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DerKrebs [107]
3 years ago
11

(Please help. I screenshot the question and choices D: )

Mathematics
2 answers:
vodka [1.7K]3 years ago
8 0
The answer is yes, 1:8
Arte-miy333 [17]3 years ago
8 0
Figure 1)
lenght₁=14
width₁=3
heigth₁=4

Figure 2)
lenght₂=112
width₂=24
height₂=32

If these figure are similar, then:
length₁/lenght₂=width₁/width₂=height₁/height₂=ratio of the smaller figure to the larger figure

Therefore:
14/112=3/24=4/32=1/8


answer: yes; 1/8
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8 0
3 years ago
A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
Fed [463]

Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

The initial conditions are : a_{0} = a_{1} = 0

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

7 0
3 years ago
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