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Afina-wow [57]
3 years ago
8

Find the Domain and Range (show work)

Mathematics
2 answers:
alexdok [17]3 years ago
8 0

see attached picture for solutions:


insens350 [35]3 years ago
7 0
Domain: {x| x ≠ -4, x ∈ R} or (-∞,-4)∪(-4,∞)
Solving: your denominator cannot equal zero so solve this x + 4 ≠ 0 

Range: (f(x) | f(x) ≥ 0} or [0, ∞)
Solving you can get zero because x = 0 gives you zero
but when you test other numbers in your domain like x = -2 and x = 2 you only get positive results. Therefore your range is all real non-negative numbers.
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6x1/4= ? (fractions)<br><br> 3x2/3 = ?
son4ous [18]
6x1/4=6/4=2/3
3x2/3=6/3=2
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3 years ago
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2. The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability th
coldgirl [10]

Answer:

0.30

Step-by-step explanation:

Probability of stopping at first signal = 0.36 ;

P(stop 1) = P(x) = 0.36

Probability of stopping at second signal = 0.54;

P(stop 2) = P(y) = 0.54

Probability of stopping at atleast one of the two signals:

P(x U y) = 0.6

Stopping at both signals :

P(xny) = p(x) + p(y) - p(xUy)

P(xny) = 0.36 + 0.54 - 0.6

P(xny) = 0.3

Stopping at x but not y

P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06

Stopping at y but not x

P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24

Probability of stopping at exactly 1 signal :

P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30

8 0
3 years ago
Harold spent 3/4 hour milked 3 groups of cows. If he spent an equal amount of time on each group of cows, what fraction of an ho
torisob [31]
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4 years ago
Z² + 2z+1-w²<br><br> Please factor.
Rus_ich [418]
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3 0
3 years ago
The table show the age in years of employees in a company
adelina 88 [10]

Answer:

A. 24 ≤ a < 26.

B. 22.5

Step-by-step explanation:

A. Determination of the modal class interval.

Mode is the class with the highest frequency.

From the table given above, the highest frequency is 8, therefore the class will the highest frequency is:

24 ≤ a < 26.

B. To obtain the mean, we must determine the class mark. This is illustrated below:

Class >>>>> class mark >>> frequency

18 – 19 >>>> 18.5 >>>>>>>>> 3

20 – 21 >>> 20.5 >>>>>>>> 2

22 – 23 >>> 22.5 >>>>>>>> 7

24 – 25 >>> 24.5 >>>>>>>> 8

26 >>>>>>>> 26 >>>>>>>>> 0

The mean is given by the summation of the product of the class mark and frequency divided by the total frequency. This is illustrated below:

Mean = [(18.5x3) + (20.5x2) + (22.5x7) + (24.5x8) + (26x0)] / (3+2+7+8+0)

Mean = (55.5 + 41 + 157.5 + 196 + 0)/20

Mean = 450/20

Mean = 22.5

Therefore, the mean age is 22.5

4 0
3 years ago
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