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mestny [16]
3 years ago
7

An athletic club charges $15 per month plus $1.75 per visit to club the function f(x)=1.75x+15 represent the monthly charge to b

elong and use in the club.Tonya budgeted $50 per month of club membership. How much more money does Tonya need to budget in order to join this athletic club and be able to meet her goal of working out 6 days per week?
Mathematics
1 answer:
marin [14]3 years ago
6 0
Assuming that a month has exactly 4 weeks, the number of days Tonya will visit the athletic club in a month is given by 6 x 4 = 24 visits.

Given that the <span>athletic club charges $15 per month plus $1.75 per visit to club, the amount Tonya will spend in the given month is given by 15 + 1.75(24) = 15 + 42 = 57.

Therefore, for </span>Tonya to <span><span>be able to meet her goal of working out 6 days per week. she</span> needs to budget $7 more.</span>
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Answer:

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Step-by-step explanation:

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maks197457 [2]
I will assume you meant D(x)=23(1.0032)^x, if so:

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vovangra [49]

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8 0
3 years ago
The shuttle bus from your parking lot and your office building operates on a 15 minute schedule. You arrive at the parking lot a
mylen [45]

Answer:

(a) The standard deviation of your waiting time is 4.33 minutes.

(b) The probability that you will have to wait more than 2 standard deviations is 0.4227.

Step-by-step explanation:

Let <em>X</em> = the waiting time for the bus at the parking lot.

The random variable <em>X</em> is uniformly distributed with parameters <em>a</em> = 0 to <em>b</em> = 15.

The probability density function of <em>X</em> is given as follows:

f_{X}(x)=\frac{1}{b-a};\ a

(a)

The standard deviation of a Uniformly distributed random variable is given by:

SD=\sqrt{\frac{(b-a)^{2}}{12}}

Compute the standard deviation of the random variable <em>X</em> as follows:

SD=\sqrt{\frac{(b-a)^{2}}{12}}

      =\sqrt{\frac{(15-0)^{2}}{12}}

      =\sqrt{\frac{225}{12}}

      =\sqrt{18.75}\\=4.33

Thus, the standard deviation of your waiting time is 4.33 minutes.

(b)

The value representing 2 standard deviations is:

X=2\times SD=2\times4.33=8.66

Compute the value of P (X > 8.66) as follows:

P(X>8.66)=\int\limits^{10}_{8.66} {\frac{1}{15-0}}\, dx\\

                    =\frac{1}{15}\times \int\limits^{10}_{8.66} {1}\, dx\\

                    =\frac{1}{15}\times |x|^{15}_{8.66}\\

                    =\frac{15-8.66}{15}\\

                    =0.4227

Thus, the probability that you will have to wait more than 2 standard deviations is 0.4227.

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Answer:

Step-by-step explanation:

4 0
3 years ago
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