the half-life of bismuth-210 is 5 days. if 25 grams of bismuth-210 are present initially, how much will remain after 10 days?
1 answer:
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Answer:
The answer to the question is;
The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.
Step-by-step explanation:
Let the sample size =n = 100
The success probability = 5 % = 0.05
Number of tickets sold = 105 tickets
In the case where there the airline has found that 5 % will not show up, then every passenger should have a seat, we have
A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails
However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus
n·p = 105×0.05 = 5.25 ≥ 5
and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5
As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution
= - 0.3358
We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307
Another way to solve the question is as follows
p = 0.95 q = 0.05
μ = np = 0.95*105 = 99.75, σ = = 2.233
P (x≤100) = P = P(z<0.34) = 0.63307.
Answer:
Width of the rectangular Park = 11 feet
Step-by-step explanation:
Given that the total length of the fencing is 62 feet. As has to be fenced around a rectangular park , it would be the perimeter of the rectangular park.
Also Shana wants the length of the run to be 20 feet. Hence the length of the park is 20 feet.
Here we will use the formula for perimeter to find the width of the run
Perimeter = 2(l+w)
62=2(l+w)
l+w =
l+w=31
20+w=31
w=31-20
w=11
Hence the width of the run for her dog in park would be 11 feet.