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Harlamova29_29 [7]
3 years ago
9

Which statement correctly describes the relationship between the graph of f(x) and g(x)=f(x+2) ?

Mathematics
2 answers:
Ahat [919]3 years ago
7 0

Adding a value to the X value shifts the graph that many units to the left.

X+2 adds 2 to x, so the graph would shift 2 unites to the left.

The answer is:

The graph of g(x) is the graph of ​f(x)​ translated 2 units left.

klemol [59]3 years ago
5 0

Answer:

Option D.

Step-by-step explanation:

The transformation of a function is defined as

g(x)=kf(x+a)+b                .... (1)

Where, k is stretch factor, a is horizontal shift and b is vertical shift.

If 0<k<1, then the graph compressed vertically by factor k and if k>1, then the graph stretch vertically by factor k.

If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.

If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.

The given relationship between two function is

g(x)=f(x+2)            .... (2)

On comparing (1) and (2) we get

h=2> 0, so the graph of g(x) is the graph of ​f(x)​ translated 2 units left.

Therefore, the correct option is D.

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Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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If you go on a road trip of 270 miles in the mountains and 7/ 10 of the trip is downhill, how many miles of the trip are not dow
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Step-by-step explanation:

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Now we have To calculate Miles that are not downhill (uphill)

But first I calculated for the number of downhill miles

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To get the number of not downhill miles

I subtracted 189 from the total number of miles on the question.

270 - 189

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Therefore 81 Miles of the trip are not downhill.

If you have anymore questions or need clarity please let me know on the comment section. Thank you and Good luck!

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