What is 203593^54/38n^7

An electric power plant uses solid waste for fuel in the production of electricity. The cost Y in dollars per hour to produce el

andrew-mc [135]

Answer:

15.64 MW

Explanation:

The computation of value of X that gives maximum profit is shown below:-

Profit = Revenue - Cost

= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2

= 14.7x - .47x^2 - 12

After solving the above equation we will get maximum differentiate for profit that is

14.7 - 0.94x = 0

So,

x = 15.64 MW

Therefore for computing the value of X that gives maximum profit we simply solve the above equation.

8 0

4 years ago

Which best describes the body in terms of simple machines?

alex41 [277]

Answer:B

Explanation:

5 0

3 years ago

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

a)

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

b)

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

c)

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

d)

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

e)

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

3 years ago

You are investigating surface hardening in iron using nitrogen gas. Two 5 mm thick slabs of iron are separately exposed to nitro

Luden [163]

Answer and Explanation:

The explanation is attached below

4 0

3 years ago

When a variable is stored in memory, it is associated with an address. To obtain the address of a variable, the & operator c

liubo4ka [24]

Answer:

Explanation:

1) C program file addressOfScalar.c

#include <stdio.h>

int main()

{

//intialize a char variable, print its address and the next address

char charvar = 'a';

printf("address of charvar = %p\n", (void *)(&charvar));

printf("address of charvar - 1 = %p\n", (void *)(&charvar - 1));

printf("address of charvar + 1 = %p\n", (void *)(&charvar + 1));

//intialize a int variable, print its address and the next address

int intvar = 1;

printf("address of intvar = %p\n", (void *)(&intvar));

printf("address of intvar - 1 = %p\n", (void *)(&intvar - 1));

printf("address of intvar + 1 = %p\n", (void *)(&intvar + 1));

}

In C programming language, an int variable takes 4 bytes of memory. So any arithmetic on integer address, always considers it as 4 bytes of data. So intvar-1 refers to a location 4 bytes before intvar's address and intvar+1 refers to 4 bytes after intvar's address.

3 0

3 years ago