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balandron [24]
4 years ago
7

The distance formula states that distance (d) is equal to the product of rate (r) and time (f). Given this problem: john at a co

nstant rate of 200 m per minute. How many minutes did it take john to run 500 m?
which equation could be used to solve the problem.

A. t = 500d/200

B. t = 200/500

C. t = 500/200

D. t = 200r/500
Mathematics
1 answer:
Alecsey [184]4 years ago
5 0
Letter a is the answer
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Answer:

\large \boxed{\text{19 visitors/mo}}

Step-by-step explanation:

The average rate of change is the change in the number of visitors divided by the change in the number of months.

1. Change in number of  visitors  

Change in visitors = 182 -125 = 57 visitors

2. Change in number of weeks

Change in time = 5 - 2 = 3 months

3. Average rate of change

\begin{array}{rcl}\text{Average rate of change} &= &\dfrac{\text{change in visitors}}{\text{change in months}}\\\\&= &\dfrac{\text{57 visitors}}{\text{3 mo}}\\\\& = & \textbf{19 visitors/mo}\\\end{array}\\\text{The average rate of change is $\large \boxed{\textbf{19 visitors/mo}}$}

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If you roll a fair 6 sided die and then flip a fair coin, what is the probability that you roll a 1, 2, 3, or 4 and get heads wi
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Step-by-step explanation: a, c, and e

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3 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

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$1,619.30

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