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statuscvo [17]
3 years ago
14

Serena invests $2500 for 3 years and 6 months and earns $328.13. What was the rate of interest? Explain how you solved.

Mathematics
1 answer:
NARA [144]3 years ago
6 0

Answer:

4.38%

Step-by-step explanation:

<em>Assuming simple interest</em>, we start by writing the formula for simple interest.

SI=Prt

Where

SI is the interest amount (328.13)

P is the principal amount (2500)

r is the rate of interest (what we need to find)

t is the time, in years (t = 3)

Since we want rate of interest, we re-arrange the formula to get r in terms of the other variables. Shown below:

SI=Prt\\r=\frac{SI}{Pt}

Now we simply plug-in the information we know and find r:

r=\frac{328.13}{(2500)(3)}\\r=0.0438

To convert into percentage, we multiply by 100:

r = 0.0438 * 100 = 4.38%

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If f(a + b) = f(a) + f(b) - 2f(ab) for all nonnegative integers a and b, and f(1) = 1, compute f(1986)
Mumz [18]

Answer:

f(1986) = 0

Step-by-step explanation:

f(a + b) = f(a) + f(b) - 2f(ab)

We need to find f(1986)

f(1986) = f(1 +1985). Using the above formula, we can write:

f(1 +1985) = f(1) + f(1985) - 2f(1 x 1985)

f(1986) = 1 - f(1985)                                         Equation 1

Applying the same formula again on f(1985), we get:

f(1985) = f(1 + 1984) = f(1) + f(1984) - 2f(1984)

f(1985) = 1 - f(1984)

Using this value in Equation 1, we get:

f(1986) = 1 - (1 - f(1984))

f(1986)= f(1984)

Continuing this, we can observe,

f(1986) = f(1984) = f(1982) = f(1980) .... = f(4) = f(2)

So,

f(1986) = f(2)

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f(2) = 1 + 1 - 2 = 0

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3 years ago
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Answer:

n ∝ (1/l) (√F/m)

Step-by-step explanation:

Since frequency, n of vibration of a stretched string is a function of its tension F, the length l and the mass per unit length m, and has dimensions [T]⁻¹ where T = time, its dimension must be equal to that of the combination of F, L and m

Since n = f(F,L,m)

n = kFᵃLᵇmˣ

dimension of n = (dimension of F)ᵃ × (dimension of L)ᵇ × (dimension of m)ˣ

Since n = frequency, dimension of n = [T]⁻¹

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Also, L = length, dimension of L = [L] and

m = mass per unit length, dimension of m = [M][L]⁻¹

So, n = FᵃLᵇmˣ

[T]⁻¹  = ([M][L][T]⁻²)ᵃ( [L] )ᵇ([M][L]⁻¹)ˣ

[T]⁻¹  = ([M]ᵃ[L]ᵃ[T]⁻²ᵃ[L]ᵇ([M]ˣ[L]⁻ˣ

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equating the exponents on both sides, we have

a + x = 0  (1 )⇒ x = -a

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n = kFᵃLᵇmˣ

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Answer:

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