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LenKa [72]
3 years ago
11

A car dealership owner creted a data table shows how many new and used cars where slide over several months

Mathematics
1 answer:
klemol [59]3 years ago
5 0

Answer:

the answer is 75

Step-by-step explanation:

i took he unit test

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Tiffany has a spinner that is split into four equal sections: red, blue, green, and yellow. Tiffany spun the spinner 680 times.
Lyrx [107]
Probability of getting Green is 1/4
1/4 ×680 = 170
so choose nearest answer will 187
4 0
3 years ago
X - y = 6, X =<br> 2x + y = 3, y =<br> solve the system of equation
nordsb [41]

Answer in pic

Step-by-step explanation:

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3 years ago
How many five-minute segments are in 1 hour?
kvasek [131]

Answer:

The correct answer is 12.

To get this you will divide 60minutes into 5.

You will get the answer "12"

Thus, there are 12, 5 minute segments, in one hour.

Hope this helps, good luck!

4 0
3 years ago
Read 2 more answers
A pole that is 3.5m tall casts a shadow that is 1.7m long. At the same time, a nearby building casts a shadow that is 36.5m long
GalinKa [24]
1.)Convert into Proportion: 3.5/1.7=x/36.5
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5 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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