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Setler79 [48]
3 years ago
5

Need help please anyone ?

Mathematics
2 answers:
lesya692 [45]3 years ago
8 0

Answer:

14.4

Step-by-step explanation:

To find the mean in math, you add all of your numbers together, the divide by how many numbers you added together. So add 17,15,14,14, and 12 together to get 72. Divide 72 by 5 to get 14.4

Tresset [83]3 years ago
5 0

Answer:

The mean would be 14.4 hours.

Step-by-step explanation:

In order to find the mean, or average, of the numbers, you would add all the numbers together.

17 + 15 +14 + 14 + 12 which equals 72

Then, you take the total which is 72, and divide it by how many numbers you added together which is 5.

72/5 = 14.4

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Answer:

36

Step-by-step explanation:

First you would solve what's in the parenthesis because the Order of Pemdas.

Second you would multiply the answer of what's in the parenthesis which is 3 to 12.

Third you would get the answer which is 36!

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If you have 6 cars, but there is only room in your driveway for 3 cars, in how many ways can you arrange the cars in your drivew
Svetradugi [14.3K]

If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

The probability that the three newest cars end up parked in the driveway is 0.167

1. If the order does not matter, combination is used

\left ({n} \atop {r}} \right.)=\frac{n!}{r!(n-r)!}

Here, n=6 r=3

using formula,we get

\left ( {{6} \atop {3}} \right)=\frac{6!}{3!3!} =5*4=20

2. If an order is important, Permutation will be applicable

^{n}P_{r} = \frac{n!}{(n-r)!} \\

∴^{6} P_{3} =\frac{6!}{3!(6-3)!}=\frac{6!}{3!}  =120

3. the probability that the three newest cars end up parked in the driveway

P=\frac{No. of possible outcomes for 3 cars}{total possibilities}

=\frac{6*4*5}{6!}

=\frac{120}{720}

=\frac{1}{6} ≈ 0.167

Hence, If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

The probability that the three newest cars end up parked in the driveway is 0.167

Learn more about probability here brainly.com/question/6077878

#SPJ4

7 0
2 years ago
What is the wrong with asking this survey question
sashaice [31]

I'm not really sure for 3, but for 4, it is not free of influence since they are waiting for a horror movie.

What would be better for question 4 is to ask people at the lobby or something before they are entering a movie theater so the results aren't "contaminated".

4 0
3 years ago
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