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Citrus2011 [14]
2 years ago
13

so i have 1,521,588 poker chips how much do i need to get 2 million (poker chips on a game called pokerstars vr)

Mathematics
2 answers:
grandymaker [24]2 years ago
6 0

Answer:

478412

Step-by-step explanation:

Elodia [21]2 years ago
6 0

Answer: 478,412

Step-by-step explanation:

In order to do this just take 2 million and subtract the number of chips you wanted to take away. For example, it'd be 2,000,00 - 1,521,588 which leaves you with 478,412.

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What is d + 6 / -2 = 5 the slash is a fraction
Charra [1.4K]

Answer:

d=8

Step-by-step explanation:

5 0
2 years ago
Eva, the owner of eva's second time around wedding dresses, currently has five dresses to be altered, shown in the order in whic
cupoosta [38]
To answer this question, an assumption must be made, that Eva spends 8 hours a day working. If this is the case, then Eva will complete jobs w, x, and v on day one, for a total of six hours. Since the next job (y) requires 4 hours, she will spend two hours working that day, leaving 2 more hours to go on that job. The next day she will spend 2 hours finishing job y, completing it, and finish the longest job z (hours) that day. This means she had 4 jobs on day one, and 2 jobs on day 2 for and average of 3 jobs per day.  
This answer assumes an 8 hour work day, and that Eva can start a job she cannot finish that day.
5 0
2 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
3 years ago
No work needed, just need answer please.
In-s [12.5K]

Answer:

1 is 27.8%

2 is 41.7%

Step-by-step explanation:

To calculate 1 there is a 10/36 chance (2, 3, 4, 5 out of possible roles of 2 - 12)

To calculate 2 there is a 15/36 chance (2, 3, 5, 7, 11 are prime)

7 0
3 years ago
Can someone explain how this is right please
Daniel [21]
Because when you add them all up that's what the answers are
4 0
3 years ago
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