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zheka24 [161]
3 years ago
7

Write the recursive rule for the nth term. Then find the 5th term. {9, 14, 19...]

Mathematics
1 answer:
Leokris [45]3 years ago
7 0

Answer:

Step-by-step explanation:

{9, 14, 19, ...} so it looks like 5 is added to the previous term to get next

The recursive rule may be

a_{1} =9

a_{n} = a_{n-1} +5

The fith term is you find by writing the arithmetic sequence formula

a_{n} = a_{1} + (n-1) d , where d is how much you add

yet for the 5th term

a_{5} = a_{1} + (5-1)* 5

a_{5} = 9 +4*5 = 9+20 =29

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3 years ago
Graph x^3 - 3x^2 + 2 = 0.1 - x What are the solutions of the equation?
liq [111]

Answer:

x=-0.6,x=1.52,x=2.08

Step-by-step explanation:

Let f(x)=x^3-3x^2+2.

and

g(x)=0.1-x

The graph of the two equations are shown in the attachment.

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3 years ago
Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B
wariber [46]

Answer:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) is the probability that given that a student has a visa card, they also have a master card

p(A/B) is the probability that given a student has a master card, they also have a visa card

e. 0.35

f. 0.31

Step-by-step explanation:

a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)

                    =0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76

b. P(AnBnC')= P(AnB)-P(AnBnC)

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c. P(B/A)= P(AnB)/P(A)

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e. P((AnB)/C))= P((AnB)nC)/P(C)

                     =P(AnBnC)/P(C)

                     =0.07/0.2= 0.35

f. P((AUB)/C)= P((AUB)nC)/P(C)

                    =(P(AnC) U P(BnC))/P(C)

                     =(0.11+0.1)/0.2

                     =0.21/0.2 = 0.31

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3 years ago
(1.2x10^5)+(2.4x10^4) what is the sum
Harman [31]

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