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%
Answer:
Step-by-step explanation:
Let Alex's age = x
Mark's age = y
Luke's age = a
x + y + z = 12
The possible combinations for their age are
1. 6, 3, 3
2. 3, 3, 6
3. 4,4, 4
4, 2, 2, 8
5. 8, 2, 2
6. 2, 8, 2
7. 9, 2, 1
Etc the age can keep going as long as you can find any three numbers and add them to give you 12
Answer:
$95.78
Step-by-step explanation:
f(t) = 300t / (2t² + 8)
t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.
$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt
$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt
$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt
$T₂ = 75 ln(t² + 4) |₅¹⁰
$T₂ = 75 ln(104) − 75 ln(29)
$T₂ ≈ 95.78
Answer: the probability that a randomly selected tire will have a life of exactly 47,500 miles is 0.067
Step-by-step explanation:
Since the life expectancy of a particular brand of tire is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = life expectancy of the brand of tire in miles.
µ = mean
σ = standard deviation
From the information given,
µ = 40000 miles
σ = 5000 miles
The probability that a randomly selected tire will have a life of exactly 47,500 miles
P(x = 47500)
For x = 47500,
z = (40000 - 47500)/5000 = - 1.5
Looking at the normal distribution table, the probability corresponding to the z score is 0.067
