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IceJOKER [234]
3 years ago
15

ABC Corp. has outstanding accounts receivable totaling $6.5 million as of December 31 and sales on credit during the year of $24

million. There is also a credit balance of $12,000 in the allowance for doubtful accounts. If the company estimates that 6% of its outstanding receivables will be uncollectible, what will be the amount of bad debt expense recognized for the year
Mathematics
1 answer:
Elden [556K]3 years ago
3 0

Answer:

$378,000

Step-by-step explanation:

The computation of the bad debt expense for the year is shown below:

Bad debt expense = Outstanding account receivable × estimated percentage given - credit balance of allowance for doubtful account

= $6,500,000 × 0.06 - $12,000

= $390,000 - $12,000

= $378,000

We simply deduct the credit balance from the estimated balance so that the correct amount could arrive

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Answer:

Option: C is the correct answer.

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Step-by-step explanation:

let A denotes the events of buying a thread.

and B denote the event of buying a needle.

Then A∩B denote the event of buying a needle and  a thread.

Also let P denote the probability of an event.

i.e. we are given:

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Also P(B|A)=0.25

As we know that:

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3 years ago
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3 years ago
At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample
omeli [17]

Given Information:

Population mean = p  = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

Answer:

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standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

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30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

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The mean of the sampling distribution will be same as population mean that is

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The standard deviation for this sampling distribution is given by

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Where p is the population mean that is proportion of female students and n is the sample size.

\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma =  0.069

Therefore, the standard deviation of the sampling distribution is 0.069.

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