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Anvisha [2.4K]
3 years ago
7

the values in the table represent a linear function. what is the common difference of the arithmetic sequence x: 1 2 3 4 5 y: 11

25 39 53 67
Mathematics
2 answers:
Andreyy893 years ago
5 0

Answer:

The common difference of the arithmetic sequence is 14.

Step-by-step explanation:

The arithmetic sequence is defined as

x:   1     2    3    4    5

y:  11   25  39  53   67

The common difference of the arithmetic sequence is

d=\frac{y_2-y_1}{x_2-x_1}

Consider any two points, i.e.,(1,11) and (2,25).

d=\frac{25-11}{2-1}

d=\frac{14}{1}

d=14

Therefore the common difference of the arithmetic sequence is 14.

GenaCL600 [577]3 years ago
3 0

Answer:

common difference: 14

Step-by-step explanation: You have to find out what number it takes to add to the number to get the following number. ex 11+14=25      25+14=39 etc


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x = -5

Explanation:

4(2x + 10) = 0

[ Simplify both sides of the equation ]

4(2x + 10) = 0

(4)(2x) + (4)(10) = 0 [Distribute]

8x + 40 = 0

[ Subtract 40 from both sides ]

8x + 40 − 40 = 0 − 40

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Let the random variable X be the number of rooms in a randomly chosen owner-occupied housing unit in a certain city. The distrib
natali 33 [55]

Answer:

a) X is a discrete random variable.

b) 2% probability of choosing a unit with 10 rooms

c) 98% probability that a unit chosen at random is not a 10-room unit

d) 30% probability that a unit chosen at random has less than five rooms

e) 7% probability that a unit chosen at random has less than five rooms

Step-by-step explanation:

The distribution means that:

7% probability that a randomly chosen owner-occupied housing unit in a certain city has 3 rooms.

23% probability that a randomly chosen owner-occupied housing unit in a certain city has 4 rooms.

36% probability that a randomly chosen owner-occupied housing unit in a certain city has 5 rooms.

19% probability that a randomly chosen owner-occupied housing unit in a certain city has 6 rooms.

5% probability that a randomly chosen owner-occupied housing unit in a certain city has 7 rooms.

5% probability that a randomly chosen owner-occupied housing unit in a certain city has 8 rooms.

3% probability that a randomly chosen owner-occupied housing unit in a certain city has 9 rooms.

?% probability that a randomly chosen owner-occupied housing unit in a certain city has 10 rooms.

(a) Is X a discrete or continuous random variable?

X is the number of rooms in a randomly selected house. The number of rooms is a countable variable, that is, only assumes values like 1,2,3,4,... You cannot have 3.5 rooms, for example.

So X is a discrete random variable.

(b) What must be the probability of choosing a unit with 10 rooms?

The sum of all probabilities must be 100%(1 decimal). So

0.07 + 0.23 + 0.36 + 0.19 + 0.05 + 0.05 + 0.03 + ? = 1

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? = 0.02

2% probability of choosing a unit with 10 rooms

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0.07 + 0.23 + 0.36 + 0.19 + 0.05 + 0.05 + 0.03 = 0.98

98% probability that a unit chosen at random is not a 10-room unit

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7% probability it has 3 rooms

23% probability it has 4 rooms. So

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(e) What is the probability that a unit chosen at random has three rooms?

7% probability that a randomly chosen owner-occupied housing unit in a certain city has 3 rooms. So

7% probability that a unit chosen at random has less than five rooms

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