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3 years ago

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Equation is y=mx+11, passes through (8'3)

2 answers:

Allisa [31]3 years ago

8 0

Y=-1x+11.

I used an online graph for this and it would be tricky on paper.

Murljashka [212]3 years ago

5 0

This has a few steps. First, solve for m by plugging in x = 8, y = 3.
3 = 8m + 11 => m = -1. This is the slope of the equation. Now just plug this into the original equation.
So the solution is y = (-1)x + 11.

y = -x + 11

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B) In 2033 the population will reach 386 million.

Step-by-step explanation:

Given : The population of a certain country in 1996 was 286 million people. In addition, the population of the country was growing at a rate of 0.8% per year. Assuming that this growth rate continues, the model $P(t) = 286(1.008 )^{t-1996}$ represents the population P (in millions of people) in year t.

To find : According to this model, when will the population of the country reach A. 306 million? B. 386 million?

Solution :

The model represent the population is $P(t) = 286(1.008 )^{t-1996}$

Where, P represents the population in million.

t represents the time.

A) When population P=306 million.

$306 = 286(1.008 )^{t-1996}$

$\frac{306}{286}=(1.008 )^{t-1996}$

$1.0699=(1.008 )^{t-1996}$

Taking log both side,

$\log(1.0699)=\log((1.008 )^{t-1996})$

$\log(1.0699)=(t-1996)\log(1.008)$

$\frac{\log(1.0699)}{\log(1.008)}=(t-1996)$

$8.479=t-1996$

$t=8.479+1996$

$t=2004.47$

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Therefore, In 2004 the population will reach 306 million.

B) When population P=386 million.

$386 = 286(1.008 )^{t-1996}$

$\frac{386}{286}=(1.008 )^{t-1996}$

$1.3496=(1.008 )^{t-1996}$

Taking log both side,

$\log(1.3496)=\log((1.008 )^{t-1996})$

$\log(1.3496)=(t-1996)\log(1.008)$

$\frac{\log(1.3496)}{\log(1.008)}=(t-1996)$

$37.625=t-1996$

$t=37.625+1996$

$t=2033.625$

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Therefore, In 2033 the population will reach 386 million.

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