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creativ13 [48]
2 years ago
8

Please help with math problem give 5 star if do

Mathematics
1 answer:
allochka39001 [22]2 years ago
7 0

Answer:

Cities A and C should be in Group 2

Step-by-step explanation:

Cities:

  • A (-16)
  • C (-5)

These should be in Group 2 because they are less than -3

Hope this helps!

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Ganezh [65]
The answer would be 4 and 2 over 3
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What are the combination of stressed and unstressed syllables for these?​
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Step-by-step explanation:

5 0
4 years ago
3. If a customer has $50.00 to spend, for how many hours can they rent the bicycle?
olya-2409 [2.1K]

Answer:

4 hours

Step-by-step explanation:

Alright, so Steve has $50 in his pocket. He wants to rent a bike for the maximum amount of time possible with his money. It costs him $12 per hour to rent a bike.

To solve this problem use simple arithmetic:

($50/$12=4.16)

So Steve is able to rent the bike for 4.16 hours, but we aren't done yet. Steve cannot purchase 16% of an hour, he can only buy FULL hours, so with his $50 , the maximum amount of time he can buy is 4 hours!

Let me know if you need any further explanations :)

7 0
3 years ago
The first terms of an infinite geometric sequence, Un are 2, 6, 18, 54... The first terms of a second infinite geometric sequenc
gladu [14]

Answer:

r = 9 and m = 112

Step-by-step explanation:

\sum_{k=1}^{225}W_{k}=\sum_{k=0}^{m}4r^{k}

Write W in terms of U and V.

\sum_{k=1}^{225}(U_{k}+V_{k})=\sum_{k=0}^{m}4r^{k}\\\sum_{k=1}^{225}U_{k}+\sum_{k=1}^{225}V_{k}=\sum_{k=0}^{m}4r^{k}

Define U and V using geometric series formula.

\sum_{k=1}^{225}2(3)^{k-1}+\sum_{k=1}^{225}2(-3)^{k-1}=\sum_{k=0}^{m}4r^{k}

Use sum of geometric series formula.

2(\frac{1-(3)^{225}}{1-3})+2(\frac{1-(-3)^{225}}{1-(-3)})=4(\frac{1-(r)^{m+1}}{1-r})

Simplify.

-1(1-3^{225})+\frac{1+3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-1+3^{225}+\frac{1}{2}+\frac{3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-\frac{1}{2}+\frac{3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3^{226}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{-1+3^{226}}{8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-3^{226}}{-8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-9^{113}}{1-9}=4(\frac{1-(r)^{m+1}}{1-r})

Therefore, r = 9 and m = 112.

8 0
3 years ago
Numbers along the outside of frequency tables, calculated from row and column totals, are __________.
Agata [3.3K]

Answer:

The answer is B.

Step-by-step explanation:

Will u give brainliest answer?

8 0
3 years ago
Read 2 more answers
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