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slamgirl [31]
3 years ago
14

To help pay for the next school event, Terry and Tony want to make a

Mathematics
1 answer:
Sloan [31]3 years ago
5 0

a) Equation: Nx=475

b)

1st case: 119 tickets

2nd case: 106 tickets

3rd case: 95 tickets

Step-by-step explanation:

a)

Terry and Tony want to make a profit of

P = $475

in total from the ticket sale.

Here we call x the price per ticket.

Then we also call N the number of tickets sold: therefore, this total profit made by Terry and Tony from the ticket sale is the product between the tickets sold, N, and the price per ticket, x:

Nx

And since they want to do a profit of P = $475, we can therefore write

Nx=475

b)

The equation of part a) can be re-arranged as

N=\frac{475}{x}

where N is the number of tickets sold and x the price per ticket.

The 1st group of student want to charge  $4.00 per ticket, so

x = 4.00

And so the number of tickets that they should sell in this case is

N=\frac{475}{4.00}=118.8 \sim 119

The 2nd group of student want to charge  $4.50 per ticket, so

x = 4.50

And so the number of tickets that they should sell in this case is

N=\frac{475}{4.50}=105.6 \sim 106

Terry and Tony want to charge $5.00 per ticket, so

x = 5.00

And so the number of tickets that they should sell in this case is

N=\frac{475}{5.00}=95

Learn more about equations:

brainly.com/question/11306893

brainly.com/question/10387593

#LearnwithBrainly

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The mode of the following data is 5 ?<br> 0, 2, 2, 5, 2, 5, 0
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3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

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The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

7 0
3 years ago
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