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il63 [147K]
3 years ago
7

A series of quarterly payments of P1,300 each at the first payment is due at 4 years, and the last payment at the end of 12 year

s. If money is worth 5 ½ % compounded quarterly:
Mathematics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

The Value of Payments is  P18,557.15

Step-by-step explanation:

The quarterly payment is an annuity payment.

Use the following formula to calculate the present value of the payments.

PV of Annuity = Annuity Payment x ( 1 - ( 1 + interest rate )^-Numbers of periods ) / Interest rates

Where

Annuity Payment = Quarterly payment = P1,300

Interest rate = 5.12% x 3/12 = 1.375%

Numbers of periods = 4 years x 12/3 = 16 quarters

PV of Annuity = Value of Payments  = ?

Placing values in the formula

Value of Payments = P1,300 x ( 1 - ( 1 + 1.375% )^-16 ) / 1.375%

Value of Payments =  P18,557.15

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nikklg [1K]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Find the product of (3x-4)(2x^2+2x-1)
lapo4ka [179]

Answer:

6x^3  - 2x^2  -11x +4

Step-by-step explanation:

(3x-4)(2x^2+2x-1)

Multiply 3x  by the second term

3x* (2x^2+2x-1)

 6x^3  +6x^2 -3x


Multiply -4  by the second term

-4*(2x^2+2x-1)

 -8x^2 -8x+4


Add these together,  lining up the terms

 6x^3  +6x^2 -3x  

            -8x^2 -8x+4

----------------------------------

6x^3  - 2x^2  -11x +4

8 0
3 years ago
Read 2 more answers
What types of solutions does 6x^2 - 20x + 1 have?​
elena55 [62]

Answer:

2 real solutions

Step-by-step explanation:

We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:

b² - 4ac

If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.

Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:

b² - 4ac

(-20)² - 4 * 6 * 1 = 400 - 24 = 376

Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.

<em>~ an aesthetics lover</em>

3 0
3 years ago
A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade
Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

P(x\geq1)=1-P(x

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly  n= 5

Number of blades in an assembly dull is M  = 10.

The probability mass function is,

P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

P(x\geq 1)=1- P(x

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly  N = 48

Number of blades selected at random from the assembly  N = 5

Number of blades in an assembly dull is  M = 10

From the information,

The probability that assembly is replaced (P)  is 0.7069.

The probability that assembly is not replaced is (Q)  is,

q=1-p\\= 1-0.7069= 0.2931

The geometric probability mass function is,

P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)

The probability that assembly is replaced no replaced until the third day of evaluation is,

P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly   N = 48

Number of blades selected at random from the assembly  n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 2.

P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 6.

P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

P(x\geq 1)=1- P(x

 

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0
3 years ago
What is the mean of this data set? 12, 17, 16, 10, 20, 13, 14, 14, 12, 12, 19, 18 0 13.50 O 14.00 O 14.75 O 15.75​
ra1l [238]
The numbers seem to be going up and down but definitely not by a pattern
8 0
2 years ago
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