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kogti [31]
3 years ago
14

Given an exponential function for compounding interest, A(x) = P(.91)^x, what is the rate of change?

Mathematics
2 answers:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

Option B = -9%

Step-by-step explanation:

Given : Given an exponential function for compounding interest, A(x) = P(.91)^x

To find : What is the rate of change?  

Solution :

The general form of the exponential function is f(x)=a(1+r)^x

Where,

r is the rate of change

if r> 1 then it is growth rate    

if r< 1 then it is decay rate.

Comparing given function with exponential function,

A(x) = P(.91)^x

1+r=0.91

r=0.91-1

r=−0.09

It is a decay rate.

Convert into percentage, multiply with 100

-0.09\times 100=-9\%

Therefore, Option B is correct.

The rate of change is -9%.      

Marina CMI [18]3 years ago
7 0

The confirmed correct answer is -9%. Trust me.

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A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
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Answer:

(a) Null Hypothesis, H_0 : \mu = $1,150  

    Alternate Hypothesis, H_A : \mu \neq $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(a)So, Null Hypothesis, H_0 : \mu = $1,150     {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, H_A : \mu \neq $1,150    {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                               T.S.  = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

(b) So, <u>test statistics</u>  =  \frac{1,200-1,150}{\frac{126}{\sqrt{81} } }  ~ t_8_0

                               =  3.571

The value of the sample test statistics is 3.571.

(c) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 3.571) = <u>Less than 0.05%</u>

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

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