Ask question

Ask question

All categories

grandymaker [24]

2 years ago

15

Mr. Abaya goy P700,000 loan for the expansion of his business payable monthly in 4years. How much os the monthly amortization if

the interest rate is 12% compounded monthly

1 answer:

Vaselesa [24]2 years ago

8 0

The monthly amortization is P23,511.63

<h2>Compound interest</h2>

The formula for calculating the compound interest is expressed as:

$A = P(1+r/n)^t$

whrere:

P is the principal = P700,000
r is the rate = 12% = 0.12
t is the time = 4 years
n = 12

Substitute the parameters

$A = 700000(1+0.12/12)^{4(12)}\\
A = 700,000(1.01)^{48}\\
A =1,128,558.25$

Calculate the monthly payment:

Monthly payment = $\frac{1,128,558.25}{48} =23,511.63$

Hence the monthly amortization is P23,511.63

Learn more on compound interest here: brainly.com/question/24924853

You might be interested in

What is the exterior angle?

olganol [36]

Answer:

angle YUT = 120

Step-by-step explanation:

in Triangle SUT

angle 70 + angle 50 + angle x = 180

70 + 50 + x = 180

120 + x = 180

x = 180 - 120

x = 60

angle SUT = 60

angle SUT + angle YUT = 180 ( straight line)

60 + x = 180

x = 120

Therefore angle YUT ( exterior angle ) = 120

7 0

3 years ago

What is the length of arc S shown below? Enter an exact expression.

aleksandrvk [35]

Answer:

S = 15π = 47.12

length of arc S = 15π

Attached is the image of the arc.

Step-by-step explanation:

Given;

Radius arc r = 10

Angle at the center of arc ⍉ = 3π/2

The length of the arc S can be derived using the formula;

S = r × ⍉

Substituting the values;

S = 10 × 3π/2

S = 30π/2

S = 15π = 47.12

length of arc S = 15π

6 0

3 years ago

If Jon has 26 apples and ate four how many does he have

netineya [11]

22 is it’s the basic

5 0

3 years ago

Read 2 more answers

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Please help thanks!!

Lostsunrise [7]

Help with what there is nothing?

4 0

3 years ago