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sammy [17]
3 years ago
11

how many ways can we award a 1st, 2nd, and 3rd place prize among eight contestants? A) 336 B.) 56 C.) none of these D.) 40320​

Mathematics
2 answers:
svetoff [14.1K]3 years ago
7 0

The ways 1st - 3rd place are ordered matter, so we aren't finding the combination. We;re finding the permutation. This means we have 8 contestants taken 3 ways. The number of permutations of n objects taken r at a time is nPr = n! / (n - r)! N is 8 and 3 is R. Substitute the values.

8P3 = 8! / (8 - 3)! → 8! / 5! → 336

(You can also input this in a scientific calculator by pressing "8", then "nPr" then "3")

alekssr [168]3 years ago
4 0

the answer is A.336

Explanation:

There are 8 choices for awarding first prize.Then there are 7 choices for awarding second prize.And there are 6 choices for awarding third prize.

Therefore, there are: 8 *7 *6 =336 ways.

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42

Step-by-step explanation:

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Ok I have a question it says, " Ella can read 2 pages in 3 minutes. How long will it take her to read 45 pages?". So I just don'
Vaselesa [24]

Answer:

67.5

Step-by-step explanation:

this gives you the basis of the problem, 2 pages every 3 minutes, so that means that you have to divide 45 by 2, this will give you 22.5

then you multiply by 3 because 22.5 is how many sets of 2 Ella has to read, and if we remember every set of 2 pages takes 3 minutes, so 22.5 times 3 is 67.5, but if you want the answer in time it would be 67 minutes and 30 seconds because a minute is 60 seconds and half of that is 30

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PLS HELP!!!!!!!!!!!!!!!!!!!! Which number is divisible by both 5 and 6? A. 123,465 B. 132,640 C. 164,780 D. 184,290
umka21 [38]

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2 years ago
Alex bought 4 hoodies. Each hoodie cost the same amount and they were on sale for $5 off each. Alex paid a total of $120. How mu
vladimir2022 [97]
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8 0
3 years ago
The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q
Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

\frac{\sum f}{4}=\frac{100}{4}=25

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF)_{p} = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

     =34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75

The value of first quartile is 34.75.

Q₃ is at the position:

\frac{3\sum f}{4}=\frac{3\times100}{4}=75

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF)_{p} = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

     =44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83

The value of third quartile is 44.83.

Then the inter-quartile range is:

IQR = Q_{3}-Q_{1}

        =44.83-34.75\\=10.08

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}

                                 =\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0
3 years ago
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