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Answer with explanation</u>
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Let
be the distance traveled by deluxe tire .
As per given , we have
Null hypothesis : 
Alternative hypothesis : 
Since
is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.
Test statistic : 
where, n= sample size
= sample mean
= Population mean
=sample standard deviation
For
, we have

By using z-value table,
P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23) [∵P(Z≤-z)=1-P(Z≤z)]
=1-0.9871=0.0129
Decision : Since p value (0.0129) < significance level (0.05), so we reject the null hypothesis .
[We reject the null hypothesis when p-value is less than the significance level .]
Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.
Monthly depreciation is calculated as follows:
65,000 (Cost) – 5,000 (Salvage Value) ÷ 60 (5 years X 12 months per year) = 1,000 in depreciation per month.
a) Accumulated depreciation from 1/1/12 to 1/1/15 is 36,000 (12 months for 2012, 2013, & 2014).
Assuming that sale was a Cash sale, the journal entry would look like this:
1/1/15
Cash (DR) 31,000
Accumulated Depreciation (DR) 36,000
Equipment (CR) 65,000
Gain on Sale of Equipment (CR) 2,000
b) Accumulated depreciation from 1/1/12 to 5/1/15 is 40,000 (12 months for 2012, 2013, 2014, & 4 months for 2015).
Assuming that sale was a Cash sale, the journal entry would look like this:
5/1/15
Cash (DR) 31,000
Accumulated Depreciation (DR) 40,000
Equipment (CR) 65,000
Gain on Sale of Equipment (CR) 6,000
c) Accumulated depreciation from 1/1/12 to 1/1/15 is 36,000 (12 months for 2012, 2013, & 2014).
Assuming that sale was a Cash sale, the journal entry would look like this:
1/1/15
Cash (DR) 11,000
Accumulated Depreciation (DR) 36,000
Loss on Sale of Equipment (DR) 18,000
Equipment (CR) 65,000
d) Accumulated depreciation from 1/1/12 to 10/1/15 is 45,000 (12 months for 2012, 2013, 2014, & 9 months for 2015).
Assuming that sale was a Cash sale, the journal entry would look like this:
10/1/15
Cash (DR) 11,000
Accumulated Depreciation (DR) 45,000
Loss on Sale of Equipment (DR) 9,000
Equipment (CR) 65,000
Not my answers, but I hope this will help you. :)
In 2010, a city's population was 1,405,233 and it was decreasing at a rate of 1.1%. At this rate when will the city's population fall below 1,200,000?
c. 2036
Answer:the answer is 2
Step-by-step explanation:
Answer:
24
Step-by-step explanation:
3(2n-8)=5n distribute 3
6n-24=5n combine like terms
-24=-n divide by -1 on both sides
24=n
n=24