I think the answer would be B
Answer:
The Value of Payments is P18,557.15
Step-by-step explanation:
The quarterly payment is an annuity payment.
Use the following formula to calculate the present value of the payments.
PV of Annuity = Annuity Payment x ( 1 - ( 1 + interest rate )^-Numbers of periods ) / Interest rates
Where
Annuity Payment = Quarterly payment = P1,300
Interest rate = 5.12% x 3/12 = 1.375%
Numbers of periods = 4 years x 12/3 = 16 quarters
PV of Annuity = Value of Payments = ?
Placing values in the formula
Value of Payments = P1,300 x ( 1 - ( 1 + 1.375% )^-16 ) / 1.375%
Value of Payments = P18,557.15
The question is incomplete, here is the complete question:
The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.
When will there be less than 1 g remaining?
<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.
<u>Step-by-step explanation:</u>
All radioactive decay processes follow first order reaction.
To calculate the rate constant by given half life of the reaction, we use the equation:
where,
= half life period of the reaction = 46 days
k = rate constant = ?
Putting values in above equation, we get:
The formula used to calculate the time period for a first order reaction follows:
where,
k = rate constant = 
t = time period = ? days
a = initial concentration of the reactant = 12.6 g
a - x = concentration of reactant left after time 't' = 1 g
Putting values in above equation, we get:
Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.
A= 26 > q
if it says 26 grader than it means the mouth opens to that 26
if it says 26 less then it means the mouth is facing to the q
but it is not so it is the mouth facing the 26
77% of the school want electronic newsletters, so 23% of the school want physical paper newsletters.
77% of 650 is 500.5.
23% of 650 is 149.5.
650/100 = 6.5
6.5x77 = 500.5
6.5x23 = 149.5
It's best to round that figure up to 150 as there will not be a parent who only needs half of the newsletter.
