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Tpy6a [65]
3 years ago
8

Sam promises to pay Sandy $2,000 in four years and another $3,000 four years later for a loan of $2,000 from Sandy today. What i

s the interest rate that Sandy is getting? Assume interest is compounded monthly. A. 14.75% B. 16.72% C. 15.10% D. 18.08%
Engineering
1 answer:
frozen [14]3 years ago
8 0

Answer:

the interest rate that Sandy is getting is (C.) 15.10%

Explanation:

Given data

in 4 year cash pay (p1)  = $2000

in 8 year cash pay (p2)  = $3000

time period (t1) = 4 years

time period (t2) =  8 years

loan value = $2000

To find out

interest rate

solution

first we know amount will be paid in first 4 year is

$2000 (1+r/100)^{12t}

$2000 (1+r/100)^{48}                  ...................1

now we calculate the next payment will paid after 4 year i.e.

$2000 (1+r/100)^{12t} - $2000  

$2000 (1+r/100)^{48}   - $2000  ..................2

after full time period of payment total amount will be paid by equation 1 and 2  i.e.

$3000 = $2000 (1+r/100)^{48} ×$2000 (1+r/100)^{48} - $2000

$3000 = $2000 ( (1+r/100)^{48} ×  (1+r/100)^{48} - 1 )   .....3

now we have solve   (1+r/100)^{48} this eqution

so we consider  (1+r/100)^{48} = A

so new equation will be by equation 3

$3000/$2000 = ( A - 1 ) × A

3/2 = A² - A

solve this equation we get  2A² - 2 A - 3 = 0 so A = 1.823

now we compute A again in  (1+r/100)^{48} = A

(1+r/100)^{48} = 1.823

so rate (r) = 1.258 % / month

and rate yearly = 1.258 ×12

the interest rate that Sandy is getting yearly 15.10 %

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A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and
Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B =  164.6 < 85.42°Ω  

C =  2.061 * 10^-3 < 90.32° s

D =  0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i  + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = \sqrt{\frac{z}{y} }  =  \sqrt{\frac{0.03 i  + j 0.35}{j4.4*10^-6 } }    

    = \sqrt{79837.128< 4.899^o}   =  282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate  gl

gl = \sqrt{zy} * d  

 d = 500

 z = 0.03 i  + j 0.35

 y = j4.4*10^-6 S/km

gl =  \sqrt{0.03 i  + j 0.35*  j4.4*10^-6}  * 500

   = \sqrt{1.5456*10^{-6} < 175.1^0} * 500

   = 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit  =  0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω  = 164.6 < 85.42°Ω  ( as calculated )

C = 1/Zc  sin h (gl) s  =  2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where :  cos h (gl)  = \frac{e^{gl} + e^{-gl}  }{2}

             sin h (gl) = \frac{e^{gl}-e^{-gl}  }{2}

     

7 0
2 years ago
Establishes general guidelines concerning licensing and vehicle
gavmur [86]

Answer:

A. National Highway Safety Act

Explanation:

The National Highway Safety Act establishes general guidelines concerning licensing, vehicle registration and inspection, and traffic laws for state regulations. The act was made in 1966 to reduce the amount of death on the highway as a result of increase in deaths by 30% between 1960 and 1965

National Traffic and Motor Vehicle Safety Act regulates vehicle manufacturers  by ensuring national safety standards and issuance recalls for defective vehicles

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3 years ago
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3 years ago
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A boy weighing 108-lb starts from rest at the bottom A of a 6-percent incline and increases his speed at a constant rate to 7 mi
baherus [9]

Answer:

88.18 W

Explanation:

The weight of the boy is given as 108 lb

Change to kg =108*0.453592= 48.988 kg = 49 kg

The slope is given as 6% , change it to degrees as

6/100 =0.06

tan⁻(0.06)= 3.43°

The boy is travelling at a constant speed up the slope = 7mi/hr

Change 7 mi/h to m/s

7*0.44704 =3.13 m/s

Formula for power P=F*v where

P=power output

F=force

v=velocity

Finding force

F=m*g*sin 3.43°

F=49*9.81*sin 3.43° =28.17

Finding the power out

P=28.17*3.13 =88.18 W

4 0
3 years ago
A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of
OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0
2 years ago
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