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sammy [17]
3 years ago
9

A company has 25 employees. One day, 8 letters that did not have names on the envelopes arrived for 8 different employees. The m

ail sorter placed each letter randomly in a different employee's mailbox.
Mathematics
1 answer:
Allisa [31]3 years ago
8 0
1. The missing question here is : "what is the probability that the mail sorter placed each letter correctly"?

2. Name the letters L1, L2, ....L8   (Letter 1, Letter 2.... Letter 8)

3. The first letter has a probability of 1/25 to be placed in the correct mailbox.

Each of the remaining 7 letters have 1/25 probability, and these probabilities are related.

4. So the probability that each letter goes to the proper mailbox is \frac{1}{25}*  \frac{1}{25}*  \frac{1}{25}*  \frac{1}{25}*  \frac{1}{25}*  \frac{1}{25}*  \frac{1}{25}*  \frac{1}{25}= \frac{1}{ 25^{8}}
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Solve for the values of x in the equation: <br> 2^(x) = 4x.
Eva8 [605]

There are two of them. 

I don't know a mechanical way to 'solve' for them.

One can be found by trial and error:

x=0 . . . . . 2^0 = 1 . . . . . 4(0) = 0 . . . . . no, that doesn't work
x=1 . . . . . 2^1 = 2 . . . . . 4(1) = 4 . . . . . no, that doesn't work
x=2 . . . . . 2^2 = 4 . . . . . 4(2) = 8 . . . . . no, that doesn't work
x=3 . . . . . 2^3 = 8 . . . . . 4(3) = 12 . . . . no, that doesn't work
<em>x=4</em> . . . . . 2^4 = <em><u>16</u></em> . . . . 4(4) = <em><u>16</u></em> . . . . Yes !  That works !       yay !

For the other one, I constructed tables of values for 2^x and (4x)
in a spread sheet, then graphed them, and looked for the point
where the graphs of the two expressions cross.

The point is near, but not exactly,         <em>x = 0.30990693...

</em>
If there's a way to find an analytical expression for the value, it must involve
some esoteric kind of math operations that I didn't learn in high school or
engineering school, and which has thus far eluded me during my lengthy
residency in the college of hard knocks.<em> </em> If anybody out there has it, I'm
waiting with all ears.<em>

</em>
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3 years ago
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The volume of a sphere is a function of its​ radius, V=4/3(pie)(r^3). Evaluate the function for the volume of a volleyball with
Travka [436]
Volume of a volleyball (a sphere) = 4/3πr³
= 4/3π(11.4)³
= 1975.392π
= 6210 cm³ (to 3 significant figures)

The volume of a volleyball with a radius of 11.4 cm will be 6210 cm³.
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Answer:

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Step-by-step explanation:

Aaden needs 40 points.

He gets 20 at the start, just for signing up, so he needs only 20 now.

If he gets 2.5 points for each visit, after 8 days he will have been given 20 more points, allowing him to get a free movie ticket.

Brainliest please?

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mr_godi [17]
The commutative property of multiplication.
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