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Sloan [31]
2 years ago
10

The values in the table represent a linear function what is the common difference of the associated arithmetic sequence x 1 2 3

4 5. y 8 23 38 53 68
a. 15
b. 7
c. 1
d. 21
Mathematics
1 answer:
Alik [6]2 years ago
3 0

The common difference of the arithmetic sequence is 15.

The correct option is (a)

<h3>what is Arithmetic Sequence?</h3>

An arithmetic sequence (also known as an arithmetic progression) is a sequence of numbers in which the difference between consecutive terms is always the same.

The arithmetic sequence is defined as

x:   1     2    3    4    5

y: 8   23   38  53  68

The common difference of the arithmetic sequence is

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

Let us take any two points: (1,8) and (2,23)

Then, d = \frac{23-8}{2-1}

          d= 15

Learn more Arithmetic sequence here:

brainly.com/question/12964865

#SPJ1

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jarptica [38.1K]

Answer:

    = -5/4

Step-by-step explanation:

The slope through two points can be found by

m = (y2-y1)/(x2-x1)

   = (-4-1)/(3--1)

   = (-4-1)/(3+1)

    = -5/4


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Answer:

3

Step-by-step explanation:

The degree of a polynomial is usually the highest degree in the function.

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You currently have a balance of $712.35 in your bank account. To avoid paying a monthly fee, you must maintain a minimum balance
vampirchik [111]

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6 0
3 years ago
Suppose you are given either a fair dice or an unfair dice (6-sided). You have no basis for considering either dice more likely
hoa [83]

Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

Number of fair dice = 1

Probability of getting a fair dice P(E₁) = \dfrac{1}{2}

Number of unfair dice = 1

Probability of getting a unfair dice  P(E₂) = \dfrac{1}{2}

Probability of getting a 3 for the fair dice P(A|E₁)= \dfrac{1}{6}

Probability of getting a 3 for the unfair dice P(A|E₂) = \dfrac{1}{3}

So, we need to find the probability that the die he rolled is fair given that the outcome is 3.

So, we will use "Bayes theorem":

P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83

Hence, our required probability is 0.83.

8 0
3 years ago
What are the x-intercepts of y=2x^2+6x-20?
NISA [10]

Answer:

the x-intercepts are -5 and 2

Step-by-step explanation:

y=2x^2+6x-20

x-intercept is when y is set equal to zero

So, y = 0

0 = 2x^2+6x-20

2x^3+6x-20 = 0\\Taking \ 2 \ common\\2(x^2+3x-10) = 0\\Dividing \ both \ sides \ by \ 2\\x^2+3x-10 = 0\\Using \ mid\ term \ break \ formula\\x^2+5x-2x-10 = 0\\x(x+5)-2(x+5)=0\\Taking \ (x+5) \ common\\(x+5)(x-2) = 0

Either,

x+5 = 0    <u><em> OR </em></u>    x-2 = 0

x = -5       <u><em> OR  </em></u>     x = 2

So, the x-intercepts are -5 and 2

7 0
2 years ago
Read 2 more answers
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