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3241004551 [841]
2 years ago
6

A company has 10 male and 11 female employees, and needs to nominate 2 men and 2 women for the company bowling team. How many di

fferent teams can be formed?
Mathematics
1 answer:
Anvisha [2.4K]2 years ago
3 0

2475 different teams can be formed.

Combination is the way in which items can be selected from a collection. If we have n total objects and r objects want to be selected, the number of combinations is:

C(n,r)=\frac{n!}{(n-r)!(r)!}

Since we need to nominate 2 men from a company of 10 males, the number of ways this can be done = C(10,2)=\frac{10!}{(10-2)!(2)!} =45\  ways

Also, we need to nominate 2 women from a company of 11 females, the number of ways this can be done =  C(10,2)=\frac{11!}{(11-2)!(2)!} =55\  ways

Therefore the total number of different teams that can be formed = 45 ways * 55 ways = 2475 ways

Find more at: brainly.com/question/8018593

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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).

We know that the population, P(t), of China, in billions, can be approximated by P(t)=1.394(1.006)^t

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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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