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timofeeve [1]
3 years ago
10

A 30-year, $1,000 strip bond was traded for $167, four years after it was issued. What was the semi-annually compounded nominal

rate at that time? Calculate percentages accurate to the nearest 0.01%.
Mathematics
1 answer:
Anna007 [38]3 years ago
4 0

Answer:

The semi-annually compounded nominal rate at that time is 7%

Step-by-step explanation:

In order to calculate the semi-annually compounded nominal rate at that time we would have use the following formula:

PV= FV/(1+r)^n

According to the given data we have the following:

PV=$167

FV=$1,000

n=30-year, and strip bond was traded four years after it was issued, hence, n=(30-4)*2 =52

Therefore, 167= $1,000/( 1+r)^52

167/$1,000 =1/(1+r)^52

0.167 =1/(1+r)^52

r =3.50%

Therefore, The semi-annually compounded nominal rate at that time=3.50%*2

The semi-annually compounded nominal rate at that time=7%

The semi-annually compounded nominal rate at that time is 7%

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3 years ago
Given a right triangle with a hypotenuse of 15 feet and a height of 6 feet, find the measurement of angle
Artist 52 [7]

Answer:

66.42° between height and hypotenuse

23.58° between base and hypotenuse

Step-by-step explanation:

Using the concept of cosine and sine, we know that cosine of an angle is equal to the adjacent divided by hypotenuse.

Similarly, sine of an angle is given by opposite divided by hypotenuse

Tangent is given by opposite divided by adjacent

In rhis case, we use cosine

\theta=cos^{-1}(\frac {6}{15})=66.4218215217982\\\theta\approx 66.42^{\circ}

Therefore, the angle between the hypotenuse and height is 66.42°

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8 0
3 years ago
A pan of warm water (46∘ Celsius) was put in a refrigerator. After 15 minutes, the water's temperature was 27∘ C; 15 minutes aft
Cloud [144]

Answer:

The refrigerator was at around 13 ∘C

Step-by-step explanation:

Newton's Law of Cooling:

The rate of change of a body temperature  (amount of heat loss/time of loss) is directly proportional to the difference between its own temperature and the surroundings.

\frac{T(t2)-T(t1)}{t2-t1} = -h (T(t2) - Tenv)

T ⇒ temperature

t ⇒ time

Tenv ⇒refrigerator temperature

\frac{T(t2)-T(t1)}{t2-t1} ⇒ rate of change of he temperature

-h ⇒ constant of proportionality (negative because the temperature is decreasing inside the refrigerator)

We have 3 points:

time (minutes) - Temperature (∘ C)

0 (when the pan was put in the refrigerator) - 46

15 (after 15 minutes) - 27

30 (15 minutes after the first 15 minutes) - 19

\frac{27 - 46}{15 - 0} = -h (27 - Tenv)

\frac{19 - 27}{15 - 0} = -h (19 - Tenv)

Now we have a system of two equations and two variables

\frac{-19}{15} = -h (27 - Tenv)\\ \frac{19}{15} = h (27 - Tenv)\\ \frac{19}{15(27 - Tenv)} = h

\frac{19 - 27}{15 - 0} = -h (19 - Tenv)\\\frac{-8}{15} = -h (19 - Tenv)\\\frac{8}{15} = h (19 - Tenv)\\\frac{8}{15} = \frac{19}{15(27 - Tenv)} (19 - Tenv)\\8(27 - Tenv) = 19 (19 - Tenv)\\216 - 8 Tenv = 361 - 19 Tenv\\19 Tenv - 8 Tenv = 361 - 216\\11 Tenv = 145 \\Tenv = 145/ 11\\Tenv = 13. 18

The refrigerator was at around 13 ∘C

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