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ollegr [7]
3 years ago
14

A code consist of a two digit number chosen from 00 to 49 followed by a letter of the alphabet. What is the probability the code

is 14z?
Mathematics
1 answer:
levacccp [35]3 years ago
4 0

Answer:

<em>The probability that code is 14z will be \frac{1}{1300}</em>

Step-by-step explanation:

From 00 to 49, there are total 50 two-digit numbers.

The total number of alphabets is 26.

So, the probability of choosing 14 from all 50 two digit numbers will be:  \frac{1}{50}

and the probability of choosing 'z' from all 26 alphabets will be: \frac{1}{26}

Thus, the probability that code is 14z is = \frac{1}{50}\times \frac{1}{26}=\frac{1}{1300}

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Maurinko [17]
Motorcyclist: 2 hours
Bus: 2.5 hours
Truck: 3 hours
Bicyclist: 7.5 hours
7 0
3 years ago
A pole that is 2.8 m tall casts a shadow that is 1.67 m long. At the same time, a nearby building casts a shadow that is 50.75 m
Veseljchak [2.6K]

Answer:

Explanation:

3.2

1.14

=

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Cross Mutliply:  

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x

Divide:  

143.2

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You get 125.614 rounded to the nearest meter is 126

Step-by-step explanation:

7 0
2 years ago
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A rare form of malignant tumor occurs in 11 children in a​ million, so its probability is 0.000011. Four cases of this tumor occ
Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

\lambda=r*t=0.000011*13,280=0.14608

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990

c) The probability of more than one case is:

P(k>1)=1-P(k\leq 1)=1-0.990=0.010

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0
3 years ago
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Shtirlitz [24]

Answer:

number of ways = 120

Step-by-step explanation:

The number of ways five children can pose for a photograph line up in a row is given by the number of permutations of 5 elements in 5 different positions (positions in the line), then

number of ways = number of permutations of 5 elements = 5! = 5 * 4 * 3 * 2 * 1 = 120

Since the first children that occupies the line can be on any of the positions (5 positions) , but then the second one can choose any of the 4 remaining positions (since the first children had already occupied one) , the third can choose 3 ... and so on.

5 0
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Here is a graph of the relationship between the number of quarters and the number of dimes when there are a total of 17 coins.
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Point a has more what she said
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