Solve and write in interval notation. 4|x+4|+5>5

2 answers:

8 0

Hi,

Solving:

$4 |x + 4| + \not5 > \not5 \\ 4 |x + 4| > 0 \\ 4 |x + 4| = 0 \: \: \: \: \: | \div 4 \\ |x + 4| = 0 \\ x + 4 = 0 \\ x = - 4 \\ x \in \: r \: \: \: \: \: ( - 4)$

Answer: The inequality is true for any value of x, except when X = -4

Hope this helps.
r3t40

pantera1 [17]2 years ago

4 0

$4|x+4|+5>5\\\\4|x+4|>0\\\\|x+4|>0\\\\x+4>0 \vee x+4-4 \vee x$

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A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c

olchik [2.2K]

The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in $E^{c}$ . Find $P(E^{c})$ .

Solution:

Part 1:

$E^{c}$ means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

$E^{c}=S-E \\ \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\ \\ 
E^{c}=(4,5,6)$

Thus the set compliment of E will contain the elements {4,5,6}.So

$E^{c}$ = {4,5,6}

Part 2)

$P(E^{c})$ means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

$P(E^{c})$ = (Number of Elements in $E^{c}$ )/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set $E^{c}$ = $n(E^{c})$ =3

So,

$P(E^{c})= \frac{n(E^{c}) }{n(S)} \\ \\ 
P(E^{c})= \frac{3}{12} \\ \\ 
P(E^{c})= \frac{1}{4}$

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To solve for the coefficient of variation you will need this formula:
CV = (SD/X) x 100
Where: CV = Coefficient of Variance
X = Mean/average
SD = Standard Deviation

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CV = (6/55) x 100 CV = (4,100/37,000) x 100
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Based on the results, salary is more variable because 10.90<11.08.

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Hope this helps a bit