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guajiro [1.7K]
3 years ago
10

A city had population 67,255 on january 1, 2000, and its population has been increasing by 2935 people each year since then. A l

inear model for the population p, where t is in years after 2000, is
a.P(t) = 2935t.
b.P(t) = 67255t.
c.P(t) = 2935 + 67255t.
d.P(t) = 67255 + 2935t.
Mathematics
2 answers:
irina1246 [14]3 years ago
4 0

We are given: On january 1, 2000 initial population   = 67,255.

Number of people increase each year = 2935 people.

Therefore, 67,255 would be fix value and 2935 is the rate at which population increase.

Let us assume there would be t number of years after year 2000 and population P after t years is taken by function P(t).

So, we can setup an equation as

Total population after t years = Number of t years * rate of increase of population + fix given population.

In terms of function it can be written as

P(t) = t * 2935 + 67255.

Therefore, final function would be

P(t) = 2935t +67255.

So, the correct option is d.P(t) = 67255 + 2935t.

Scrat [10]3 years ago
4 0

Answer:

P(t) = 67255 + 2935t is correct. The figure 67255 does not change, while 2935 is added each time t increases.

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3 0
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6 0
2 years ago
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4 0
2 years ago
The Key Club is having a fundraising dinner for charity. The venue that holds a maximum of 500 people will
Alexus [3.1K]

Answer:

\dfrac{\$1000+\$20x}{x} \leq \$25

Minimum 200 people other than the 2 charity representatives.

Step-by-step explanation:

Given that:

The venue can hold a maximum of 500 people.

Cost of venue = $1000

Per person cost for food = $20

Two charity representatives get to attend the dinner for free.

To find:

The inequality and to determine how many people must come to keep costs at most $25.

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Cost per person = Total cost divided by Number of people attending the dinner.

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\dfrac{\$1000+\$20x}{x} \leq \$25\\\Rightarrow 1000+20x\leq25x\\\Rightarrow 1000 \leq 5x\\\Rightarrow x\geq 200

Therefore, the answer is:

Minimum 200 people other than the 2 charity representatives should attend the dinner.

8 0
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