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uysha [10]
3 years ago
12

Suppose f and g are two functions defined on some neighbourhood of c such that lim f(x)=l and lim g(x)=m.Show that lim f(x)+g(x)

=l+m
Mathematics
1 answer:
zavuch27 [327]3 years ago
7 0

Trivial if you use limit laws.

\lim f(x) + g(x) = \lim f(x) + \lim g(x) = l + m

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Show me how to work the equation x÷16=3​
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In your sock drawer you have 5 blue, 7 gray, and 2 black socks. Half asleep one morning you grab 2 socks at random and put them
pentagon [3]

The question is incomplete! The complete question along with answers and explanation is provided below!

In your sock drawer you have 5 blue, 7 gray, and 2 black socks. Half asleep one morning you grab 2 socks at random and put them on. Find the probability you end up wearing the following socks. (Round your answers to four decimal places.)

a) 2 blue socks

b) no gray socks

c) at least 1 black sock

d) a green sock

e) matching socks

Answer:

a) P(2 blue socks) = 0.1099 = 10.99%

b) P(no gray socks) = 0.2307 = 23.07%

c) P(at least 1 Black sock) = 0.2748 = 27.48%

d) P(green sock) = 0%

e) P(Matching socks) = 0.3516 = 35.16%

Step-by-step explanation:

Given Information:

5 Blue socks

7 Gray socks

2 black socks

Total socks = 5 + 7 + 2 = 14

a) The probability of wearing 2 blue socks

P(2 blue socks) = P(B1 and B2)

P(B1) = no. of blue socks/total no. of socks

P(B1) = 5/14 = 0.3571

Now there are 4 blue socks remaining and total 13 socks remaining

P(B2|B1) = 4/13 = 0.3077

P(B1 and B2) = 0.3571*0.3077 = 0.1099 = 10.99%

b) The probability of wearing no gray socks

5 Blue socks + 2 black socks = 7 socks are not gray

P(no gray socks) = P(Not G1 and Not G2)

P(Not G1) = no. socks that are not grey/ total no. of socks

P(Not G1) = 7/14 = 0.5

Now there are 6 socks remaining that are not gray and total 13 socks remaining

P(Not G2 | Not G1) = 6/13 = 0.4615

P(Not G1 and Not G2) = 0.5*0.4615 = 0.2307 = 23.07%

c) The probability of wearing at least 1 black sock

5 Blue socks + 7 Gray socks = 12 socks are not black

P(at least 1 Black) = 1 - P(Not B1 and Not B2)

P(Not B1) = no. socks that are not black/ total no. of socks

P(Not B1) = 12/14 = 0.8571

Now there are 11 socks remaining that are not black and total 13 socks remaining

P(Not B2 | Not B1) = 11/13 = 0.8461

P( Not B1 and Not B2) = 0.8571*0.8461 = 0.7252

P(at least 1 Black) = 1 - P( Not B1 and Not B2)

P(at least 1 Black) = 1 - 0.7252 = 0.2748 = 27.48%

d) The probability of wearing a green sock

There are 0 green socks, therefore

P(Green) = 0/14 = 0%

e) The probability of wearing matching socks

P(Matching socks) = P(2 Blue socks) + P(2 Gray socks) + P(2 Black socks)

P(2 Blue socks) already calculated in part a

P(2 Blue socks) = P(B1 and B2) = 0.1099

For Gray socks

P(G1) = no. of gray socks/ total no. of socks

P(G1) = 7/14 = 0.5

Now there are 6 gray socks remaining and total 13 socks remaining

P(G2 | G1) = 6/13 = 0.4615

P(2 Gray socks) = P(G1 and G2) = 0.5*0.4615 = 0.2307

For Black socks

P(B1) = no. of black socks/ total no. of socks

P(B1) = 2/14 = 0.1428

Now there is 1 black sock remaining and total 13 socks remaining

P(B2 | B1) = 1/13 = 0.0769

P(2 Black socks) = P(B1 and B2) = 0.1428*0.0769 = 0.0110

P(Matching socks) = P(2 Blue socks) + P(2 Gray socks) + P(2 Black socks)

P(Matching socks) = 0.1099 + 0.2307 + 0.0110 = 0.3516 = 35.16%

7 0
2 years ago
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