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lilavasa [31]
3 years ago
13

If $6500 is invested at a rate of 6% compounded continuously, find the balance in the account after 3 years

Mathematics
1 answer:
Alex787 [66]3 years ago
6 0

\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6500\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ t=years\dotfill &3 \end{cases} \\\\\\ A=6500e^{0.06\cdot 3}\implies A=6500e^{0.18}\implies A\approx 7781.91

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A recent study found that 51 children who watched a commercial for Walker Crisps (potato chips) featuring a long-standing sports
hichkok12 [17]

Answer:

1

The claim is that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

2

The kind of test to use is a t -test because a t -test is used to check if there is a difference between means of a population

3

t  =  3.054

4

The p-value  is   p-value  =  P(Z >  3.054) = 0.0011291

5

The conclusion is  

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

The test statistics is  

Step-by-step explanation:

From the question we are told that

   The first sample size is  n_1  =  51

    The first sample  mean is \mu_1  =  36

    The second sample size is  n_2  =  41

    The second sample  size is  \mu_2  =  25

     The first standard deviation is  \sigma _1  =  21.4 \  g

    The second standard deviation is  \sigma _2  =  12.8 \  g

  The  level of significance is  \alpha =  0.05

The  null hypothesis is  H_o  :  \mu_1 = \mu_ 2

The  alternative hypothesis is  H_a :  \mu_1 > \mu_2

Generally the test statistics is mathematically represented as

    t  =  \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2}{n_1}  + \frac{s_2^2}{n_2}  } }

=>   t  =  \frac{ 36 - 25}{ \sqrt{ \frac{ 21.4^2}{51}  + \frac{ 12.8^2}{41}  } }

=> t  =  3.054

The  p-value is mathematically represented as

     p-value  =  P(Z >  3.054)

Generally from the z table  

             P(Z >  3.054) =  0.0011291

=>   p-value  =  P(Z >  3.054) = 0.0011291

From the values obtained  we see that p-value  < \alpha so  the null hypothesis is rejected

Thus the conclusion is  

  There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

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