + 2y = 97.50
25x + 2y - 2y = 97.50 - 2y
25x = 97.50 - 2y
25x / 25 = 97.50/25 - 2y/25
x = 3.9 - (2/25)y

In equation (2), let's replace x by its algebraic value
32x + 3y = 127
32(-2/25y + 3.9) + 3y = 127
11/25y + 124.8 = 127
11/25y + 124.8 - 124.8 = 127 - 124.8
11/25y = 2.2
(11/25y) / (11/25) = 2.2 / (11/25)
y = 5

x = -2/25y + 3.9
x = -2/25 * 5 + 3.9
x = 3.5

So the cost of each student is $3.5, and the cost of each teacher is $5.

Hope this helps! :)




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We are given that the Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age.

It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level.

(a) <u>Margin of error formula is given by;</u>

             Margin of Error =  Z_(_\frac{\alpha}{2}_)  \times \sqrt{\frac{\hat p(1-\hat p)}{n} }  

where, \alpha = level of significance = 1 - 0.95 = 0.05 or 5%

Standard of error =  \sqrt{\frac{\hat p(1-\hat p)}{n} }

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The critical value of z for level of significance of 2.5% is 1.96.

So, <em>Margin of Error </em>=  Z_(_\frac{\alpha}{2}_)  \times \sqrt{\frac{\hat p(1-\hat p)}{n} }  

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(b) Now, the pivotal quantity for 95% confidence interval for the population proportion who think that licensed drivers should be required to retake their road test once they turn 65 is given by;

                   P.Q. =  \frac{\hat p-p}{ \sqrt{\frac{\hat p(1-\hat p)}{n} }}  ~ N(0,1)

So, <u>95% confidence interval for p</u> =  \hat p \pm \text{Margin of error}

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