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Deffense [45]
3 years ago
5

Show that the credit card number 4306 0912 4887 1457 is valid/invalid. Explain how you know.

Mathematics
1 answer:
elena-s [515]3 years ago
7 0
Step 1a. For a card number with an even number of digits (e.g., Visa or MasterCard), double alternating digits starting with the first digit in the sequence.
Step 1b. For a card with an odd number of digits (e.g., American Express), double alternating digits starting with the second digit in the sequence.
Step 2. If the doubling resulted in a number with two digits, add them together to get a single digit number
Step 3. Now go back to the original credit number and replace the digits that you doubled with the new value — either the doubled value, or the doubled value with the digits added together — and add it all up.
Step 4. Check to see if the sum is evenly divisible by 10 (you can simply look to see whether or not it ends with a zero).
If the card number does not pass this check, then it is not a valid number. If, on the other hand, it does pass, then it may be a valid number with valid credit report.
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Q1 - What is the possibility that the offspring will have white flowers? Express your answer as a ratio.

A: White flowers only appear if the offspring is homozygous for b because this phenotype is recessive, so they have to present bb phenotype. If both parents are heterozygous (Bb), each one gives one gene at a time per individual after crossing, so 1/4 would be BB, 2/4 would be Bb and 1/4 would be bb.

Q2 - If one parent is also heterozygous for smooth peas, does this mean that all of the offspring with blue flowers will also have smooth peas? Explain your answer without using a Punnett square.

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Q3 - Suppose both parents are also heterozygous for smooth peas (S), with wrinkled peas (s) as the recessive trait. Use a Punnett square to predict the possibility that the offspring will be homozygous for both flower color and pea texture. Make sure to include a Punnett square and express your prediction as a ratio.

A: Parental genotypes are BbSs x BbSs. Given that the genes segregate independently, we will have the Punnett diagram as follows:

        BS        /    Bs    /     bS   /   bs

BS    BBSS   / BBSs  / BbSS  / BbSs

Bs    BBSs   / BBss  / BbSs   / Bbss

bS    BbSS  / BbSs  / bbSS  / bbSs

bs     BbSs / Bbss  / bbSs  / bbss

Homozygous genotypes are the ones that both genes are the same when expressed, either BB,bb,SS or ss. The probability the offspring is homozygous for both flower color and pea texture is 4/16, or 1/4.

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