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2 years ago

Consider the recursively defined function below.

Mathematics

2 answers:

ddd [48]2 years ago

6 0

Answer:

-5.25, -3.5, -1.75, 0, 1.75

The recursive relation tells you this is an arithmetic sequence with a common difference of 1.75. Each term is 1.75 more than the one before.

To find the 2nd term, add 1.75 to the first term: -5.25 + 1.75 = -3.5

To find the 3rd term, add 1.75 to the second term: -3.5 + 1.75 = -1.75

To find the 4th term, add 1.75 to the third term: -1.75 + 1.75 = 0

To find the 5th term, add 1.75 to the forth term: 0 + 1.75 = 1.75

Dvinal [7]2 years ago

5 0

Answer:

-5.25, -3.5, -1.75, 0, 1.75

Step-by-step explanation:

The recursive relation tells you this is an arithmetic sequence with a common difference of 1.75. Each term is 1.75 more than the one before.

To find the 2nd term, add 1.75 to the first term: -5.25 + 1.75 = -3.5

To find the 3rd term, add 1.75 to the second term: -3.5 + 1.75 = -1.75

To find the 4th term, add 1.75 to the third term: -1.75 + 1.75 = 0

and so on ...

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Select all the expressions that are equivalent to 4x + 8

RoseWind [281]

Answer:

2(2x+4) 4(x+2) 2(2x)+2(4)

Step-by-step explanation:

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3 years ago

What is the equation of the line that is parallel to the given line and passes through the point (2, 3)?

KonstantinChe [14]

$\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{We have the points (-4, 0) and (4, -4)}\\\\\text{substitute:}\\\\m=\dfrac{-4-0}{4-(-4)}=\dfrac{-4}{8}=-0.5\\\\\text{therefore we have the equation of the given line in slope-intercept form:}\\y=-0.5x+b.\\\\\text{Parallel lines have the same slope.}\\\\\text{Therefore the line equation from question is y = -0.5x + b.}$

$\text{It passes through the point (2, 3).}\\\\\text{Substitute the coordinates of the given point to the equation of the line}:\\\\3=-0.5(2)+b\\3=-1+b\qquad|+1\\4=b\to b=4\\\\\text{therefore}\ y=-0.5x+4\qquad|+0.5x\\\\0.5x+y=4\qquad|\cdot2\\\\\boxed{x+2y=8\to B.)}$

7 0

3 years ago

Read 2 more answers

Meigs hourly salary is $7.75 last week she worked a 30 hour week how much did she earn

s2008m [1.1K]

Since $7.75 is the hourly salary, and last week she worked 30 hours, she earned $7.75 * 30 which is $232.50.
Meig earned $232.50

7 0

3 years ago

Read 2 more answers

Y=5x^2-8x-12 <br><br><br> linear , exponential , or quadratic and why

Assoli18 [71]

Answer:

Quadratic

Step-by-step explanation:

This is a quadratic equation because it's in the form of y=ax^2+bx+c

If it were linear, the graph would be a straight line and wouldn't contain a second degree term

If it were exponential there would be a growth or decay factor

4 0

3 years ago

In a widget factory, machines A, B, and C manufacture, respectively, 20, 30, and 50 percent of the total. Of their output 6, 3,

steposvetlana [31]

Answer:

38.71% probability it was manufactured by machine A.

29.03% probability it was manufactured by machine B.

32.26% probability it was manufactured by machine C.

Step-by-step explanation:

We have these following probabilities:

A 20% probability that the chip was fabricated by machine A.

A 30% probability that the chip was fabricated by machine B.

A 50% probability that the chip was fabricated by machine C.

A 6% probability that a chip fabricated by machine A was defective.

A 3% probability that a chip fabricated by machine B was defective.

A 2% probability that a chip fabricated by machine C was defective.

The question can be formulated as:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

What are the probabilities that it was manufactured by machines A, B, and C?

Machine A

What is the probability that the widget was manufactures by machine A, given that it is defective?

P(B) is the probability that it was manufactures by machine A. So $P(B) = 0.20$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine A. So $P(A/B) = 0.06$

P(A) is the probability that a widget is defective. This is the sum of 6% of 20%, 3% of 30% and 2% of 50%. So

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.2*0.06}{0.031} = 0.3871$

38.71% probability it was manufactured by machine A.

Machine B

What is the probability that the widget was manufactures by machine B, given that it is defective?

P(B) is the probability that it was manufactures by machine B. So $P(B) = 0.30$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine B. So $P(A/B) = 0.03$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.03}{0.031} = 0.2903$

29.03% probability it was manufactured by machine B.

Machine C

What is the probability that the widget was manufactures by machine C, given that it is defective?

P(B) is the probability that it was manufactures by machine C. So $P(B) = 0.50$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine C. So $P(A/B) = 0.02$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.02}{0.031} = 0.3226$

32.26% probability it was manufactured by machine C.

6 0

2 years ago