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2 years ago

Calculate the mean and median of the following

Engineering

1 answer:

AfilCa [17]2 years ago

8 0

Answer:

mean:24/5

ascending order:1,2,3,4,5,5,6,6,7,9

median:5

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OlgaM077 [116]

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3 years ago

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Describe the engineering design process in your own words.

MrMuchimi

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3 years ago

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A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro

Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= $\frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt$

$= \int\limits^a_b {(5-3t)} \, dt$

$5t - \frac{3t^2}{2} +c$

v2 = 5x2 - 3x2 + c

= 10-6+c

= 4+c

$s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c$

S2 - S1

$=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)$

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

5 0

2 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

so here

volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

8 0

2 years ago

I gave 15 min to finish this java program

lisov135 [29]

Answer:

class TriangleNumbers

{

public static void main (String[] args)

{

for (int number = 1; number <= 10; ++number) {

int sum = 1;

System.out.print("1");

for (int summed = 2; summed <= number; ++summed) {

sum += summed;

System.out.print(" + " + Integer.toString(summed));

}

System.out.print(" = " + Integer.toString(sum) + '\n');

}

Explanation:

We need to run the code for each of the 10 lines. Each time we sum numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.

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3 years ago