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kotykmax [81]
2 years ago
5

Find the savings plan balance after 18 months with an APR of 3% and monthly payments of $200.

Mathematics
1 answer:
ArbitrLikvidat [17]2 years ago
4 0

The saving plan balance is $3,677.53.

Given that,

  • The no of months is 18.
  • The annual percentage rate is 3% so the monthly rate percentage is 0.25%.
  • And, the monthly payment is $200.

Based on the above information, the following formula should be used:

FVA = PMT×  [(1+r)^n - 1] ÷ r

= $200 × [(1+ 0.0025)^18 - 1] ÷ 0.0025

= $3,677.53

Therefore we can conclude that the saving plan balance is $3,677.53.

Learn more: brainly.com/question/14191332

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Linda plans to put in wall-to-wall carpet in her bedroom. She measures the dimensions of her bedroom and skectches up the floor
Sunny_sXe [5.5K]

The image of the floor dimensions is missing, so i have attached it.

Answer:

$88.92

Step-by-step explanation:

Looking at the attached floor plan, we have to divide it into two parts to solve easily.

To get the smaller side of the bigger triangle; 3 - 0.6 = 2.4m

Thus,bigger triangle measures; 3.6m x 2.4m

For the smaller triangle, dimensions are; 0.6m x 1.2m

Area of bigger triangle = 3.6 × 2.4 = 8.64 m²

Area of smaller triangle = 0.6 × 1.2 = 0.72 m²

Total area = 8.64 + 0.72 = 9.36 m²

Since the carpet costs $9.50 per sq.m.

Thus;

Total spent for wall to wall carpet is = 9.36 × 9.5 = $88.92

8 0
3 years ago
For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger
11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = p=\frac{1}{120}.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X </em>is:

P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419

6 0
3 years ago
The sum of three consecutive integers is 141. what is the smallest integer
Dafna1 [17]
Integer 1 = n
integer 2 = n + 1
integer 3 = n + 2

n + (n + 1) + (n + 2) = 141
3n + 3 = 141
3n = 141 - 3
3n = 138
n = 138/3
n = 46 --(1st integer)

n + 1 = 46 + 1 = 47 --(2nd integer)
n + 2 = 46 + 2 = 48 --(3rd integer)

so the smallest integer is 46
3 0
3 years ago
I need someone to answer the question. the answer choices are below
Brums [2.3K]
The answer is b, 615.44
3 0
2 years ago
Read 2 more answers
Is 5 x pi a rational number
lyudmila [28]

yes because 5 times pi jis just basically a decimal times a whole number and pi is never ending

4 0
3 years ago
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