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Alexxx [7]
3 years ago
7

Where would (0,12.5) be plotted on the graph ? Please answer this !!

Mathematics
1 answer:
bezimeni [28]3 years ago
6 0

On the y-axis. Right in the middle of 12 and 13.

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Can you help please no links or point stealing or I will report
valentina_108 [34]

1.A school assembly has 179 students in it. Nine teachers escort "k" number of

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Kinsley's town had an election last month. 4,800 people voted in the election and 3,200
ryzh [129]

Answer:55

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Is the GCF of any two odd numbers always odd?
vampirchik [111]
Probably true

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4 years ago
What is the value of Fraction 1 over 2x3 + 3.4y when x = 3 and y = 4?
Luden [163]
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In another state, all license plates consist of from four to six symbols chosen from the 26 letters of the alphabet together wit
fredd [130]
A)
First note that the plates can have between 4 and 6 symbols so we will need to find the number of plates with 4 symbols, 5 symbols and 6 symbols. We add these to get the total. In this part repetition of symbols is allowed. Since there are 26 + 10 =36 possible symbols we look at each position on the plate and think of how many choices there are. We multiply the number f choices using the counting principal since the choices are each independent -- one symbol does not affect another. There are 36 choices for the first symbol, 35 for the second and so on. The number of plates is:
4-symbols = (36)(36)(36)(36)=36^4
5 symbols = (36)^5
6 symbols = 36^6
So the total here is: 36^4+36^5+36^6

B) Here we do not repeat symbols so there are 36 choices for the first symbol but only 35 for the next and 34 for the one after and so on.
4-symbols = (36)(35)(34)(33)
5 symbols = (36)(35)(34)(33)(32)
6 symbols = (36)(35)(34)(33)(32)(31)
So the total here is: (36)(35)(34)(33)+(36)(35)(34)(33)(32)+(36)(35)(34)(33)(32)(31)

c)
In order for there to be a repeated symbol we have 36 choices for the first symbol, 36 for the next and so on. However, for the last symbol we have to pick from one of the ones already selected so there are 3, 4 or 5 choices respectively.
4-symbols = (36)(36)(36)(3)
5 symbols = (36)(36)(36)(36)(4)
6 symbols = (36)(36)(36)(36)(36)(5)
So the total here is: (36^3)(3)+(36^4)(4)+(36^5)(5)

D)
The probability is given by (the number of plates with at least one repeated symbol)/(the total number of plates if repetitions are allowed) = (the answer to c) / (the answer to a)
3 0
3 years ago
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