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SVETLANKA909090 [29]
3 years ago
10

Write six hundred millions, two hundred thousands, one thousand, nine tens, four ones

Mathematics
1 answer:
Kipish [7]3 years ago
4 0

In number form it sounds like it would be 600,201,094

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What’s 8 squared into a decimal
ycow [4]

Answer:

2.828 i think

Step-by-step explanation:

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A sample of n = 5 scores has a mean of m = 8. one new score is added to the sample and the new mean is calculated to be m = 9. w
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If the 5 scores have a mean of 8, then their total sum would be 8*5 = 40

Now if one score if added (let's call this score x), there are 6 scores and the mean changes to 9, thus:
(40 + x)/6 = 9
40 + x = 54
x = 14
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Select the equivalent expression.
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Step-by-step explanation:

a^-4 x 8^14

the solution is C

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Round 62.17354 to the nearest hundredth
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The answer is 62.17

Step-by-step explanation:

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1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,
crimeas [40]

Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222

3 0
3 years ago
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