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ddd [48]
3 years ago
10

Suppose Rainforest sells 2 comma 000 books on account for $ 19 each​ (cost of these books is $ 22 comma 800​) on October ​10, 20

18 to ABC Store. One hundred of these books​ (cost $ 1 comma 140​) were damaged in​ shipment, so Rainforest later received the damaged goods from ABC Store as sales returns on October ​13, 2018. ​(Assume both companies use a perpetual inventory system and that sales are recorded at the net​ amount.) Read the requirements LOADING.... Requirement 1. Journalize ABC Store​'s October 2018 transactions. ​(Record debits​ first, then credits. Exclude explanations from journal​ entries.) Oct. ​10: ABC Store purchased 2 comma 000 books on account for $ 19 each from Rainforest.
Business
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

Journal entries for ABC Store's

inventory   38,000

  account payable 38,000

to record purchase of 2,000 books

account payable 1,900

   inventory                  1,900

to record return of 100 damaged books

Explanation:

Requirement 1 journalize ABC Store's

We need to journalize base on ABC store. Assuming perpetual inventory.

ABC purchased 2,000 books at $19 each total 38,000

we increase our inventory for the amount purchased and also declare the liability, as those book were not paid right away

later it return 100 books the cost is $19 each total 1,900

this decrease the ammount due to Rainforest and also decrease the inventory

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a. The point estimate of the married people that sleep well is = 0.6299. The confidence interval for the mean hours of sleep for those that have never been married but report good sleep is 0.6186, 0.6412

b. The point estimate of the number of hours for people that have never been married = 7.2257

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a. The point estimate

P=\frac{Xp}{n}

Xp = successes\\n = sample size\\

Those who have never been married n = 7044

Xp = those who sleep for 7 hours at least

To get Xp, the <u>excel function </u>

Countif(range, criteria)

Countif(A:A, >=7)

This gives Xp = 4437

Point estimate =

\frac{4477}{7044}

= 0.6299

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At 95% interval of confidence,

α = 1-0.95 = 0.05

α/2 = 0.05/2 = 0.025

Z0.025 = 1.96 ( Use the standard probability normal table)

Margin of Error

=M.E = 1.96\sqrt \frac{0.6299(1-0.6299)}{7044} \\\\= 1.96\sqrt{0.000033} \\\\1.96*0.0057

= 0.0113

The 95% confidence interval

= 0.6299 - 0.0113 , 0.6299 + 0.0113

= 0.6186, 0.6412

The confidence interval for the mean hours of sleep for those that have never been married is 0.6186, 0.6412

b. Sample mean

x = ∑x/n

x = number of sleep of the never been married

Mean is calculated using excel function

Average(range)

Average(a:a)

This gives the sample mean = 7.2257

The point estimate of the number of hours for people that have never been married = 7.2257

<u>Standard deviation</u> using excel =

Stdev(A:A) = 0.8211

Margin of error =1.96*\frac{0.8211}{\sqrt{7044} }

= 0.0192

The<u> confidence interval</u> =

= 7.2257-0.0192, 7.2257+0.0192

= 7.2065, 7.2447

95% confidence interval for the mean number of  hours of sleep for those who have never been married = 7.2065, 7.2447

c. From the available data, the mean number of sleep per day that is gotten from the 4437 people that are unmarried and sleep healthily for at least 7 hours = 34295.1

Mean = 34295.1/4437

= 7.729

the estimated number of sleep for those that have never been married = 7.729 hours

Read more on brainly.com/question/15601189?referrer=searchResults

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