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4 years ago

What is the percentage of error in weight if measured weight is 22 pounds and actual weight is 25 pounds

1 answer:

7 0

The percent error is by 12%

<span>The percent error is calculated by subracting the actual weight from the estimated weight and then dividing it by the actual value and multiplying it by 100% in absolute value(Which means to remove any negative sign) </span>

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Simplify 1.61 to 2.3

andrew-mc [135]

Answer:

The answer is 37.03

Step-by-step explanation:

Hope this is correct

HAVE A GOOD DAY!

7 0

4 years ago

) In the first week of her school's fundraiser, Julie sold 6 items. By the end of the

VashaNatasha [74]

Answer:

6+x > 20

Step-by-step explanation:

solving the inequality-

subtract 6 from sides.

it is x> 14

Julie sold more than 14 items after the first week of the fundrai.

have a great Day!

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3 years ago

100 POINTS NO JOKE

Alenkasestr [34]

Answer: y=18x

Step-by-step explanation:

18$ each. You start off with no profit so.

5 0

3 years ago

Find the slope plz help help plz

BaLLatris [955]

Answer:

-2

Step-by-step explanation:

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3 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

4 years ago