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Ira Lisetskai [31]
3 years ago
5

a country's population in 1990 was 134 million. in 2000 it was a 139 million. estimate the population in 2014 using the exponent

ial growth formula. round to the nearest million
Mathematics
1 answer:
yuradex [85]3 years ago
6 0
F=ir^t

139=134r^10

139/134=r^10

r=(139/134)^(1/10) then:

f=134(139/134)^(t/10)  so in 2014, t=24 so

f=134(139/134)^(2.4)

f≈146 million  (to nearest million)

Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...

A=Pe^(kt)

139=134e^(10k)

139/134=e^(10k)

ln(139/134)=10k

k=ln(139/134)/10 so

A=134e^(t*ln(139/134)/10)  when t=24

A=134e^(2.4*ln(139/134))

A≈146 million (to nearest million)

The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...
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a student either knows the answer or guesses. Let 3434 be the probability that he knows the answer and 1414 be the probability t
Arlecino [84]

Answer:    \dfrac{12}{13}

Step-by-step explanation:

Let A = he known the answer then A' = he guess the answer.

B = he answered it correctly

As per given , we have

P(A)=\dfrac{3}{4}\ \ ,\ \ P(A')=\dfrac{1}{4}

P(B|A)=1

P(B|A')=\dfrac{1}{4}

By Bayes theorem , we have

P(A|B)=\dfrac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A')P(A')}\\\\ P(A|B)=\dfrac{1\times\dfrac{3}{4}}{1\times\dfrac{3}{4}+\dfrac{1}{4}\times\dfrac{1}{4}}\\\\= \dfrac{12}{13}

The probability that the student knows the answer given that he answered it correctly is  \dfrac{12}{13} .

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3 years ago
Let c be the curve which is the union of two line segments, the first going from (0, 0) to (4, 4) and the second going from (4,
victus00 [196]
First of all we need to find a representation of C, so this is shown in the figure below.

So the integral we need to compute is this:

I=\int_c 4dy-4dx

So, as shown in the figure, C = C1 + C2, so:

I=\int_{c_{1}} (4dy-4dx)+\int_{c_{2}} (4dy-4dx)=I_{1}+I_{2}

Computing first integral:

c_{1}: y-y_{0}=m(x-x_{0}) \rightarrow y=x

Applying derivative:

dy=dx

Substituting this value into I_{1}

I_{1}=\int_{c_{1}} (4dx-4dx)=\int_{c_{1}} 0 \rightarrow \boxed{I_{1}=0}

Computing second integral:

c_{2}: y-y_{0}=m(x-x_{0}) \rightarrow y-0=-(x-8) \rightarrow y=-x+8

Applying derivative:

dy=-dx

Substituting this differential into I_{2}

I_{2}=\int_{c_{2}} 4(-dx)-4dx=\int_{c_{2}} -8dx=-8\int_{c_{2}}dx

We need to know the limits of our integral, so given that the variable we are using in this integral is x, then the limits are the x coordinates of the extreme points of the straight line C2, so:

 I_{2}= -8\int_{4}^{8}}dx=-8[x]\right|_4 ^{8}=-8(8-4) \rightarrow \boxed{I_{2}=-32}

Finally:

I=\int_c 4dy-4dx=0-32 \rightarrow \boxed{I=-32}
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