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Alekssandra [29.7K]
3 years ago
6

What is the APY for money invested at each​ rate? ​(A) 13​% compounded quarterly ​(B) 12​% compounded continuously ​(A) APY_____

​(Round to three decimal places as​ needed.)
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
7 0
13% compounded quarterly=(1+.13/4)^4-1=.13647=13.647%
12% compounded continuously=e^.12 -1=.1275=12.75%
☺☺☺☺
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The undefined terms line and plane are needed to precisely define which mathematical term?
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The undefined terms line and plane are needed to precisely define which mathematical term?

A) line segment

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4 0
4 years ago
A. What is the probability that this couple spends 45 dollars or more?
guapka [62]

Using the probability table, it is found that:

  • a) There is a 0.25 = 25% probability that this couple spends 45 dollars or more.  
  • b) The expected amount the couple actually has to pay is $36.85.

Item a:

To find the probabilities involving the total cost, we have to <u>add the variables X and Y</u> from the table, then:

P(X = 30) = P(X = 15|Y = 15) = 0.2

P(X = 35) = P(X = 15|Y = 20) + P(X = 20|Y = 15) = 0.15 + 0.15 = 0.3

P(X = 40) = P(X = 15|Y = 25) + P(X = 20|Y = 20) + P(X = 25|Y = 15) = 0.05 + 0.15 + 0.05 = 0.25

P(X = 45) = P(X = 20|Y = 25) + P(X = 25|Y = 20) = 0.1 + 0.1 = 0.2

P(X = 50) = P(X = 25|Y = 25) = 0.05

The probability involving <u>values of 45 or more</u> is:

P(X \geq 45) = P(X = 45) + P(X = 50) = 0.2 + 0.05 = 0.25

0.25 = 25% probability that this couple spends 45 dollars or more.

Item b:

For a <em>discrete distribution</em>, the expected value is the <u>sum of each outcome multiplied by it's respective probability</u>, hence, involving the 10% discount for prices above $45:

E(X) = 0.2(30) + 0.3(35) + 0.25(40) + 0.9[0.2(45) + 0.05(50)] = 36.85

The expected amount the couple actually has to pay is $36.85

A similar problem is given at brainly.com/question/25782059

8 0
2 years ago
What is the remainder to x^3+4x^2+x-6 divided by (x-1)
serg [7]

x^3=x^2\cdot x, and x^2(x-1)=x^3-x^2. This gives a remainder of

(x^3+4x^2+x-6)-(x^3-x^2)=5x^2+x-6

5x^2=5x\cdot x, and 5x(x-1)=5x^2-5x. This gives a new remainder of

(5x^2+x-6)-(5x^2-5x)=6x-6

6x=6\cdot x, and 6(x-1)=6x-6. This gives a new remainder of

(6x-6)-(6x-6)=0

and so there is no remainder.

###

Quicker method: Use the polynomial remainder theorem, which says the remainder upon dividing a polynomial p(x) by x-c is p(c). Here we have

p(x)=x^3+4x^2+x-6

c=1\implies p(c)=1+4+1-6=0

5 0
4 years ago
I don't know how to do this
Dvinal [7]
a_1=2;\ r=-\dfrac{3}{2}\\\\a_2=a_1r\to a_2=2\cdot\left(-\dfrac{3}{2}\right)=-3\\\\a_3=a_2r\to a_3=-3\cdot\left(-\dfrac{3}{2}\right)=\dfrac{9}{2}\\\\a_4=a_3r\to a_4=\dfrac{9}{2}\cdot\left(-\dfrac{3}{2}\right)=-\dfrac{27}{4}\\\\a_5=a_4r\to a_5=-\dfrac{27}{4}\cdot\left(-\dfrac{3}{2}\right)=\dfrac{81}{8}
3 0
4 years ago
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