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belka [17]
4 years ago
8

Andrea was watching her brother in the ocean and noticed that the waves were coming into the beach at a frequency of 0.367347 Hz

. How many waves hit the beach in 49 s? Answer in units of waves.
Mathematics
1 answer:
anygoal [31]4 years ago
8 0

Answer:

<u>18 waves</u> hit the beach in 49 s.

Step-by-step explanation:

Given:

The frequency of the waves that were coming into the beach is 0.367347 Hz.

Now, to find the number of waves that hit the beach in 49 s.

Let the number of waves be n.

The frequency of waves (f) = 0.367347 Hz.

The time it takes to hit the beach (t) = 49 s.

Now, we put formula to get the number of waves:

f=\frac{n}{t} \\\\0.367347=\frac{n}{49} \\\\Multiplying\ both\ sides\ by\ 49\ we\ get:\\\\18.000003=n\\\\n=18.000003.

<u><em>The number of waves = 18.</em></u>

Therefore, 18 waves hit the beach in 49 s.

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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

3 0
3 years ago
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