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DochEvi [55]
3 years ago
7

Suppose that the number of times that Vampire Bill gets attacked by Hep V in a given year is a Poisson random variable with para

meter λ = 5. Now, he discovered a new drug (based on large quantities of vitamin C) that has just been marketed to protect from Hep V. It claims to reduce the Poisson parameter to λ = 3 for 75% of the population. For the rest of the population, the drug has no appreciable effect on Hep V. If Bill tries the drug for a year and only had 2 Hep V attacks in that time, how likely is it that the drug was beneficial for him?
Engineering
1 answer:
Masteriza [31]3 years ago
6 0

Answer:

Explained

Explanation:

let probability of getting attached by Hep(V) without benefit of new drug =P(A)

probability of getting attached by Hep(V) with new drug =P(B)

hence probability that Hep V attack are 2 given he is without benefit of drug =P(C|A) =\small e^{-\lambda }\lambda ^{x}/x! where \small \lambda=5 ,x=2

=0.084

and probability that Hep V attack are 2 given he is with new drug =P(C|B) =\small e^{-\lambda }\lambda ^{x}/x! where \small \lambda=3 ,x=2

=0.224

hence probability of Hep V attack are 2 =P(C) =P(A)*P(C|A) +P(B)*P(C|B) =0.25*0.084+0.224*0.75=0.189

from above probability that drug is beneficial given he had 2 Hep V attacks =P(B)*P(C|B)/P(C) =0.224*0.75/0.189 =0.8889

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Air is pumped from a vacuum chamber until the pressure drops to 3 torr. If the air temperature at the end of the pumping process
malfutka [58]

Answer:

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr = 3\times 0.133\ kPa

At the end of pumping, temperature of air, T = 5^{\circ}C = 278 K

After the rise in the air temperature, T' = 20^{\circ}C = 293 K

Now, we know the ideal gas eqn:

PV = mRT

So

P = \frac{m}{V}RT

P = \rho_{a}RT          (1)

where

P = Pressure

V = Volume

\rho_{a} = air\ density

R = Rydberg's constant

T = Temperature

Using eqn (1):

P = \rho_{a}RT

\rho_{a} = \frac{P}{RT}

\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3}

Now, at constant volume the final pressure, P' is given by:

\frac{P}{T} = \frac{P'}{T'}

P' = \frac{P}{T}\times T'

P' = \frac{3}{278}\times 293 = 3.16 torr

7 0
4 years ago
What is the difference between a Datum and a Datum Feature? a) A Datum and Datum Feature are synonymous. b) A Datum is theoretic
aliina [53]

Difference between Datum and Datum feature is<em> 'Datum is theoretical and Datum feature is real'. </em>

Option: (b)

<u>Explanation:</u>

A Datum is a perfect plane, line, point or surface but only occurs theoretically.

However a Datum Feature is fully based on a tangible surface, axis or point on a part where that theoretical datum is located.

The reason behind in this is they are not equal to each other because the 'part surface' is never 100% perfect.

The important functional features of the Datum is controlled during measurements.

7 0
3 years ago
Jack has been concerned about the rapidly changing green regulations in his state and his ability as a mechanical engineer to ke
uysha [10]

Answer:

Option A, B and D

Explanation:

Jack can easily convince boss if he focus around two major aspects of the company

a) Revenue enhancement - Jack must outline the benefits of his research that can be used to improvise customer offerings and  hence can be further used to devise more energy-efficient options to customer

b) Reduction in mistakes - Issues such as poor implementation can be avoided with better approach and understanding.

Hence, option A, B and D are correct

3 0
3 years ago
In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery
Dima020 [189]

Answer:

  248.756 mV

  49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

  voltage: (100 V)(1000/(1000 +1000) = 50 V

  impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

  voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

  impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

__

The open-circuit voltage is the difference between these terminal voltages:

  (50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

__

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

  (50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0
3 years ago
Assume we have already defined a variable of type String called password with the following line of code: password' can have any
omeli [17]

Answer:

The Java code is given below with appropriate comments for better understanding

Explanation:

import java.util.Scanner;

public class ValidatePassword {

  public static void main(String[] args) {

      Scanner input = new Scanner(System.in);

          System.out.print("Enter a password: ");

          String password = input.nextLine();

          int count = chkPswd(password);

          if (count >= 3)

              System.out.println("Secure");

          else

              System.out.println("Not Secure");

  }

  public static int chkPswd(String pwd) {

      int count = 0;

      if (checkForSpecial(pwd))

          count++;

      if (checkForUpperCasae(pwd))

          count++;

      if (checkForLowerCasae(pwd))

          count++;

      if (checkForDigit(pwd))

          count++;

      return count;

  }

  // checks if password has 8 characters

  public static boolean checkCharCount(String pwd) {

      return pwd.length() >= 7;

  }

  // checks if password has checkForUpperCasae

  public static boolean checkForUpperCasae(String pwd) {

      for (int i = 0; i < pwd.length(); i++)

          if (Character.isLetter(pwd.charAt(i)) && Character.isUpperCase(pwd.charAt(i)))

              return true;

      return false;

  }

  public static boolean checkForLowerCasae(String pwd) {

      for (int i = 0; i < pwd.length(); i++)

          if (Character.isLetter(pwd.charAt(i)) && Character.isLowerCase(pwd.charAt(i)))

              return true;

      return false;

  }

  // checks if password contains digit

  public static boolean checkForDigit(String pwd) {

      for (int i = 0; i < pwd.length(); i++)

          if (Character.isDigit(pwd.charAt(i)))

              return true;

      return false;

  }

  // checks if password has special char

  public static boolean checkForSpecial(String pwd) {

      String spl = "[email protected]#$%^&*(";

      for (int i = 0; i < pwd.length(); i++)

          if (spl.contains(pwd.charAt(i) + ""))

              return true;

      return false;

  }

}

5 0
3 years ago
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