The next number is 38. The pattern is 7, 16, 8, 27, 9; if you look at the first 4 numbers, you notice that it counts to 7, 8, 9. Then you have 16 and 27, if the pattern continues; the next number is 38.
Answer:
% Po lost = 100[1 - e^(-0.005t)] %; 73.0 g
Step-by-step explanation:
p(t) = 100e^(-0.005t)
Initial amount: p(0) = 100
Amount remaining: p(t) = 100e^(-0.005t)
Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]
% of Po lost = amount lost/initial amount × 100 %
= [1 - e^(-0.005t)] × 100 % = 100[1 - e^(-0.005t)] %
p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g
The mass of polonium remaining after 63 days is 73 g.
Answer:
=0.1587 or 15.87%
So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.
Step-by-step explanation:
In order to find the percent of the invoices paid within 5 days of receipt we have to find the value of Z first.

where:
X is the random varable which in our case is 15 days
u is the mean or average value which is 20 days
S is the standard deviation which is 5 days

Z=-1.0
We have to find Probability at Z less than -1
P(Z<-1.0) which can be written as:
=1-P(Z>1.0)
From Cumulative distribution table:
=1-(0.3413+0.5)
=0.1587 or 15.87%
So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.