With all the stuff added together, the sum is 2.03 kg
Answer:
The sweat chloride reference value is less than 30 mmol/L. A value of more than 60 mmol/L of chloride in the sweat is consistent with a diagnosis of cystic fibrosis. The values of 30-60 mEq/L may represent heterozygous carriers, these carriers cannot be accurately identified with a sweat chloride test.
At the end of the zeroth year, the population is 200.
At the end of the first year, the population is 200(0.96)¹
At the end of the second year, the population is 200(0.96)²
We can generalise this to become at the end of the nth year as 200(0.96)ⁿ
Now, we need to know when the population will be less than 170.
So, 170 ≤ 200(0.96)ⁿ
170/200 ≤ 0.96ⁿ
17/20 ≤ 0.96ⁿ
Let 17/20 = 0.96ⁿ, first.
log_0.96(17/2) = n
n = ln(17/20)/ln(0.96)
n will be the 4th year, as after the third year, the population reaches ≈176
Answer:
0.01024
Step-by-step explanation:
Assume marking was done at random : Hence, each of the 5 time slots have equal Chamves of being marked ;
Number of time slots, n = 5
Required to mark, number of preferred timeslot x = 2
P(x) = 2 /5 = 0.4
Probability of 0.4 that an interviewee gets one of his preferred timeslot.
Probability that each of the 5 interviewees gets one of their preferred time slots :
Using the multiplication rule of independence :
0.4 * 0.4 * 0.4 * 0.4 * 0.4 = 0.4^5 = 0.01024
0.01024 * 100% = 1.024%
