Your question doesn't say what are the options, but we can make some reasoning.
The average daily balance method is based, obviously, on the <span>average daily balance, which is the average balance for every day of the billing cycle. Therefore, in order to calculate the average daily balance, you need to sum the balance of every day and then divide it by the days of the billing cycle.
In your case:
ADB = (9</span>×2030 + 21×1450) / 30 = 1624 $
Now, in order to calculate the interest, you should first calculate the daily rate, since APR is usually defined yearly, and therefore:
rate = 0.23 ÷ 365 = 0.00063
Finally, the expression to calculate the interest could be:
interest = ADB × rate × days in the billing cycle
or else:
<span>interest = ADB × APR ÷ 365 × days in the billing cycle
In your case:
interest = 1624 </span>× 0.23 ÷ 365 × 30
= 30.70 $
Answer:
a. r=d/t b. 800000/4=200,000 c.800/2=200 d.not sure km are used more often as theyre simpler to use but there is not much that i am basing this off of
Step-by-step explanation:
Answer:
Hello! Your answer is D: 75.
Step-by-step explanation:
First we need to find the volume of pyramid A in order to solve this problem:
The formula for the pyramid is: V=
(B)(h)
Now we plug in the numbers:
V=
(81)(4)
V=
(324)
V=108in³
Now we divide pyramid B's volume by pyramid A's:
8,100/108 = 75
So the volume pyramid B is 75 times bigger than the volume of pyramid A!
Hope this helps!
Answer:
The engineer must verify and verify through a statistical inference that estimates and possible parameters may emerge, as well as determine what hypothesis tests should be performed to draw the most accurate conclusion.
Step-by-step explanation:
The engineer must assume that there may be more than one reasonable estimator for a different event or experiment; Something that could help you would be to perform an estimation of parameters, in that estimation it is required to know the properties of the estimators; that is to say that the closer the value of an estimator is to the real value of the parameter, it could be said that it is the most efficient or exact extimator.