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Elis [28]
2 years ago
13

Assume a bank loan requires an interest payment of $85 per year and a principal payment of $1,000 at the end of the loan's eight

-year life. What would be the present value of this loan if it carried a 10% interest rate?
Mathematics
1 answer:
lapo4ka [179]2 years ago
3 0

Answer:

The present value is  PV  = \$ 396,987

Step-by-step explanation:

From the question we are told that

   The  interest payment per year is  C =  \$ 85

    The principal payment is  P =  \$ 1000

     The  duration is  n =  8   years

      The  interest rate is  r =  10\%  =  0.10

The present value is  mathematically represented as

      PV  =  [ \frac{C}{r} *  [1 - \frac{1 }{ (1 +r)^n} ] + \frac{P}{(1 + r)^n}  ]

substituting values

      PV  =  [ \frac{85}{0.10} *  [1 - \frac{1 }{ (1 +0.10)^8} ] + \frac{1000}{(1 + 0.10)^ 8}  ]

      PV  = \$ 396,987

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2h x (l+b)

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A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of
notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.01 = 0.99, so Z = 2.327.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

M = z\frac{\sigma}{\sqrt{n}}

0.0001 = 2.327\frac{0.0003}{\sqrt{n}}

0.0001\sqrt{n} = 2.327*0.0003

\sqrt{n} = \frac{2.327*0.0003}{0.0001}

(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2

n = 48.73

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49 measurements are needed.

7 0
2 years ago
Sitting on a park bench, you see a swing that is 100 feet away and a slide that is 80 feet away. The angle between them is 30 de
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We are given : Distance of the swing = 100 feet.

Distance of slide = 80 feet.

Angle between swing and slide = 30 degrees.

We need to find the distance between the swing and the slide.

Distance of swing, distance of slide and distance between the swing and the slide form a triangle.

We can apply cosine law to find the distance between the swing and the slide.

c^2 = a^2 +b^2  - 2ab cos C

c^2 = 100^2 +80^2 - 2(100)(80) cos 30°

c^2 = 10000 + 6400 -2* 8000 (\frac{\sqrt{3}} 2)}

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<h3>Therefore, the approximate distance between the swing and the slide is 50 feet.</h3>
5 0
2 years ago
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class consists of 55% boys and 45% girls. It is observed that 25% of the class are boys and scored an A on the test, and 35% of
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6 0
2 years ago
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KonstantinChe [14]
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