Show that 4128 0012 4389 0110 is an invalid VISA number card

1 answer:

7 0

visa cards begin with 4 and then have 13 or 16 digits but here we have a number beginning with 4 and then it has the 15 digits.

Step-by-step explanation:

Each card has a different first digit number. The maximum length of this type of credit card number can be up to 19 digits.
In the visa card starting 6 digits are always said to be the issuer identifier.
The visa cards will have the starting digits as 4 whereas in Mastercard the starting digit is 5 but it has 16 digits numbers.

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Somebody please help me :)

aleksandrvk [35]

Answer:

(5,7.5) (12,18) (18,27)

Step-by-step explanation:

It costs $1.50 per cupcake.

y = 1.5x, where x is the amount of cupcakes and y is the amount of money in dollars.

7.5 = 1.5 times 5

18 = 1.5 times 12

27 = 1.5 times 18

6 0

3 years ago

Let x be a random variable representing the amount of sleep each adult in New York City got last night. Consider a sampling dist

dangina [55]

Answer:

Step-by-step explanation:

Given that X is a random variable representing the amount of sleep each adult in New York City got last night.

a) Normal distribution (by central limit theorem)

b) Sample mean will approach population mean mu

c) the standard deviation σx of the sampling distribution approach to σ/√n

d) The means are the same.

. The standard deviations are σ / √50 and σ / √100, respectively.

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3 years ago

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Help mi plzz.........

faust18 [17]

Tan3A=tan(2A+A)

We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))

tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)

We know that , tan2x=2tanx/(1 - tan^2x)

So, by substituting tan2A in (1),we get,

=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]

=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]

=[3tanA - tan^3A]/[1-3tan^2A]

Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]