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dezoksy [38]
3 years ago
5

A $7,000 motorcycle depreciates at 10% each year. Which value would represent the base of the exponential function that models t

his situation?
0.90
0.10
1.10
7,000
Mathematics
1 answer:
stira [4]3 years ago
3 0

Answer:

The base of the exponential function is 0.90

Step-by-step explanation:

we know that

The exponential function that models this situation is equal to  

V(x)=P(1-r)^{x}  

where  

V is the depreciated value  

P is the original value  

r is the rate of depreciation  in decimal  

x  is Number of Time Periods  

in this problem we have  

P=\$7,000\\r=10\%=0.10

substitute

V=7,000(1-0.10)^{x}

V=7,000(0.90)^{x}

therefore

The base of the exponential function is 0.90

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Answer:

1/5, 4/5, 5/5

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Answer:

{52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in {52 \choose 7} ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in {7 \choose 2} ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in {45 \choose 7}. Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in {7 \choose 2} ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in {38 \choose 7}. And again, we have to choose which ones are the hidden ones, which can be chosen in {7 \choose 2} ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in {31 \choose 7} ways. And choosing which ones are the hidden ones for this player can be done in {7 \choose 2} ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

{52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

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