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NNADVOKAT [17]
3 years ago
10

Describe a real world scenario that is an arithmetic series. Show your work as you find the 7th term of your scenario.

Mathematics
1 answer:
Iteru [2.4K]3 years ago
7 0

A new night club opens. The first night 100 patrons attend. For each successive night for a month, 20 more attend. How many attend on the 7th night?

Arithmetic sequence formula finding the nth number:

<span>a</span><span>n</span><span> = a</span><span>1</span><span> + (n – 1)d</span>

<span>A(7) = 100 + (7-1) 20</span>

<span>A(7) = 100 + (6)20</span>

<span>A(7) = 100 + 120</span>

<span>A(7) = 220</span>

<span>
</span>

<span>So, on the 7th night, there 220 patrons.</span>

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17. When a single set of values is randomly divided into two equal groups, explain how the means of these two groups may be very
joja [24]

If you take 2 groups of equal cardinality, it could happen that, for example, many of the higher values go to the first group and therefore, many of the low values go to the second group, making their respective means quite different and different from the original sample mean. This could even go worse due to the possible existence of outliers, that is, values that are far different than the sample mean. An outlier tend to disrup the mean of a sample, but for smaller samples the result is much dramatic.

For example, let X be {1,2,3,4,5,6,7,8,9,100}

The elements of X sum 145, hence the mean of X is 14,5. Let divide X in two groups

Y = {1,2,3,5,9}

Z = {4,6,7,8,100}

The elements of Y sum 20, so its mean is 4

The elements of Z sum 125, so its mean is 25

Both means are quite different from each other and quite different from the mean of X. Note that if we take the mean of the means the result is 4+25/2 = 14,5 which is equal to the mean of X.

7 0
3 years ago
How many 3 digit numbers are possible when a) the leading digit cannot be zero and the number must be a multiple of 4?
guajiro [1.7K]

Step-by-step explanation:

I assume the digits can be repeated.

so, e.g. 555 is a valid number for this problem, right ?

that means we start with permutations with repetition :

n^r

n = the total number of items to pick from.

r = the number of items being picked per result.

we have 10 digits (0,1,2,3,4,5,6,7,8,9), and we pick 3 of them.

that gives us (with very little surprise, I hope)

10³ = 1000 different possible numbers from 000 to 999.

from these numbers we eliminate all with leading 0.

as we handled all digits the same way and with the same priority, there is the same amount of numbers for every digit in the leading position.

that means 1/10 of the total amount of numbers has a leading 0, or a leading 1, or a leading 2, ...

so, we need to subtract 1/10 × 1000 from 1000 :

1000 - 1000×1/10 = 1000 - 100 = 900

that would be the numbers 100 to 999.

and we have one more condition : the number must be a multiple of 4.

how many are there ?

well, that's the funny thing about numbers : from all numbers 1/2 of them are multiples of 2 (or divisible by 2), 1/3 of them are multiples of 3 (or divisible by 3), and ... you guessed it, 1/4 of them are multiples of 4 (or divisible by 4). and so on.

and so, 1/4 of our 900 numbers are multiples of 4 :

1/4 × 900 = 225

so, there are 225 possible 3-digit numbers that are multiples of 4 and do not start with a 0.

6 0
2 years ago
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