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3 years ago

What are the first three terms of the sequence defined by the recursive function an = an-1-(an-2-4), when a5 =-2 and a6 = 0?

Mathematics

2 answers:

hodyreva [135]3 years ago

7 0

Answer:

C. 6, 10, 8

Step-by-step explanation:

i took the test

Wittaler [7]3 years ago

6 0

Answer:

6, 10 and 8

Step-by-step explanation:

Given the recursive function an = an-1-(an-2 - 4), when a5 =-2 and a6 = 0?

a6 = a5 - (a4 - 4)

0 = -2 - (14 - 4)

2 = - (a4 - 4)

-2 = a4 - 4

a4 = -2 + 4

a4 = 2

a5 = a4 - (a3 - 4)

-2 = 2 - (a3 - 4)

-2-2 = - (a3 - 4)

-4 = -(a3 - 4)

4 = a3 - 4

a3 = 4+4

a3 = 8

Also

a4 = a3 - (a2 - 4)

2 = 8 - (a2 - 4)

2-8 = - (a2 - 4)

-6 = -(a2 - 4)

6 = a2 - 4

a2 = 6+4

a2 = 10

a3 = a2 - (a1 - 4)

8 = 10 - (a1 - 4)

8-10 = - (a1 - 4)

-2 = -(a1 - 4)

2 = a1 - 4

a1 = 2+4

a1 = 6

Hence the first 3 terms are 6, 10 and 8

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On average, Tanner has noticed that 22 trains pass by his house daily (24 hours) on the nearby train tracks. What is the probabi

blondinia [14]

Answer:

0.11069

Step-by-step explanation:

We will assume that the trains pass by his house following a uniform distribution with values between 0 and 24. The probability of a train passing on a 9-hour time period is 9/24 = 3/8 = 0.375. Lets call Y the amount of trains passing by his house during that 9-hour period. Y follows a Binomail distribution with parameters 22 and 0.375.

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${22 \choose 0} * 0.375^0*(1-0.375)^{22} + {22 \choose 1}*0.375^1*(1-0.375)^{21} +\\\\{22 \choose 2} * 0.375^2*(1-0.375)^{20} + {22 \choose 3}*0.375^3*(1-0.375)^{19} + \\{22 \choose 4} * 0.375^4*(1-0.375)^{18} + {22 \choose 5}*0.375^5*(1-0.375)^{17} = 0.11069$

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3 years ago

Question 4 A family has two cars. The first car has a fuel efficiency of miles per gallon of gas and the second has a fuel effic

lesya [120]

Complete question is;

A family has two cars. The first car has a fuel efficiency of 35 miles per gallon of gas and the second has a fuel efficiency of 40 miles per gallon of gas. During one particular week, the two cars went a combined total of 1850 miles, for a total gas consumption of 50 gallons. How many gallons were consumed by each of the two cars that week?

Answer:

Car 1 consumed 30 gallons of gas.

Car 2 consumed 20 gallons of gas.

Step-by-step explanation:

Let x be number gallons consumed by car 1

Let y be number of gallons consumed by car 2

We are told that total gas consumption of 50 gallons. Thus;

x + y = 50 - - - - - eq(1)

We are told that the first car has a fuel efficiency of 35 miles per gallon of gas and the second has a fuel efficiency of 40 miles per gallon of gas and they went a combined total of 1850 miles. Thus;

35x + 40y = 1850 - - - - eq2

Let's make y the subject in eq(1),

y = 50 - x - - - - eq(3)

Putting 50 - x for y in eq 2 gives;

35x + 40(50-x) = 1850

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-5x + 2000 = 1850

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x = 150/5

x = 30 gallons

Substitute 30 for x into eq 3,we have;

y = 50 - 30

y = 20 gallons

Thus;

Car 1 consumed 30 gallons of gas.

Car 2 consumed 20 gallons of gas.

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4 years ago

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