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d1i1m1o1n [39]
4 years ago
8

Could someone please help me on 2 and 3 been struggling for hours

Mathematics
1 answer:
Aneli [31]4 years ago
3 0

Answer:

  2a: amount = 24.9(442/249)^(4t-1)

  2b: 14.03 billion

  2c: 1.38 trillion

  2d: 1.158 hours

  2e: 0.302 hours

__

  3a: 2200 mL

  3b: 2450 mL

  3c: 500 mL

  3d: 5 s

  3e: 1/5 Hz or 12 cycles per minute

Step-by-step explanation:

2.

(a) I find it easiest to write an exponential equation as follows:

  define ...

     a0 = initial given amount (not necessarily at time 0)

     t0 = time corresponding to a0

     a1 = second given amount (at a later time)

     t1 = time corresponding to a1

  write the equation as ...

     amount = a0 · (a1/a0)^((t-t0)/(t1-t0))

This is usually a good place to start. It can be converted to the form ...

     amount = a0·e^(kt)

if necessary or desirable.

__

In terms of the above, the problem statement tells us ...

  a0 = 24.9 billion

  t0 = 1/4 hour

  a1 = 44.2 billion

  t1 = 1/2 hour

so our exponential equation is ...

  amount = 24.9 · (44.2/24.9)^((t-1/4)/(1/2-1/4))

  amount = 24.9(442/249)^(4t-1)

Please be aware there are many other ways to write this. Writing the equation this way will make it match the given numbers exactly. Most other forms will be an approximation. For example, you could write ...

  amount ≈ 14.03(9.9287^t)

__

(b) Filling in t=0, we find the amount to be ...

  amount = 24.9(442/249)^-1 = 24.9(249/442) ≈ 14.0274 . . . . billion

__

(c) Filling in t=2, we find the amount to be ...

  amount = 24.9(442/249)^(4·2-1) = 24.9(442/249)^7 ≈ 1382.8 . . . billion

__

(d) We want to find t for amount = 200. Filling in the number, we get ...

  200 = 24.9(442/249)^(4t-1)

  200/24.9 = (442/249)^(4t-1) . . . . . . divide by 24.9

  log(200/24.9) = (4t -1)log(442/249) . . . . take logarithms

  log(200/24.9)/log(442/249) = 4t -1 . . . . . divide by log(442/249)

Finishing the solution for t, we have ...

  t = (1/4)(log(200/24.9)/log(442/249) +1) ≈ 1.1577 . . . hours

__

(e) When td is added to t, the value of the exponential formula is doubled:

  24.9(442/249)^(4(t+td)-1) = 2(24.9(442/249)^(4t-1)

Dividing by half the expression on the left side, we have ...

  (442/249)^(4td) = 2

Taking logarithms, we get ...

  4t·log(442/249) = log(2)

and dividing by the coefficient of t gives ...

  t = log(2)/(4·log(442/249)) ≈ 0.3019686 . . . hours

_____

3. This question involves figuring out the amplitude and average value of a sine function. The amplitude is the multiplier of sin( ), so is 250 (mL). The average value is the value added to the sin( ) function, so is 2450 (mL).

At the maximum of the sine function, the amplitude is added to the average value. At the minimum, the amplitude is subtracted from the average value.

(a) The minimum lung capacity is 2450 -250 = 2200 mL.

__

(b) The average lung capacity is 2450 mL.

__

(c) The tidal volume is the difference between the maximum and minimum lung capacity. It is twice the amplitude, so is 2·250 mL = 500 mL.

__

(d) The period of the function is the value of t that makes the argument of the sine function be 2π. That value is t=5. The period is 5 seconds. This is the time required for one full cycle of respiration.

__

(e) The frequency in Hz is the reciprocal of the period in seconds. It is ...

  1/(5 s) = 1/5 Hz

In the context of this problem, it means 1/5 of a cycle of respiration is completed each second. (In terms of cycles per minute, it is 12 cycles per minute. That is, the owner of these lungs completes about 12 full cycles of respiration per minute.)

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Substitute:

a_n=-22\cdot(-4)^{n-1}=-22\cdot(-4)^n\cdot(-4)^{-1}=-22\cdot(-4)^n\cdot\left(-\dfrac{1}{4}\right)=5.5(-4)^n

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4 years ago
A meat inspector has randomly selected 30 packs of 95% lean beef. The sample resulted in a mean of 96.2% with a sample standard
kirza4 [7]

Answer:

a

 The upper bound of the 99% prediction level is 98.2  

b

 The  95% confidence interval is 9.7383 <  \mu < 10.2617

Step-by-step explanation:

Considering first question

From the question we are told that

   The sample size is  n  =  30  

   The sample mean is  \= x  =  96.2\%

   The standard deviation is s  = 0.8\%

Generally the degree of freedom is mathematically represented as

        df  =  n - 1

=>      df  =  30 - 1

=>      df  =  29

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the t distribution table the critical value  of   at a degree of freedom of is  

   t_{\alpha , 29} = 2.462

Generally the  99%  prediction level is mathematically represented as

      \= x \pm [(t_{\alpha  , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]

Generally the upper bound of the 99%  prediction level is mathematically represented as

      \= x + [(t_{\alpha  , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]  

=>    96.2 + (2.462 ) * 0.8 * (\sqrt{1 + \frac{1}{ 30} } )}]  

=>    98.2  

Considering second question

 Generally the sample is mathematically represented as

             \= x  = \frac{\sum x_i}{n}

=>           \= x  = \frac{ 9.8 + 10.2 + \cdots +9.6 }{7}  

=>           \= x  =  10    

Generally the standard deviation  is mathematically represented as

           \sigma =  \sqrt{ \frac{ \sum ( x_ i - \= x)}{n-1} }

=>        \sigma =  \sqrt{ \frac{ ( 9.8  -10)^2 +  ( 10.2  -10)^2 + \cdots + ( 9.6  -10)^2  }{7-1} }

=>        \sigma = 0.283

Generally the degree of freedom is mathematically represented as

      df =  n- 1

=>    df =  7- 1

=>    df =  6

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the t distribution table the critical value  of   at a degree of freedom of  is  

   t_{\frac{\alpha }{2} , 6 } =  2.447

Generally the margin of error is mathematically represented as  

      E = t_{\frac{\alpha }{2} , 6 } *  \frac{\sigma }{\sqrt{n} }

=>    E =2.447*    \frac{0.283 }{\sqrt{7} }

=>    E =0.2617

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>   10 -0.2617 <  \mu < 10 + 0.2617

=>   9.7383 <  \mu < 10.2617

7 0
4 years ago
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