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bulgar [2K]
3 years ago
12

600 is deposited in an account that pays 7% annual interest,compunded continuosly.what is the balance after 5 years?

Mathematics
2 answers:
netineya [11]3 years ago
8 0
<span><span><span>Year 1  $642.00
Year 2  $686.94
Year 3  $735.03
Year 4  $786.48
Year 5  $841.53


</span></span></span>
SVEN [57.7K]3 years ago
4 0
Hi there

The formula of compounded continuously is
A=pe^rt
A future value?
P present value 600
R interest rate 0.07
T time 5 years
E constant
So
A=600×e^(0.07×5)
A=851.44

Hope it helps
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Answer:

c is the answer

Step-by-step explanation:

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Two bags of apples weigh 8.4 pounds.How many bags of apples will fit in a box with a 50 pound weight limit?
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A consumer group is testing camp stoves. To test the heating capacity of a stove, they measure the time required to bring 2 quar
kotykmax [81]

Answer:

a

The decision rule  is  

Reject the null hypothesis

  The conclusion is  

There is sufficient evidence to show that there is a difference between the performances of these two models

b

The  95% confidence interval is  0.224   <  \mu_1 - \mu_2  < 2.776

Step-by-step explanation:

From the question we are told that

    The sample size is  n  =  36

    The first sample mean is  \= x_1   =  11.4

    The first standard deviation is  s_1 =  2.5

    The second sample mean is   \= x_2 =  9.9

     The second standard deviation is  s_2 =  3.0

      The level of significance is  \alpha  =  0.05

The null hypothesis is  H_o  :  \mu_1 - \mu_2 = 0

The alternative hypothesis is H_a :  \mu_1 - \mu_2 \ne 0

Generally the test statistics is mathematically represented as

      z =  \frac{ (\= x_1 - \= x_2 ) - (\mu_1 - \mu_2 ) }{ \sqrt{ \frac{s_1^2 }{n} + \frac{s_2^2 }{ n}  } }

=>    z =  \frac{ ( 11.4  - 9.9) - 0  }{ \sqrt{ \frac{2.5^2 }{36} + \frac{ 3^2 }{36 }  } }

=>     z = 2.3

From the z table  the area under the normal curve to the left corresponding to  2.3 is  

       P( Z >  2.3 ) =  0.010724

Generally the p-value is mathematically represented as

      p-value =  2 * P( Z >  2.3 )

=>    p-value  =  2 * 0.010724

=>    p-value  =  0.02

From the value obtained we see that  p-value  <  \alpha hence  

The decision rule  is  

Reject the null hypothesis

  The conclusion is  

There is sufficient evidence to show that there is a difference between the performances of these two models

Considering question b

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 normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \sqrt{ \frac{s_1^2 }{n } + \frac{s_2^2}{n}}

 => E = 1.96  *    \sqrt{ \frac{2.5^2 }{ 36 } + \frac{ 3^2}{36}}

  => E = 1.276

Generally 95% confidence interval is mathematically represented as  

      ( \= x_1 - \= x_2) -E <  \mu_1 - \mu_2  < ( \= x_1 - \= x_2) + E

=>  ( 11.4 - 9.9 ) -1.276  <  \mu_1 - \mu_2 < ( 11.4 - 9.9 ) + 1.276

=>  0.224   <  \mu_1 - \mu_2  < 2.776

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1/5k=10
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