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marissa [1.9K]
3 years ago
9

If you apply the changes below to the absolute value parent function,

Mathematics
1 answer:
QveST [7]3 years ago
6 0

Answer:

F(x) = |x|

And we need to apply some transformations. The first part is hift 8 units left. And we need to do this:

F(x) = |x+8|

Then we need to apply the second transformation Shift 3 units down. and we can do this:

F(x) = |x+8|-3

Step-by-step explanation:

For this case we have the following function given:

F(x) = |x|

And we need to apply some transformations. The first part is hift 8 units left. And we need to do this:

F(x) = |x+8|

Then we need to apply the second transformation Shift 3 units down. and we can do this:

F(x) = |x+8|-3

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2(8r+5)-3=4(4r-1)+11 is it an only solution or a no solution or infinite solution
pogonyaev

Answer:

Infinitely many solutions.

Step-by-step explanation:

Let's begin by carrying out the indicated multiplications, which must be done before any addition or subtraction:

2(8r+5)-3=4(4r-1)+11  becomes  16r + 10 - 3 = 16r - 4 + 11.

Subtracting 16r from both sides, we get 10 - 3 = - 4 + 11, or 7 = 7

This is always true, so we can conclude that this equation has infinitely many solutions.

3 0
2 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

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

3 0
3 years ago
What is the solution to the system of linear equations
Wittaler [7]
The answer is c. (0,2)
3 0
3 years ago
Read 2 more answers
Write 3 1/8 as a decimal.
Alenkinab [10]
No it is not. It is a fraction
5 0
3 years ago
Read 2 more answers
sally has 20 coins in her piggy bannk, all dimes and quarters. the total amount of money is $3.05. how many of each coin does sh
VMariaS [17]
D=dimes
q=quarters

d + q = 20
.10d + .25q = 3.05

d=-q + 20

.10(-q + 20) +.25q = 3.05
-.10q + 2 + .25 = 3.05
.15q + 2 = 3.05
.15q + 2 - 2 = 3.05 - 2
.15q = 1.05
.15q/.15 = 1.05/.15
q = 7

d + 7 = 20
d + 7 - 7 = 20 - 7
d = 13

Sally has 7 quarters and 13 dimes.
Hope this helps!
5 0
3 years ago
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