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3 years ago

7

You have just graduated from college and purchased a car for $8000. Your credit limit is $11,000. Assume that you make no paymen

ts and purchase nothing more and that there are no other fees. The monthly interest rate is 2%. What is your initial credit card balance? a. $19,000 c. $8000 b. $3000 d. $11,000

2 answers:

Angelina_Jolie [31]3 years ago

8 0

<span>assume the graduate put the purchase on his card.thenInitial balance = $8000 on the first statement (+fees and interest charges, if any)That means he owes the card issuer $8000.Credit balance is what the issuer owes the card holder, which is zero
</span> option "c. $8000" is your answer

Svetlanka [38]3 years ago

6 0

Answer:

d. $11,000

Step-by-step explanation:

PV=8000

FV=11000

FV = PVI(1+i)^n

Calculate n

And calculate monthly payments.

Payment Capital Interest Loan

8.000

341 181 160 7.819

341 185 156 7.635

341 188 153 7.446

341 192 149 7.254

341 196 145 7.059

341 200 141 6.859

341 204 137 6.655

341 208 133 6.447

341 212 129 6.236

341 216 125 6.019

341 220 120 5.799

341 225 116 5.574

341 229 111 5.345

341 234 107 5.111

341 239 102 4.872

341 243 97 4.628

341 248 93 4.380

341 253 88 4.127

341 258 83 3.868

341 264 77 3.605

341 269 72 3.336

341 274 67 3.062

341 280 61 2.782

341 285 56 2.497

341 291 50 2.206

341 297 44 1.909

341 303 38 1.607

341 309 32 1.298

341 315 26 983

341 321 20 662

341 328 13 334

341 334 7 0

10.908 8.000 2.908

total loan capital interest

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3 years ago

The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and

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Answer:

1a

$P(39 < X < 48 ) = 0.8767$

1b

95% of all sample means will fall between $40.1 < \mu < 49.9$

1c

$\= x = 41. 795$

2

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Step-by-step explanation:

From the question we are told that

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The population standard deviation is $\sigma = 10$

The sample size is n = 16

Generally the standard error of the mean is mathematically represented as

$\sigma_{x} = \frac{ \sigma}{\sqrt{n} }$

=> $\sigma_{x} = \frac{ 10 }{\sqrt{16 } }$

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Generally the probability that the sample mean will be between 39 and 48 minutes is

$P(39 < X < 48 ) = P( \frac{ 39 - 45}{ 2.5} < \frac{X - \mu }{\sigma } < \frac{ 48 - 45}{ 2.5} )$

=> $P(39 < X < 48 ) = P(-2.4 < Z< 1.2 )$

=> $P(39 < X < 48 ) = P( Z< 1.2 ) - P(Z < -2.4)$

From the z table the area under the normal curve to the left corresponding to 1.2 and -2.4 is

=> $P( Z< 1.2 ) = 0.88493$

and

$P( Z< - 2.4 ) = 0.0081975$

So

$P(39 < X < 48 ) = 0.88493 -0.0081975$

=> $P(39 < X < 48 ) = 0.8767$

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 is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * 2.5$

=> $E =4.9$

Generally the 95% of all sample means will fall between

$\mu -E < and \mu +E$

=> $45 -4.9\ and \ 45 + 4.9$

Generally the value which 90% of sample means is greater than is mathematically represented

$P( \= X > \= x ) = 0.90$

=> $P( \= X > \= x ) = P( \frac{\= X - \mu }{ \sigma_x} > \frac{\= x -45 }{ 2.5} ) = 0.90$

=> $P( \= X > \= x ) = P( Z > z ) = 0.90$

Generally from the z-table the critical value of 0.90 is

$z = -1.282$

$\frac{\= x -45 }{ 2.5} = -1.282$

=> $\= x = 41. 795$

Considering question 2

Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

$s = \frac{1}{5} \sigma$

Generally the standard deviation of the sample mean is mathematically represented as

$s = \frac{\sigma }{ \sqrt{n} }$

=> $\frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }$

=> $n = 5^2$

=> $n = 25$

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