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stiv31 [10]
2 years ago
10

The data shows that recently the size of the raccoon population decreased. How will the decrease in the raccoon population affec

t the other populations?
Engineering
1 answer:
LenaWriter [7]2 years ago
3 0

Answer: Increase in the population of lower organisms.

Explanation: Raccoons are mesopredators found in the middle levels of food webs and have critical impacts on the dynamics of many other species. Thus, they greatly contribute to the functioning of the ecosystem.

They are naturally equipped to checkmate the population size of other organism that are below them (i.e., their preys) in the food web. They are; predators, pathogen carriers (such as rabies), and competitors, as they compete with some specialist in the food web.

Hence, the decrease in the population of raccoons will lead to the increase in the population size of lower organisms they prey on in the food web or chain.

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La probabilidad de que un nuevo producto tenga éxito es de 0.85. Si se eligen 10 personas al azar y se les pregunta si compraría
liq [111]

Answer:

La probabilidad pedida es 0.820196

Explanation:

Sabemos que la probabilidad de que un nuevo producto tenga éxito es de 0.85. Sabemos también que se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto. Para responder a la pregunta, primero definiremos la siguiente variable aleatoria :

X: '' Número de personas que adquirirán el nuevo producto de 10 personas a las que se les preguntó ''

Ahora bien, si suponemos que la probabilidad de que el nuevo producto tenga éxito se mantiene constante (p=0.85) y además suponemos que hay independencia entre cada una de las personas al azar a las que se les preguntó ⇒ Podemos modelar a X como una variable aleatoria Binomial. Esto se escribe :

X ~ Bi(n,p) en donde ''n'' es el número de personas entrevistadas y ''p'' es la probabilidad de éxito (una persona adquiriendo el producto) en cada caso.

Utilizando los datos ⇒ X ~ Bi(10,0.85)

La función de probabilidad de la variable aleatoria binomial es :

p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}    con x=0,1,2,...,n

Si reemplazamos los datos de la pregunta en la función de probabilidad obtenemos :

P(X=x)=\left(\begin{array}{c}10&x\end{array}\right)(0.85)^{x}(0.15)^{10-x} con x=0,1,2,...,10

Nos piden la probabilidad de que por lo menos 8 personas adquieran el nuevo producto, esto es :

P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)

Calculando P(X=8), P(X=9) y P(X=10) por separado y sumando, obtenemos que P(X\geq 8)=0.820196

7 0
2 years ago
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat
zlopas [31]

Answer: 10.631\times 10^3\ N/m^2

Explanation:

Given

Discharge is Q=12.5\ L

Diameter of pipe d=150\ mm

Distance between two ends of pipe L=800\ m

friction factor f=0.008

Average velocity is given by

\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s

Pressure difference is given by

\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow  \Delta P=10.631\ kPa

8 0
2 years ago
Given a matrix, clockwise-rotate elements in it. Please add code to problem3.cpp and the makefile. Use the code in p3 to test yo
rusak2 [61]

Answer:

/* C Program to rotate matrix by 90 degrees */

#include<stdio.h>

int main()

{

int matrix[100][100];

int m,n,i,j;

printf("Enter row and columns of matrix: ");

scanf("%d%d",&m,&n);

 

/* Enter m*n array elements */

printf("Enter matrix elements: \n");

for(i=0;i<m;i++)

{

 for(j=0;j<n;j++)

 {

  scanf("%d",&matrix[i][j]);

 }

}

 

/* matrix after the 90 degrees rotation */

printf("Matrix after 90 degrees roration \n");

for(i=0;i<n;i++)

{

 for(j=m-1;j>=0;j--)

 {

  printf("%d  ",matrix[j][i]);

 }

 printf("\n");

}

 

return 0;

 

}

5 0
3 years ago
Describe the three primary duties for engineer
Wewaii [24]

Answer:Prepare plans with detailed drawings that include project specifications and cost estimates.

Design and execute engineering experiments to create workable solutions.

Develop engineering calculations, diagrams and technical reports.

Explanation:

4 0
3 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
2 years ago
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