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PolarNik [594]
3 years ago
11

Suppose the expression a(b)n models the approximate number of people who visited an aquarium each day since an aquarium opened,

where a is the initial number of people who visited, b is the rate of increase in the number of people who visited each day, and n is the number of days since the aquarium opened.
If the expression below models the number of visitors of a particular aquarium, what is the correct interpretation of the second factor?
54(1.3)^7


A.
There were 9.1 times as many people who visited the aquarium on the 7th day as on the first day.
B.
There were 1.3 times as many people who visited the aquarium on the 7th day as on the first day.
C.
There were 6.27 times as many people who visited the aquarium on the 7th day as on the first day.
D.
There were 10.2 times as many people who visited the aquarium on the 7th day as on the first day.
Mathematics
1 answer:
Viefleur [7K]3 years ago
8 0

Answer:

C. There were 6,27 times as many people who visited the aquarium on the seventh day as on the first day.

Explanation:

\displaystyle 1,3^7 = 6,2748517 ≈ 6,27

I am joyous to assist you anytime.

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20. A ball is dropped from a height of a little over 5 feet, and the height is measured at small intervals. The table below show
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Step 1:

a) Write the quadratic model

a = -15.64

b = -1.24

c = 5.23

P(t)=-15.64t^2\text{ - 1.24t + 5.2}3

b) t = 0.30

\begin{gathered} P(t)=-15.64t^2\text{ - 1.24t + 5.2}3 \\ =\text{ -15.64 }\times0.3^2\text{ - 1.24}\times\text{ 0.3 + 5.2}3 \\ =\text{ -1.4076 - 0.372 + }5.23 \\ =\text{ 3.45} \end{gathered}

c) t = 0.52 seconds

\begin{gathered} p(t)=-15.64t^2\text{ - 1.23t + 5.23} \\ =\text{ -15.64}\times0.52^2\text{ - 1.23}\times0.52\text{ + 5.23} \\ =\text{ -4.23 - 0.64 + 5.23} \\ =\text{ 0.36} \end{gathered}

d) 0.30 is more likely to be relaible.

e)

\begin{gathered} p(t)\text{ = 1} \\ p(t)=-15.64t^2\text{ - 1.23t + 5.23} \\ 1=-15.64t^2\text{ - 1.23t + 5.23} \\ -15.64t^2\text{ - 1.23t + 4.23 = 0} \\ t\text{ = 0.48222} \\ \text{t = 0.48 seconds} \end{gathered}

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