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3 years ago

5) How did Nadia overcome her stage fright on the night of the

SAT

1 answer:

slamgirl [31]3 years ago

3 0

Answer:

hint: all you have to do is read the story lol.. but the one that makes the most sense c idk.

Explanation:

WE DON"T HAVE THE STORY

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5. If a standard 6-sided die

Olenka [21]

The probability would be 1/36

7 0

3 years ago

The box of candies is shared equally.

masha68 [24]

Answer:

hmm I guess you want to complete ur question..,m

4 0

2 years ago

Point's here you go okay byeb

Art [367]

Thank you kind sir or miss

7 0

2 years ago

saddle inc. has two types of handbags: standard and custom. the controller has decided to use a plantwide overhead rate based on

aleksklad [387]

a. Computation for the overhead rate using the traditional plantwide approach

Plantwide overhead rate = 303,000/145,000

Plantwide overhead rate= 208.97%

(43,000 + 102,000 =145,000)

b. Computation for the overhead rates using the activity-based casting approach

Machining = 196,000/2,580

Machining= 75.97 per machine hour

Machine setup = 107,000/490

Machine setup = 218.37 per setup

c. Calculation to determine the difference in allocation between the two approaches.

Overhead allocated :

Traditional costing

Standard= (43,000×208.97%)

Standard=89,857

Custom=(102,000×208.97%)

Custom=213,149

Activity based costing

Standard= (1,310×75.97)+(110×218.37)

Standard=99,520.7+24,020.7

Standard=123,541

Custom=(1,270×75.97)+(380×218.37)

Custom=96,482+82,981

Custom=179,463

Learn more here:

brainly.com/question/16565524?referrer=searchResults

5 0

2 years ago

In a study of helicopter usage and patient survival, among the 56,074 patients transported by helicopter, 235 of them

xxMikexx [17]

Using the hypergeometric distribution, it is found that the probability is 0.845 = 84.5%.

--------------------

The patients are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

The parameters are:

x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.

Combination formula:

is the number of different combinations of x objects from a set of n elements, given as:

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

--------------------

56074 patients, thus $N = 56074$
Sample of 40, thus, $n = 40$
235 left against medical advice, thus, $k = 235$ .

The probability that none left against medical advice is P(X = 0), so:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,56074,40,235) = \frac{C_{235,0}*C_{55839,40}}{C_{56074,40}} = 0.845$

The probability is 0.845 = 84.5%.

A similar problem is given at brainly.com/question/24008577

6 0

2 years ago