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astraxan [27]
2 years ago
7

At a jeans factory, a random quality check found 8 pairs defective and 292 pairs with no problems. how many defective pairs can

be expected in the inventory of 3,000 pairs of jeans? 80 pairs 82 pairs 292 pairs 375 pairs
Mathematics
1 answer:
anygoal [31]2 years ago
6 0
Hello!

to solve this equation, you have to find the percentage of defects out of the total amount of jeans. if there are 300 pairs of jeans with 8 faulty pairs, then the percentage of defects would be 2.666(etc). 2.666(etc) percent of 3000 is roughly 80 pairs of defective jeans.

therefore your final answer would be 80 pairs of jeans!
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Margarita [4]

Answer:

output = input^3 - 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
How do you do this for expanding and factoring expressions 8(4x-12)
Snezhnost [94]

Answer:

32x - 96

Step-by-step explanation:

Apply the distributed property:

  1. 8 × 4x = 32x
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I hope this helps!

7 0
3 years ago
The undefined terms line and plane are needed to precisely define which mathematical term?
vladimir1956 [14]

One <em><u>possible answer</u></em> is:

Parallel lines

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6 0
4 years ago
After 4 days, a plant is measured to be 10 inches tall. Later, after a total of 20 days, then height is measured to be 18 inches
vesna_86 [32]

Given :

After 4 days, a plant is measured to be 10 inches tall.

Later, after a total of 20 days, then height is measured to be 18 inches.

To Find :

The rate of growth for the plant.

Solution :

We know, rate is given by :

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Therefore, rate of increase of height is 0.5 inch/day .

Hence, this is the required solution.

7 0
3 years ago
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