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mestny [16]
3 years ago
13

The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the o

rganism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5556 years. Suppose C(t) is the amount of carbon-14 present at time t. The exponential decay differential equation that models this scenario is C'=-kC . Solve the differential equation to answer the following questions.
(a) Find the value of the constant k in the differential equation.
Mathematics
1 answer:
poizon [28]3 years ago
6 0
<span>As we know that the
 k = ln(2)/t½
so putting values
 = ln(2)/5556 years 
</span><span> =0.6931471/5556
</span> =0.0001247565yr⁻
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Write down the equation of a straight line that is parallel to y = 4x+2
ELEN [110]

Answer:

y = 4x + 5

Step-by-step explanation:

anything with the slope of 4x is parallel to the given equation. therefore, there are an infinite amount of answers.

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4 0
2 years ago
Steve is buying apples for the fifth grade each bag holds $12 if there are 75 students total how many bags of apples will Steve
schepotkina [342]
Steve will have to buy "6.25" bags of apples if he wants to give one apple to each student.
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8 0
3 years ago
An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha
TEA [102]

Answer:

The probability is   P(X  >300 ) = 0.97219

Step-by-step explanation:

From the question we are told that

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 The sample size n =  320 passengers

  The probability the a randomly selected passenger shows up on to the airport

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Generally the mean is mathematically represented as

    \mu  =  n*  p

  => \mu  =  320 *  0.96

    => \mu  = 307.2

Generally the standard deviation is  

    \sigma =  \sqrt{n *  p *  (1 -p ) }

=>  \sigma =  \sqrt{320  *  0.96 *  (1 -0.96 ) }

=> \sigma =3.50

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

  P(X  > k ) =  P( \frac{ X -\mu }{\sigma }  >  \frac{k - \mu}{\sigma } )

Here \frac{ X -\mu }{\sigma }  =Z (The \ standardized \  value \  of  \ X )

=>P(X  >300 ) =  P(Z >  \frac{300 - 307.2}{3.50} )

Now applying  continuity correction we have

    P(X  >300 ) =  P(Z >  \frac{[300+0.5] - 307.2}{3.50} )    

=>    P(X  >300 ) =  P(Z >  \frac{[300.5] - 307.2}{3.50} )

=>    P(X  >300 ) =  P(Z >  -1.914 )

From the z-table  

    P(Z >  -1.914 ) =  0.97219

So

    P(X  >300 ) = 0.97219

8 0
3 years ago
If m is 7 and x is 7 so what is m​
olchik [2.2K]
Can u just post the whole question or elaborate further
6 0
3 years ago
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ivann1987 [24]
For this case the first thing we must observe is that the mass increases 0.4 grams when the diameter increases 1 millimeter.
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6 0
3 years ago
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