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jok3333 [9.3K]
2 years ago
11

Write a statement that increases numPeople by 5. Ex: If numPeople is initially 10, the output is: There are 15 people.

Engineering
1 answer:
BartSMP [9]2 years ago
8 0

Answer:

import java.util.Scanner;

public class AssigningNumberToVariable {

public static void main (String [] args) {

int numPeople;

int numPeople_inc;

numPeople = 10;

/* solution */

numPeople_inc=numPeople+5;

System.out.print("There are ");

System.out.print(numPeople_inc);

System.out.println(" people.");

}

}

Explanation:

1. Declare a new variable called: numPeople_inc

2. Add 5 each time: numPeople_inc=numPeople+5;

3. Print the answer:  numPeople_inc=numPeople+5;

Optional, just write the line in the place where says: " your solutions goes here":

numPeople = numPeople+5;

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The present worth of income from an investment that follows an arithmetic gradient is projected to be $475,000. The income in ye
Nikitich [7]

Answer:

G = $37,805.65

Explanation:

I found this on another site:

475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)

475,000 = 25,000(4.3553) + G(9.6842)

9.6842G = 366,117.50

G = $37,805.65

4 0
3 years ago
An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60
belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F              hi = 43.43 kJ/kg

WBT = 60 F             wi = 0.00874 kJ/kg

                                u1 = 0.8445 m/kg

oulet condition,

DBT = 70 F        RH = 100.1

h1 = 83.504kJ/kg

Wo = 0.222kJ/kg

check the attached file for complete solution

3 0
2 years ago
An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°
Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

D =  7.5 \times 10^{-11} m^2/s

for 0.225% carbon concentration following formula is used

\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})

where, erf stand for error function

\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248

0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})

erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248

erf(\frac{x}{2\sqrt{DT}}) = 0.751

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

\frac{x}{2\sqrt{DT}} = 0.815

x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)

x = 2.39\times 10^{-3} m

x = 0.002395 mm

8 0
3 years ago
A battery is connected to a resistor. Increasing the resistance of the resistor will __________. A battery is connected to a res
belka [17]

Answer: the increase in the external resistor will affect and decrease the current in the circuit.

Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is

E = IR + Ir = I(R + r)........(1)

Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)

I = E/(R + r)

As R increases, and E, r remain constant, the value (R + r) increases, hence the value of current, I, in the external circuit decreases.

8 0
3 years ago
The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm
Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0
2 years ago
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