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Sliva [168]
3 years ago
5

Questions:

Mathematics
2 answers:
strojnjashka [21]3 years ago
7 0

Answer:

Step-by-step explanation:

Q1 .  

Money saved every month = $200  

Time period until you retire = 35 years

Rate of Interest = 8%

The money is compounded monthly.

a). Amount for retirement = Saving annuity

Saving or making  a series of payments at regular intervals is called annuity.

Annuity formula is :

P_{n} = d((1 + r/k)^{nk} - 1)/(r/k)

PN is the balance in the account after N years.

d is the regular deposit (the amount you deposit each month)

r is the annual interest rate in decimal form(R/100)

k is the number of compounding periods in one year

Here d= $200 , r = 0.08 , n = 35 years , k = 12 (compounded monthly)

P_{n} = 200((1 + 0.08/12)^{(35*12)} - 1)/(0.08/12)

= $ 458776.50

b). Looking at the formula used in above option we can -

Increase number of years to increase retirement amount

Increase the monthly amount deposited

Increasing rate of interest also increases the amount

Compounding Annually , half yearly or quaterly will decrease the amount

c). Let us change the amount from $200 to $300 . Then amount will be

P_{n} = d((1 + r/k)^{nk} - 1)/(r/k)

P_{n} = 300((1 + 0.08/12)^{(35*12)} - 1)/(0.08/12)

= $ 688164.75

sveta [45]3 years ago
6 0

Answer:

A

hope this helps

Step-by-step explanation:

12 x 200 = 2,400

2,400 x 35 = 84,000

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Ivahew [28]

Answer:

(23 times 1/3) -5  = 2 2/3

Step-by-step explanation:

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3 years ago
Customers arrivals at a checkout counter in a department store per hour have a Poisson distribution with parameter λ = 7. Calcul
IgorLugansk [536]

Answer:

(1)14.9% (2) 2.96% (3) 97.04%

Step-by-step explanation:

Formula for Poisson distribution: P(k) = \frac{\lambda^ke^{-k}}{k!} where k is a number of guests coming in at a particular hour period.

(1) We can substitute k = 7 and \lambda = 7 into the formula:

P(k=7) = \frac{7^7e^{-7}}{7!}

P(k=7) = \frac{823543*0.000911882}{5040} = 0.149 = 14.9\%

(2)To calculate the probability of maximum 2 customers, we can add up the probability of 0, 1, and 2 customers coming in at a random hours

P(k\leq2) = P(k=0)+P(k=1)+P(k=2)

P(k\leq2) = \frac{7^0e^{-7}}{0!} + \frac{7^1e^{-7}}{1!} + \frac{7^2e^{-7}}{2!}

P(k \leq 2) = \frac{0.000911882}{1} + \frac{7*0.000911882}{1} + \frac{49*0.000911882}{2}

P(k\leq2) = 0.000911882+0.006383174+0.022341108 \approx 0.0296=2.96\%

(3) The probability of having at least 3 customers arriving at a random hour would be the probability of having more than 2 customers, which is the invert of probability of having no more than 2 customers. Therefore:

P(k\geq 3) = P(k>2) = 1 - P(k\leq2) = 1 - 0.0296 = 0.9704 = 97.04\%

4 0
3 years ago
Please help asap 25 pts
beks73 [17]
y=ax^2+bx+c\\\\\text{Axis of symmetry:}\ x=\dfrac{-b}{2a}


\text{We have:}\\\\y=4x^2+16x-18\\\\a=4;\ b=16;\ c=-18\\\\x=\dfrac{-16}{2\cdot4}=\dfrac{-16}{8}=-2

Answer: A.
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3 years ago
Whats 9x9-18+18-20+2?
zhuklara [117]

Answer:

9x^9-18

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Pls solve the simultaneous equation in the attachment.
siniylev [52]

Answer:

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Step-by-step explanation:

Part a) we have

\frac{x}{2}-\frac{y}{5}=1 ----> equation A

y-\frac{x}{3}=8 ----> equation B

Multiply equation A by 10 both sides to remove the fractions

5x-2y=10 ----> equation C

isolate the variable y in equation B

y=\frac{x}{3}+8 ----> equation D

we have the system of equations

5x-2y=10 ----> equation C

y=\frac{x}{3}+8 ----> equation D

Solve the system by substitution

substitute equation D in equation C

5x-2(\frac{x}{3}+8)=10

solve for x

5x-\frac{2x}{3}-16=10

Multiply by 3 both sides

15x-2x-48=30

15x-2x=48+30

Combine like terms

13x=78

x=6

<em>Find the value of y</em>

y=\frac{x}{3}+8

y=\frac{6}{3}+8

y=10

The solution is the ordered pair (6,10)

Part b) we have

xy=21 ---> equation A

x+5y=22 ----> equation B

isolate the variable x in the equation B

x=22-5y ----> equation C

substitute equation C in equation A

(22-5y)y=21

solve for y

22y-5y^2=21

5y^2-22y+21=0

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

<em>Find the values of x</em>

For y=1.4

x=22-5(1.4)=15

For y=3

x=22-5(3)=7

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)

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4 years ago
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