Answer:1)Number of different license plates can be made =17576×100×260=456976000
2)Probability of getting a plate with abc123 or 123 abc=0.0000004
Given that :-A license plate has six characters.
As three characters are letters ∴ ways of these 3 letters into the plate with repetition = 26×26×26=17576 ways
and two characters are numbers (0-9 - total 10 characters)∴ways of these 2 numbers into the plate with repetition=10×10 =100 ways
and one character is a letter or a number=26×10=260
So number of different license plates can be made =17576×100×260=456976000
Now Probability of getting a plate with abc123 or 123 abc
=P(abc123 or 123abc)=P(abc123)+P(123abc)-P(abc123)×P(123abc)=1/456976000+1/456976000-1/456976000×1/456976000
=1/456976000(1+1-1/456976000)
=1/456976000(2-0.0000002)=2/456976000=0.0000004
Probability of getting a plate with abc123 or 123 abc=0.0000004
Answer:
Answer is option b and c and d

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Answer:
Cuadrilátero A: 
Cuadrilátero B: 
Step-by-step explanation:
Existen dos formas distintas de determinar las áreas de cada cuadrilátero:
(i) <em>Obtener el área de cada cuadrado y sumar los resultados.</em>
(ii) <em>Calcular los lados del cuadrilátero grande y determinar el área. </em>
Cuadrilátero A
Método (i)


Método (ii)


Cuadrilátero B
Método (i)


Método (ii)


Answer:
14.63% probability that a student scores between 82 and 90
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that a student scores between 82 and 90?
This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So
X = 90



has a pvalue of 0.9649
X = 82



has a pvalue of 0.8186
0.9649 - 0.8186 = 0.1463
14.63% probability that a student scores between 82 and 90