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coldgirl [10]
2 years ago
5

Which of the following statements is true? A) A counting semaphore can never be used as a binary semaphore. B) A binary semaphor

e can never be used as a counting semaphore. C) Spinlocks can be used to prevent busy waiting in the implementation of semaphores. D) Counting semaphores can be used to control access to a resource with a finite number of instances.
Mathematics
1 answer:
saveliy_v [14]2 years ago
5 0

Answer:

C

Step-by-step explanation:

Statement C of the following statements is true

C). Spin-locks can be used to prevent busy waiting in the implementation of semaphore.

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Consider the lengths of stay at a hospital’s emergency department. Hours Count Percent 1 18 3.44 2 55 10.50 3 81 15.46 4 109 20.
Vladimir [108]

Answer:

Probability = 0.502

Step-by-step explanation:

We are given the following data :

 Hours         Count            Percent

  1                    18               3.44

  2                    55              10.50

  3                    81               15.46

  4                    109             20.80

  5                     88              16.79

  6                     66              12.60

  7                     39               7.44

  8                     17                3.24

  9                     17                3.24

  10                   19                3.63

   15                  15                2.86

We need to calculate the probability

P(Length of stay of exactly 1 is less than or equal to 4)

P(Y \leq 4) = P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)

P(Y \leq 4) = 0.0344 + 0.1050 + 0.1546 + 0.2080 = 0.502

We convert the percent into probabilities by dividing them with 100. This gave us the required probabilities.

8 0
2 years ago
A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints
Sliva [168]

Answer:

a)0.099834

b) 0

Step-by-step explanation:

To solve for this question we would be using , z.score formula.

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.

a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.

Standard Deviation = √variance

= √0.16 = 0.4

Standard deviation = 0.4

Mean = 21.37

x = 20.857

z = (x-μ)/σ

z = 20.857 - 21.37/0.4

z = -1.2825

P-value from Z-Table:

P(x<20.857) = 0.099834

b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.

z score formula used = (x-μ)/σ/√n

x = 20.857

Standard deviation = 0.4

Mean = 21.37

n = 100

z = 20.857 - 21.37/0.4/√100

= 20.857 - 21.37/ 0.4/10

= 20.857 - 21.37/ 0.04

= -12.825

P-value from Z-Table:

P(x<20.857) = 0

c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.

5 0
2 years ago
A heap of rubbish in the shape of a cube is being compacted into a smaller cube. Given that the volume decreases at a rate of 3
inna [77]

Answer:

-1/4 meter per minute

Step-by-step explanation:

Since, the volume of a cube,

V=r^3

Where, r is the edge of the cube,

Differentiating with respect to t ( time )

\frac{dV}{dt}=3r^2\frac{dr}{dt}

Given, \frac{dV}{dt}=-3\text{ cubic meters per minute}

Also, V = 8 ⇒ r = ∛8 = 2,

By substituting the values,

-3=3(2)^2 \frac{dr}{dt}

-3=12\frac{dr}{dt}

\implies \frac{dr}{dt}=-\frac{3}{12}=-\frac{1}{4}

Hence, the rate of change of an edge is -1/4 meter per minute.

4 0
2 years ago
There are 252 pieces of candy in a 14-ounce jar. How many pieces of candy are there per ounce in the jar?
djyliett [7]

"Per" essentially means "divided by." To find pieces per ounce, divide pieces by ounces.


(252 pieces)/(14 ounces) = (252/14) pieces/ounce = 18 pieces/ounce

5 0
2 years ago
Read 2 more answers
A soda factory has a special manufacturing line to fill large bottles with 2 liters of their beverage. Every process is computer
Drupady [299]

Answer:

z= \frac{2.1-1.98}{0.08}= 1.5

And we can use the normal standard table and the complement rule and we got:

P(z>1.5)= 1-P(Z

And the best answer would be:

C 0.07

Step-by-step explanation:

Let X the random variable who represent the amount of soda filled in large bottles and we know this:

\mu = 1.98, \sigma =\sqrt{0.0064}= 0.08

And we want to find this probability:

P(X> \mu +1.5 \sigma = 1.98 +1.5*0.08 =2.1)

And for this case we can use the z score formula given by:

z=\frac{X -\mu}{\sigma}

And replacing we got:

z= \frac{2.1-1.98}{0.08}= 1.5

And we can use the normal standard table and the complement rule and we got:

P(z>1.5)= 1-P(Z

And the best answer would be:

C 0.07

7 0
2 years ago
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