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faust18 [17]
3 years ago
7

The revenue each season from tickets at the theme park is represented by t(c)=5x. The cost to pay the employees each season is r

epresented by r(x)=(1.5)^x. Examine the graph of the combined function for total profit and estimate the profit after four seasons

Mathematics
1 answer:
butalik [34]3 years ago
3 0

Answer:

The profit after 4 seasons is 14.938 units

Step-by-step explanation:

To quickly solve this problem, we can use a calculator or any graphing tool

to examine both graphs

Revenue

t(x)=5x

Cost

r(x)=(1.5)^x

Profit

P = Revenue - costs = t(x) -r(x)

P = 5x - (1.5)^x

The profit after 4 seasons is 14.938 units

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Given that a=4, b=4 and c=-2 work out b(squared) -2ac
Nimfa-mama [501]

Answer:

32

Step-by-step explanation:

substitute the given values of a, b and c into the expression:

b² - 2ac = 4² - (2 x 4 x -2)

             = 16 - (-16)

             = 16 + 16

             = 32

3 0
2 years ago
PLEASE HELP/ANSWER! From 1980 to 1990, Lior’s weight increased by 25%. If his weight was k kilograms in 1990, what was it in 198
Elena-2011 [213]

The weight in 1980 is \frac{4k}{5} kilograms

<em><u>Solution:</u></em>

From 1980 to 1990, Lior’s weight increased by 25%

His weight is "k" kilograms in 1990

<em><u>To find: weight in 1980</u></em>

This is a percentage increase problem

Let "x" be the weight in kilograms in 1980

<em><u>The percentage increase is given by formula:</u></em>

\text{Percentage increase } = \frac{\text{Final value - initial value}}{\text{initial value}} \times 100

Here,

Initial value in 1980 = x

Final value in 1990 = k

Percentage increase = 25 %

<em><u>Substituting the values in formula,</u></em>

25 = \frac{k-x}{x} \times 100\\\\25x = 100(k-x)\\\\x = 4(k-x)\\\\x = 4k - 4x\\\\5x = 4k\\\\x = \frac{4k}{5}

Thus the weight in kilograms in 1980 is \frac{4k}{5}

7 0
3 years ago
An SRS of size 20 is drawn from a population that has a normal distribution. The sample has a mean of 122.5 and a standard devia
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Answer:

The value is  \sigma_{\= x} = 1.789

Step-by-step explanation:

From the question we are told that

   The sample size is  n =  20  

    The sample mean is \= x = 122.5

    The standard deviation is  s =  8

Generally the standard error of the mean is mathematically represented as

       \sigma_{\= x} =  \frac{s}{\sqrt{n} }

=>    \sigma_{\= x} =  \frac{8}{\sqrt{20 } }

=>    \sigma_{\= x} = 1.789

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30 300 000 is 30.3 million expanded.

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