I really dont get it what does it mean

Multiply the expression 2(x+1)(2x-5)

Sedbober [7]

This gives you the answer, look at it tells you how to solve this and get your answer

8 0

2 years ago

Allowance method entries

Feliz [49]

Using the Allowance Method, the relevant transactions can be completed in the books of Wild Trout Gallery as follows:

1. Allowance for Doubtful Accounts

Accounts Debit Credit

Jan. 1 Beginning balance $53,800

Jan. 19 Accounts Receivable 2,560

Apr. 3 Accounts Receivable $14,670

July 16 Accounts Receivable 19,725

Nov. 23 Accounts Receivable 4,175

Dec. 31 Accounts Receivable 25,110

Dec. 31 Ending balance $56,500

Dec. 31 Bad Debts Expenses $55,470

Totals $116,005 $116,005

Accounts Receivable

Accounts Debit Credit

Jan. 1 Beginning balance $2,290,000

Jan. 19 Allowance for Doubtful 2,560

Jan. 19 Cash $2,560

Apr. 3 Allowance for Doubtful 14,670

July 16 Allowance for Doubtful 19,725

July 16 Cash 6,575

Nov. 23 Allowance for Doubtful 4,175

Nov. 23 Cash 4,175

Dec. 31 Allowance for Doubtful 25,110

Dec. 31 Sales Revenue 8,020,000

Dec. 31 Cash $8,944,420

Dec. 31 Ending balance $1,299,500

Totals $10,316,735 $10,316,735

3. Expected net realizable value of the accounts receivable as of December 31 = $1,243,000 ($1,299,500 - $56,500)

Allowance for Doubtful Accounts ending balance = $40,100 ($8,020,000 x 0.5%)

Allowance for Doubtful Accounts

Accounts Debit Credit

Jan. 1 Beginning balance $53,800

Jan. 19 Accounts Receivable 2,560

Apr. 3 Accounts Receivable $14,670

July 16 Accounts Receivable 19,725

Nov. 23 Accounts Receivable 4,175

Dec. 31 Accounts Receivable 25,110

Dec. 31 Ending balance $40,100

Dec. 31 Bad Debts Expenses $39,070

Totals $99,605 $99,605

4. a. Bad Debt Expense for the year = $39,070

4.b. Balance for Allowance Accounts = $40,100

4.c. Expected net realizable value of the accounts receivable = $1,259,400 ($1,299,500 - $40,100)

Data Analysis:

Jan. 19 Accounts Receivable $2,560 Allowance for Uncollectible Accounts $2,560

Jan. 19 Cash $2,560 Accounts Receivable $2,560

Apr. 3 Allowance for Uncollectible Accounts $14,670 Accounts Receivable $14,670

July 16 Cash $6,575 Allowance for Uncollectible Accounts $19,725 Accounts Receivable $26,300

Nov. 23 Accounts Receivable $4,175 Allowance for Uncollectible Accounts $4,175

Nov. 23 Cash $4,175 Accounts Receivable $4,175

Dec. 31 Allowance for Uncollectible Accounts $25,110 Accounts Receivable $25,110

Accounts Receivable ending balance = $1,299,500

Allowance for Uncollectible Accounts ending balance = $56,500

Learn more: brainly.com/question/22984282

4 0

2 years ago

Are y = 1/4 -2 and y = 1/4 + 2 parallel?

zhenek [66]

Answer:

yes because the slope is equal. so these two lines are parallel.

3 0

2 years ago

does the assumption of a poisson distribution with a mean of 6.0 seem appropriate as a probability model for these data? use 0.0

Alenkasestr [34]

The Poisson distribution with a mean of 6.0 is an appropriate model.

In mathematics and statistics, the concept of mean is crucial. The most typical or average value among a group of numbers is called the mean.
It is a statistical measure of a probability distribution's central tendency along the median and mode. It also goes by the name "expected value."
There are different ways of measuring the central tendency of a set of values. There are multiple ways to calculate the mean. Here are the two most popular ones:
Arithmetic mean is the total of the sum of all values in a collection of numbers divided by the number of numbers in a collection.

Hence, The Poisson distribution with a mean of 6.0 is an appropriate model.

learn more about mean click here:

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6 0

10 months ago

2. The time between engine failures for a 2-1/2-ton truck used by the military is

OLEGan [10]

Answer:

A truck "will be able to travel a total distance of over 5000 miles without an engine failure" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a random variable normally distributed (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The normal distribution is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

$\\ \mu = 6000$ miles.

$\\ \sigma = 800$ miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

We will use the concept of the standard normal distribution, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding z-score.
A z-score is a kind of standardized value which tells us the distance of a raw score from the mean in standard deviation units. The formula for it is: $\\ z = \frac{x - \mu}{\sigma}$ . Where x is the value for the raw score (in this case x = 5000 miles).
The values for probabilities for the standard normal distribution are tabulated in the standard normal table (available in Statistics books and on the Internet). We will use the cumulative standard normal table (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

$\\ P(x>5000)$ miles.

The z-score for x = 5000 miles is:

$\\ z = \frac{5000 - 6000}{800}$

$\\ z = \frac{-1000}{800}$

$\\ z = -1.25$

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations below the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

$\\ P(z$

So

$\\ P(z$

Consulting a standard normal table available on the Internet, we have

$\\ P(z$

Then

$\\ P(z1.25)$

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

$\\ P(z>-1.25) = 1 - P(z$

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

$\\ P(z>-1.25) = 1 - 0.10565$

$\\ P(z>-1.25) = 0.89435$

In words, a truck "will be able to travel a total distance of over 5000 miles without an engine failure" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of 12 trucks, and this is a problem of the sampling distribution of the means.

In this case, we have samples from a normally distributed data, then, the sample means are also normally distributed. Mathematically:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the samples means are normally distributed with the same mean of the population mean $\\ \mu$ , but with a standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ .

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z \sim N(0, 1)$

Then

The "average time-between-failures of 5000" is $\\ \overline{x} = 5000$ . In other words, this is the mean of the sample of the 12 trucks.

Thus

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}$

$\\ z = \frac{-1000}{230.940148}$

$\\ z = -4.330126$

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

$\\ P(z$

The complement of P(z<-4.33) is:

$\\ P(z>-4.33) = 1 - P(z$ or practically 1.

In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

7 0

2 years ago