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Svet_ta [14]
4 years ago
7

1) Which equation could be used to estimate the population of Oak City in 10 years?

Mathematics
1 answer:
DaniilM [7]4 years ago
6 0

Answer:

1) A

2) B

Step-by-step explanation:

For this problem we use exponential growth formulas

y = y_0(1 + r) ^ t if the population increases when t increases

y = y_0(1-r) ^ 2 If the population decreases when t increases

Where and is the population as a function of time, y_0 is the initial population, r is the growth rate, t is the time in years.

1) In this situation we have that y_0 = 48000, the growth is 2.4%, then r = 0.024 and t = 10 and the equation is:

y = 48000(1 + 0.024) ^ {10}

y = 48000(1.024) ^ {10} -------- Option A)

2) In this situation we have that y_0 = 82000, the decrease is 1.7%, then r = 0.017 and t = 5 and the equation is:

y = 82000(1-0.017) ^ 5

y = 82000(0.983) ^ 5 -------- Option B)

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4 years ago
The given tables each show the number of stories completed in the construction of four different high-rise buildings and the num
pentagon [3]

<u>Answer:</u>

The correct answer option is B. Number of Days 120 240 360 480 600 Number of Stories 5 10 15 20 25.

<u>Step-by-step explanation:</u>

We are given some tables showing the number of stories completed in the construction of four different high-rise buildings and the number of days spent working on the building.

We are to determine whether which table represents a linear relationship.

Table B represents a linear relationship since its ratio of number of days to number of stories completed is constant.

Number of Days    120  240  360  480  600  

Number of Stories   5      10     15    20     25

Ratio                        24    24    24    24     24

3 0
3 years ago
Read 2 more answers
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Rina8888 [55]
To solve this we are going to use the formula for speed: S= \frac{d}{t}
where
S is the speed
d is the distance 
t is the time 

Let S_{l} be the speed of the boat in the lake, S_{a} the speed of the boat in the river, t_{l} the time of the boat in the lake, and t_{a} the time of the boat in the river. 

We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span>S_{a}=S_{l}-2
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span>t_{l}=t_{a}+1
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span>S_{a}= \frac{6}{t_{a} }
But we know that S_{a}=S_{l}-2, so:
S_{l}-2= \frac{6}{t_{a} } equation (1)

Speed of the boat traveling in the lake:
S_{l}= \frac{15}{t_{l} }
But we know that t_{l}=t_{a}+1, so:
S_{l}= \frac{15}{t_{a}+1} equation (2)

Solving for t_{a} in equation (1):
S_{l}-2= \frac{6}{t_{a} }
t_{a}= \frac{6}{S_{l}-2} equation (3)

Solving for t_{a} in equation (2):
S_{l}= \frac{15}{t_{a}+1}
t_{a}+1= \frac{15}{S_{l}}
t_{a}=\frac{15}{S_{l}}-1
t_{a}= \frac{15-S_{l}}{S_{l}} equation (4)

Replacing equation (4) in equation (3):
t_{a}= \frac{6}{S_{l}-2}
\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}

Solving for S_{l}:
\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}
(15-S_{l})(S_{l}-2)=6S_{l}
15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}
S_{l}^2-11S_{l}+30=0
(S_{l}-6)(S_{l}-5)=0
S_{l}=6 or S_{l}=5

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.
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Round 12,406.4578 to the nearest thousands
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