For accounting purposes, the value of assets (land, buildings, equipment) in a business are depreciated at a set rate per year.
2 answers:
Answer: $ 125987.80
Step-by-step explanation:
Given: The value, V(t) of $393,000 worth of assets after t years, that depreciate at 15% per year, is given by the formula
, here
is the initial asset value and b is the multiplicative decay factor.
The exponential decay function is given by ;-
, where A is the initial value , x is the times period and b is the multiplicative decay factor.
where b = 1-r, r is the rate of decay.
Since r = 15%=0.15
Therefore, b = 1-0.15=0.85
Now ,for 7 years , the value of assets is given by :-
Hence, the assets valued at after 7 years = $ 125987.80
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Answer:
Null hypothesis:
Alternative hypothesis:
Step-by-step explanation:
1) Data given and notation
Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4
We can calculate the sample mean and deviation for this data with the following formulas:
The results obtained are:
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :
Null hypothesis:
Alternative hypothesis:
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
4) P-value
First we need to find the degrees of freedom for the statistic given by:
Since is a two sided test the p value would given by:
5) Conclusion
If we compare the p value and the significance level assumed we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.