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inessss [21]
3 years ago
5

Many credit card companies charge a compound interest rate of 1.8% per month on a credit card balance. Nelson owes $850 on a cre

dit card. If he makes no purchases or payments, he will go deeper and deeper into debt.
Mathematics
2 answers:
Finger [1]3 years ago
8 0

The sequence will be


850, 850 \times (1.018)^1, 850 \times (1.018)^2, 850 \times (1.018)^3, 850 \times (1.018)^4, ...


That's


850, 865.30, 880.875, 896.731, 912.872, ...



Angelina_Jolie [31]3 years ago
4 0

Answer:

We will do 850(1.018)^{x} to know how much Nelson will go into debt with each passing month.

Step-by-step explanation:

Many credit card companies charge a compound interest rate of 1.8% per month on a credit card balance.

Means each month the rate increases exponentially.

Given is - Nelson owes $850 on a credit card and makes no purchases or payments, he will get into debt in the following way:

We will do 850(1.018)^{x}

Here x represents the time.

$850 was for the first month.

850(1.018)^{2} = $880.87 is for the 2nd month.

850(1.018)^{3} = $896.73 is for the 3rd month.

850(1.018)^{4} = $912.87 is for the fourth month.

850(1.018)^{5} = $929.30 for fifth month and so on.

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A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv
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Answer:

Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

            of  novels that seniors read during their time in college. An exit

            survey was conducted with a random sample of 9 seniors. The

            sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

              been met, which of the following is a 94.645% confidence

              interval for the population mean number of novels read by

              all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

                                = n - 1

                                = 9 - 1

                                = 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

   =1 - 0.94645

  =0.05355 ≈ 0.05

⇒α = 0.5

Now,

\frac{\alpha}{2} = \frac{0.05}{2}

  = 0.025

Then,

t_{\frac{\alpha}{2}, df }  = 2.306

∴ we get

Population mean = x⁻ ± t_{\frac{\alpha}{2}, df } ×\frac{s}{\sqrt{n} }

                           = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

⇒Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

3 0
3 years ago
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