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3 years ago

7

Six students are being grouped into three pairs to work on a science lab. how many different combinations of three pairs are pos

sible

1 answer:

pychu [463]3 years ago

4 0

We want to select 3 pairs of people from a group of 6.

We use the formula:

n! / (r! * (n -r)!)

where n = 6 and r = 2

combinations = 6! / (2! * 4!) =

combinations = 6 * 5 / 2

combinations = 15

Source:

1728.com/combinat.htm

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2 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

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The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

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$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

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Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

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So, we have:

$A = Ce^{kt}$

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$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

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$t = 6\ A = 500000$

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$500000= 1000000e^{k*6}$

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$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

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$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

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The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

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