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Sedaia [141]
3 years ago
6

NEED TO KNOW IMMEDIATELY the Revenue each season from tickets at the theme part is represented by t(x) = 3x the cost to pay the

employees each season is represented by r(x) = (1.25)^x examine the graph of the combined function for total profit and estimate after five seasons
Mathematics
1 answer:
prohojiy [21]3 years ago
6 0

Answer:

r(5)=3.05 t(5)=15

Step-by-step explanation:

x=5

r(5)=(1.25)^5

r(5)=3.05

t(5)=3(5)

t(5)=15

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Decimals from 7.0 to 8.4 with an interval of 0.2 between each pair of decimals
Vitek1552 [10]

Decimals from 7.0 to 8.4 with an interval of 0.2 between each pair of decimals

We start with 7.0  and keep adding 0.2 till we get 8.4

7.0 + 0.2 = 7.2

7.2 + 0.2 = 7.4

7.4 + 0.2 = 7.6

7.6 + 0.2 = 7.8

7.8 + 0.2 = 8.0

8.0 + 0.2 = 8.2

8.2 + 0.2 = 8.4

So, the decimals from 7.0 to 8.4 with an interval of 0.2 are

7.2, 7.4, 7.6, 7.8, 8.0, 8.2

3 0
3 years ago
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Write 9 decimals with three decimal places that when rounded to the nearest tenth round to 1.3
zmey [24]
1.291
1.251
1.341
1.331
1.321
1.311
1.281
1.271
1.261
6 0
3 years ago
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A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal
GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean income = \frac{\sum X}{n} = $4.70

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = $1.83

            n = sample of parking meters = 10

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.262 < t_9 < 2.262) = 0.95  {As the critical value of t at 9 degrees of

                                            freedom are -2.262 & 2.262 with P = 2.5%}  

P(-2.262 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.262) = 0.95

P( -2.262 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu < 2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.262 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-2.262 \times {\frac{s}{\sqrt{n} } } , \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 4.70-2.262 \times {\frac{1.83}{\sqrt{10} } } , 4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } } ]

                                         = [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error  =  2.262 \times {\frac{s}{\sqrt{n} } }

                                          =  2.262 \times {\frac{1.83}{\sqrt{10} } }  = <u>1.31</u>

4 0
3 years ago
A line parallel to $3x-7y = 65$ passes through the point $(7,4)$ and $(0,K)$. What is the value of K?
Zepler [3.9K]

Answer:

1

Step-by-step explanation:

Hi, its you again lol.

I am in elementary btw.

I don't want to put the equation down.

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3 years ago
How many km/hr is 28 feet per minute
Marina86 [1]
28 ft per min = 0.512064 km per hour
7 0
3 years ago
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