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IrinaK [193]
3 years ago
8

An Aibo costs $2,899.99. This includes a toy ball, a charging station, and several years of AI cloud service, which helps it lea

rn. For how long is that service free? How do you know?

Mathematics
1 answer:
ivann1987 [24]3 years ago
3 0

Answer:

See Below

Step-by-step explanation:

The attached graph shows the cumulative costs of owning an AIBO.

$2899.99 (about $2900) includes fixed cost of Toy Ball, Charging Station

Several years of AI cloud service is a variable cost because this is changing and it helps the robot dog "learn". It is going to cost after some time. But how long?

Looking at the cost of an AIBO, we see that until 3 years, the cost line is veyr close to 3000 dollars, we can say 2899.99 dollars. After 3 years, the cost rises with a slope. Now, this is the cost of AI cloud service being IMPOSED after first 3 years of being "free".

Thus,

We can say that service is free for first 3 years. We know by looking at the graph shown.

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A family is taking a picture, with everyone in the family standing in a row. In the family, there are two sets of twins. In how
Wittaler [7]

Answer:

376320.

Step-by-step explanation:

Given:

There are two sets of twins in a family.

Total family member = 9

Solution:

Let family members = AA BB C D E F G

<u>AA and BB are two set of twins .</u>

First of we will find the arrangements of 5 family member   ( 9 - 4 = 5 )

No of ways 5 family member can seat =^{5}P_{5}

Now, we will find the arrangements of BB.

As given that nobody is sitting next to their twin, means A twins cannot seat together.

Therefore, they can seat in these blank places given below:

......B.......B......C......... D...... E........ F........ G.......... in ^{8}P_{2}

Similarly,  B twins cannot seat together.

Therefore, they can seat in these blank places given below:

......A.......A......C......... D...... E........ F........ G.......... in ^{8}P_{2}

Total  number of arrangements of all family members = ^{5}P_{5}\times ^{8}P_{2} \times^{8}P_{2}

                                                                                         =\frac{5!}{5-5!} \times\frac{8!}{8-2!} \times\frac{8!}{8-2!}\\=\frac{5!}{0!}\times\frac{8!}{6!} \times\frac{8!}{6!}\\\\=5!\times\frac{8\times7\times6!}{6!} \frac{8\times7\times6!}{6!}

                                                                                           =5\times4\times3\times2\times1\times56\times56\\=376320

Therefore, Total  number of arrangements of all family members is 376320.

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