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3 years ago

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Manuel Fraser’s bank granted him a single-payment loan of $9,650. He agreed to repay the loan in 146 days at an ordinary interes

t rate of 7.75%. What is the maturity value of the loan?

2 answers:

bonufazy [111]3 years ago

7 0

Answer:

Maturity Value=$ 9953.305

Step-by-step explanation:

Manuel Fraser’s bank granted him a single-payment loan of $9,650.

He agreed to repay the loan in 146 days at an ordinary interest rate of 7.75%.

i.e. we have:

P=$ 9650

R=7.75

T=146 days= 146/360 year.

Now the interest(I) on the loan is given by the formula as:

$I=\dfrac{P\times R\times T}{100}$

Hence, on putting the values of P, R and T in the formula of the interest we obtain:

$I=\dfrac{9650\times 7.75\times \dfrac{146}{360}}{100}\\\\\\I=\dfrac{9650\times 7.75\times 146}{360\times 100}\\\\I=\dfrac{10918975}{36000}\\\\I=303.305$

Hence, the maturity value is:

Maturity value=P+I

= 9650+303.305

= $ 9953.305

Hence,

Maturity Value=$ 9953.305

sesenic [268]3 years ago

3 0

A=9650(1+0.0775*146/360)
A=9953.30

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Read 2 more answers

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

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3 years ago

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 42 and σ = 5.5.

alexandr1967 [171]

Answer:

a) $P( X$

We want this probability:

$P( X >64)$

And using the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

We got:

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b) For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.674$

And if we solve for a we got

$a=42 +0.674*5.5=45.707$

So the value of height that separates the bottom 75% of data from the top 25% is 45.707.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(42,25.5)$

Where $\mu=42$ and $\sigma=5.5$

And we want this probability:

$P( X$

And using the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

We got:

$P( X$

We want this probability:

$P( X >64)$

And using the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

We got:

$P( X >64) =P(Z> \frac{64-42}{5.5}) =P(Z>4)=0.0000316$

Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.674$

And if we solve for a we got

$a=42 +0.674*5.5=45.707$

So the value of height that separates the bottom 75% of data from the top 25% is 45.707.

8 0

3 years ago