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alekssr [168]
2 years ago
11

Money can be easily converted into goods or services. This is an example of which "function of money"?

Mathematics
1 answer:
EastWind [94]2 years ago
4 0

Answer:

medium of exchange

Step-by-step explanation:

just took the test

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Please help me!! ✍ <br> Thank you!! ♛<br> Answers are included! ♒
tino4ka555 [31]
THat would be A , B, C and E
4 0
3 years ago
Angie and Kenny play online video games. Angie buys 1 software package and 3 months of game play. Kenny buys 1 software package
andrey2020 [161]

Answer:The cost of one month of game-play =$20


Step-by-step explanation:

Let the cost of one month of gameplay be x

Then cost of game-play bought by Angie =3x.....(1)

and Then cost of game-play bought by Kenny=4x......(2)

Cost of each software package =$50......(3)

The the total cost =240= sum of costs of software bought by both of them and game-play)=50+50+3x+4x

⇒240=100+7x.......→(by adding like terms)

⇒140=7x⇒x=20.....→( dividing both sides by 7 )

∴the cost of one month of game-play =$20



8 0
3 years ago
Read 2 more answers
Which number line repersents the sum of 1.25+(-0.75
aleksklad [387]

Answer: See explanation

Step-by-step explanation:

You didn't give the options relating to the question.

To solve this first, you need to know that we have to note that (+) × (-) = (-). Therefore, 1.25+(-0.75) will be thesame as writing 1.25 - 0.75. This will then be:

= 1.25 - 0.75

= 0.50

3 0
3 years ago
Read 2 more answers
The higher price of a commodity is ¢565.00​
EastWind [94]

Answer:

There is no question to answer. This is simply just a statement. If you need help, feel free to put the question under this comment because I don't understand what anyone is supposed to be answering from this "question."

5 0
2 years ago
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
2 years ago
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