we know that
The probability that "at least one" is the probability of exactly one, exactly 2, exactly 3, 4 and 5 contain salmonella.
The easiest way to solve this is to recognise that "at least one" is ALL 100% of the possibilities EXCEPT that none have salmonella.
If the probability that any one egg has 1/6 chance of salmonella
then
the probability that any one egg will not have salmonella = 5/6.
Therefore
for all 5 to not have salmonella
= (5/6)^5 = 3125 / 7776
= 0.401877 = 0.40 to 2 decimal places
REMEMBER this is the probability that NONE have salmonella
Therefore
the probability that at least one does = 1 - 0.40
= 0.60
the answer is
0.60 or 60%
4000+7000= 11,000 nearest thousands
4281+7028= 11,309 actual number
Answer:
Remaining credit = $11.36 [11 dollars and 36 cents]
9514 1404 393
Answer:
a) P(t) = 6.29e^(0.0241t)
b) P(6) ≈ 7.3 million
c) 10 years
d) 28.8 years
Step-by-step explanation:
a) You have written the equation.
P(t) = 6.29·e^(0.0241·t)
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b) 2018 is 6 years after 2012.
P(6) = 6.29·e^(0.0241·6) ≈ 7.2686 ≈ 7.3 . . . million
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c) We want t for ...
8 = 6.29·e^(0.0241t)
ln(8/6.29) = 0.0241t
t = ln(8/6.29)/0.0241 ≈ 9.978 ≈ 10.0 . . . years
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d) Along the same lines as the calculation in part (c), doubling time is ...
t = ln(2)/0.0241 ≈ 28.7613 ≈ 28.8 . . . years
