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Natasha2012 [34]
3 years ago
10

Sean has a collection of coins. One tenth of the coins are from Europe. Thirty-two hundredth are from Asia. The rest are from Af

rica. Write a decimal to show the total part of the coins that are from Europe or Asia.
Mathematics
2 answers:
Tema [17]3 years ago
7 0
Lets write each part, the total will be x:
(1/10)x = Europe
3200 = Asia
x - <span>(1/10)x - 3200 = Africa

Europe + Asia = </span><span>(1/10)x + 3200 = x/10 + 3200
= (320x + 1)/3200
that would be expressed as a fraction, and depends on the total of coins, x</span>
Elodia [21]3 years ago
5 0

Answer: 0.42

Step-by-step explanation:

Let the total number of coins be x.

The part of coins are  from Europe = \frac{1}{10}

Then, the number of coins from Europe = \frac{1}{10}x=0.10x

The part of coins are from Asia =0.32

The number of coins from Asia=0.32x

Now, the number of coins from Europe or Asia =0.10+0.32x=0.42

Hence, the total part of the coins that are from Europe or Asia=0.42

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3 years ago
The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th
vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

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To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

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\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t

To find the population growing at the start of 2014 we say t = 0

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year

To find the population growing at the start of 2015 we say t = 1

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year

To convert billion to million you multiple by 1000

P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year

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