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Bingel [31]
3 years ago
6

The Rotokas of Papua New Guinea have twelve letters in their alphabet. The letters are: A, E, G, I, K, O, P, R, S, T, U, and V.

Suppose license plates of five letters utilize only the letters in the Rotoka alphabet. How many license plates of five letters are possible that begin with either G or K, end with T, cannot contain S, and have no letters that repeat?
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
3 0

12 letters in all but less "S" which is not allowed. That leave 11 usable letters, of which 5 distinct letters are used for each plate.

begins with G or K (2 choices). That leaves 10 letters.

Ends with T (1 choice). That least 9 for the middle letters.

Second letter (9 choices)

Third letter (8 choices)

Fourth letter (7 choices)

Total number of licence plates

=product of choices of letters for each position

= 2*9*8*7*1

=1008

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A company that produces fine crystal knows from experience that 17% of its goblets have cosmetic flaws and must be classified as
Alex73 [517]

Answer:

0.3891 = 38.91% probability that only one is a second

Step-by-step explanation:

For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

17% of its goblets have cosmetic flaws and must be classified as "seconds."

This means that p = 0.17

Among seven randomly selected goblets, how likely is it that only one is a second

This is P(X = 1) when n = 7. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891

0.3891 = 38.91% probability that only one is a second

7 0
3 years ago
What’s the correct answer for this ?
icang [17]

Answer:

D.

Step-by-step explanation:

Since both the triangles as kept separately are similar, so we'll take proptionality of their sides to find one side

MP/ML=MN/MK

20/28=35/MK

CROSS MULTIPLYING

20×MK=28×35

MK=980/20

MK=49

5 0
3 years ago
Find the slope of the line through each pair of points. <br> (19, −16), (−7, −15)
erik [133]

Point-Slope:

y+16=-1/26*(x-19)

Finding the regular slope:

m=1/-26=0.03846

7 0
3 years ago
Read 2 more answers
Simplify (x^6)^2 • x^3
givi [52]

Answer:

x^15

Step-by-step explanation:

Recall these rules of exponents:

(a^m)^n = a^mn

a^m • a^n = a^(m + n)

(x^6)² • x³ = x^(2 • 6) • x³ = x^12 • x³ = x^(12 + 3) = x^15

8 0
2 years ago
Please help me!? Explain why i got this wrong ? how to do it right?
IgorLugansk [536]
Y - y1 = m(x - x1)
slope(m) = -3/7
(5,8)...x1 = 5 and y1 = 8
now we sub
y - 8 = -3/7(x - 5) <===
3 0
4 years ago
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