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Advocard [28]
3 years ago
6

If $4000 is borrowed at a rate of 16% interest

Mathematics
1 answer:
Gemiola [76]3 years ago
6 0

Answer:

(a) $7492

(b) $10,253

(c) $14,032

Step-by-step explanation:

As we know, the final Amount can be calculated with the formula for compound interest,

A = P(1 + \frac{r}{n} )^{nt}

where,

A = Final Amount due

P = Initial principal amount borrowed

r = rate of interest in decimal

n = number of times applied per time period

t = total time period

Now, according to the given data,

(a) in 4 years ;-

⇒ A = 4000(1 + \frac{0.16}{4} )^{4(4)}

⇒ A = 7492

(b) in 6 years ;-

⇒ A = 4000(1 + \frac{0.16}{4} )^{4(6)}

⇒ A = 10,253

(c) in 8 years ;-

⇒ A = 4000(1 + \frac{0.16}{4} )^{4(8)}

⇒ A = 14,032

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Bentley is going to invest $98,000 and leave it in an account for 7 years. Assuming
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The rate of interest for compounded daily is 2.1 6

Step-by-step explanation:

Given as :

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<u>From compounded method</u>

Amount = Principal × (1+\dfrac{rate}{365\times 100})^{365\times Time}

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4 0
3 years ago
The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
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Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

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                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

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(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

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Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

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The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

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