which of the following inequalities has a solution set that , when graphed on the number line , is a single line segment of finite length

a x^4 = 1
b x^3 = 27
c x^2 = 16
d 2<=|x| <=5
e 2<= 3x + 4 <= 6

For a certain examination , a score of 58 was 2 standard deviations below the mean and a score of 98 was 3 standard deviations above the mean . what was the nean score
74
76
78
80
82

1) I think you got the question wrong. Anyway, lets try

a) x^4 = 1
b) x^3 = 27
c) x^2 = 16
d) 2<=|x| <=5
e) 2<= 3x + 4 <= 6

a) X = +/- 1, which means X is two points and not a line
b) X = +3, which means X is a point at 3 and not a line
c) X = +/- 4, which means X is two points and not a line
d) If you open the mod the value will be

2<=X<=5 for positive X (which means X lies between 2 and 5)


and for negative X

-5<=X <= -2 (which means X lies between -5 and -2)

As you can see, there are two lines to X and not single line as the question asked

e) At this point you know its E, so save time and move on BUT lemme go ahead and explain

Subtract 4 from the entire equation

-2<=3x<=3

Divide by 3

-2/3 <=x<=1

There's your answer mate

QUESTION 2)

For a certain examination , a score of 58 was 2 standard deviations below the mean and a score of 98 was 3 standard deviations above the mean . what was the nean score
74
76
78
80
82

Let X be the mean and D be the standard deviation

You can write 58 = X - 2D --- equation 01
and

98 = X + 3D -- equation 02

equation 02 minus equation 01

40 = 5D

D = 8

In equation 01: X = 58 + 2D = 74

In equation 02: X = 98 - 24 = 74

Thanks !

i started solving question number 2 , by applying Standard deviation formulaes n all
It didn even strike me that i can solve it using two equations

hi -

from now on, please post questions in separate threads. the poster above has been nice enough to respond to this post despite its containing multiple questions, but that's generally not the way things are done around here.

A tiny correction for question 1 e):
2<= 3x + 4 <= 6

Subtracting 4 from the enire equation gives:
-2<= 3x <=2

Thus,
-2/3 <= x <= 2/3

---------------------------------

Moty Keret

yeah, this isn't the right question. none of these choices should have an equals sign in it; they should all be "<".

(by the way, it's easy to type "<"; just type "<" with an underline. this is better for our readers; no matter how hard i try, for instance, i can't help seeing "<=" as an arrow pointing to the left.)

moty has the right idea in the post above:
when you have a 3-sided "sandwich" inequality with the variable only in the middle, such as that in choice (e) of this problem, then you can operate on all 3 sides of the inequality at once.
so, as moty did, you can subtract 4 from all three sides of the inequality, and then divide all three sides of the inequality by 3.

if there are variables on two, or all three, sides of the inequality, though, you can't do this anymore; you have to separate the inequality into two separate, "normal", two-sided inequalities and solve them separately.
for instance, if you have 2x - 3 < x + 1 < 3x - 1, it's impossible to operate on all three sides at once, so you split it up into left- and right-hand inequalities:
2x - 3 < x + 1 --> solve: x < 4
x + 1 < 3x - 1 --> solve: 2 < 2x -> 1 < x
combining these gives the ultimate solution, which is 1 < x < 4.

--

the solution to the correct version of (a), x^4 < 1, is -1 < x < 1.
the solution to the correct version of (c), x^2 < 16, is -4 < x < 4.

there are two ways to get these solutions:

(1) MEMORIZE their general form.
this is probably what you should do if you don't like wasting time.
whenever you have an INEQUALITY INVOLVING AN EVEN POWER, such as the ones in choices (a) and (c), you should know that your solution is going to contain both positive and negative roots of the number that appears in the inequality.
if you see x^2 < 16, for instance, the solution is -4 < x < 4.
if you see x^2 > 16, the solution is x > 4 OR x < -4. (there's no way to write this is a "sandwich" inequality.)

(2) use the fact that √(x^2) = |x| (and its equivalents for 4th, 6th, ... powers).
this is an extra step, and is sort of pointless - i.e., it will just yield the same result as the aforementioned memorization, but at greater effort - but, if you're the type who absolutely must know the justification for everything, then you can use this.
for instance,
x^2 < 16
take √ of both sides --> |x| < 4
solve (probably via prior memorization) --> -4 < x < 4
etc.

i'm not going to address the second question, because the forum policy is to post each problem in a separate thread.
please post the second problem in a separate thread.
thanks.

you're also supposed to use a title consisting of the first several words of the problem statement, too.