If 17 ounces of cocoa are needed for 6 cookies, how many ounces of cocoa is needed for 8 cookies?

vladimir1956 [14]

Answer:

17.82

Step-by-step explanation:

since this is a right triangle you can use one of the trigonometric ratios, in this case it would be cosine because you have values for the adjacent and hypotenuse sides

cos(27°) = x/20

x = 20 · cos(27°)

x = 17.82

5 0

2 years ago

Software filters rely heavily on ""blacklists"" (lists of known ""phishing"" URLs) to detect fraudulent e-mails. But such filter

crimeas [40]

Answer:

a) $X \sim Binom(n=16, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The expected value is given by this formula:

$E(X) = np=16*0.2=3.2$

b) $P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Part a

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=16, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The expected value is given by this formula:

$E(X) = np=16*0.2=3.2$

Part b

For this case we want this probability:

$P(X=0)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And using this function we got:

$P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815$

6 0

3 years ago

Question: Researchers in Pakistan wanted to better understand the effects of anthracycline (a chemotherapeutic drug) on the hear

Sedbober [7]

Using the normal distribution, it is found that:

1. His z-score was of Z = -1.88.

2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of $\mu = 1.35$ .

The standard deviation is of $\sigma = 0.33$ .

Item 1:

Considering his ratio, we have that X = 0.73, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.73 - 1.35}{0.33}$

$Z = -1.88$

His z-score was of Z = -1.88.

Item 2:

The probability is the <u>p-value of Z = -1.88</u>, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

Item 3:

Score of X = 1.96, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.96 - 1.35}{0.33}$

$Z = 1.85$

The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

Item 4:

Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

More can be learned about the normal distribution at brainly.com/question/24663213

7 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. <u>Allowance for Doubtful Accounts</u>

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

<u>Accounts Receivable</u>

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%)

<u>Allowance for Doubtful Accounts</u>

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

There are 5 roses for every 9 carnations in a bouquet. Mr.Rodriguez buys a bouquet with 42 flowers in it. How many carnations ar

Novosadov [1.4K]

Answer:

I think its eight but I could be wrong

3 0

3 years ago

Read 2 more answers