A sequence is defined by the recursive formula f (n + 1) = f(n) – 2. If f(1) = 18, what is f(5)?

Answers

Answer 1
Answer: For this case we have the following recursive formula:
 f (n + 1) = f (n) - 2
 What we must do in this case is to evaluate different values of n until we find f (5)
 We have then:
 For n = 1:
 
f (1 + 1) = f (1) - 2 f (2) = f (1) - 2 f (2) = 18 - 2 f (2) = 16
 For n = 2:
 
f (2 + 1) = f (2) - 2 f (3) = f (2) - 2 f (3) = 16 - 2 f (3) = 14
 For n = 3:
 
f (3 + 1) = f (3) - 2 f (4) = f (3) - 2 f (4) = 14 - 2 f (4) = 12
 For n = 4:
 
f (4 + 1) = f (4) - 2 f (5) = f (4) - 2 f (5) = 12 - 2 f (5) = 10
 Answer:
 
The value of the recursive formula for f (5) is given by:
 
f (5) = 10
Answer 2
Answer: f(2)=18-2=16\n f(3)=16-2=14\n f(4)=14-2=12\n \boxed{f(5)=12-2=10}

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Answers

Answer:

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Step-by-step explanation:

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Solve "P x 87 + 67 x 87 + (33 - P) x 87 = ?"

Answers

P\cdot87+67\cdot87+(33-P)\cdot87=87\cdot(P+67+33-P)\n\n=87\cdot100=8700

Answer:

P x 87 + 67 x 87 + (33 - P) x 87=8700

Step-by-step explanation:

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Answers

Step-by-step explanation:

Speed is independent and distance is dependent

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Answers

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Answers

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Step-by-step explanation:

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MDM4U11. Avery and Bradley work at a large electronics manufacturer that produces DVD players. The
defective rate on the assembly line has gone up to 12% and the manager wants to know the
probability that a skid of 50 DVD players will contain at least 3 defective units.
a) Help Avery use the binomial distribution P(x)=, C.pqrs to answer this question

Answers

Answer:

0.9487

Step-by-step explanation:

a) The probability of having a defective product = p = 12% = 0.12

The probability of not having a defective product = q = 1 - p = 1 - 0.12 = 0.88

The number of DVD players = n = 50

X is the number of defective products.

The  probability that a skid of 50 DVD players will contain at least 3 defective units = P(X ≥ 3) = 1 - P(X ≤ 2)

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

Using binomial:

P(X = 0) = C(n,x)p^xq^(n-x)=C(50,0)p^0q^(50-0)=(50!)/((50-0)!0!)*0.12^0*0.88^(50-0)=0.0017

P(X = 1) = C(n,x)p^xq^(n-x)=C(50,1)p^1q^(50-1)=(50!)/((50-1)!1!)*0.12^1*0.88^(50-1)=0.0114

P(X = 2) = C(n,x)p^xq^(n-x)=C(50,2)p^2q^(50-2)=(50!)/((50-2)!2!)*0.12^2*0.88^(50-2)=0.0382

P(X ≤ 2)= 0.0017 + 0.0114 + 0.0382 = 0.0513

P(X ≥ 3) = 1 - P(X ≤ 2) = 1 - 0.0513 = 0.9487 = 94.87%