To solve this inequality, you must first isolate the term that includes the variable, which is -1/2x. This can be done by first subtracting 1/3 from both sides.
Before you can subtract 1/3 from 3/5, however, you must first convert the fractions to have common denominators.
A common denominator for 1/3 and 3/5 is 15. This is the LCM, least common multiple, of 3 and 5, making it the lowest possible common denominator.
Using this common denominator, 1/3 changes to 5/15 and 3/5 changes to 9/15.
Now we can subtract these equivalent fraction for 1/3, which is 5/15, from the fraction equivalent to 3/5, which is 9/15.
9/15 - 5/15 = 4/15
This fraction can't be simplified any further so this step is done.
Now the inequality is -1/2x > 4/15.
The next step is to isolate x by dividing both sides by -1/2.
An important note to remember when doing this step is that whenever dividing by a negative in inequalities, you must flip the inequality symbol.
In this case, that means dividing both sides by -1/2 and changing the greater than sign (>) to a less than sign (<).
-1/2x ÷ -1/2 = x
4/15 ÷ -1/2
When dividing fractions, find the reciprocal of the second fraction then multiply.
The reciprocal of -1/2 is -2/1, or -2 when simplified.
4/15 • -2 = -8/15
This means x < -8/15.
This has x isolated and the inequality simplified as far as possible.
That means this is the answer.
Answer:
x < -8/15
Hope this helps!
Answer:
7
Step-by-step explanation:
Hope this helps. Pls give brainliest.
Answer:
1 and A, Solution = (3,14): 3 and B, Solution = (-4.5,-19): 2 and C Solution = (-10,7)
Step-by-step explanation:
The histogram is especially useful in comparing mean and median values of a variable. We have that 5.5+6+7+10+7.5+8+9.5+9+8.5+8+7+7.5+6+6.5+5.5=111.5 Since there are 15 values, their mean is 111.5/15=7.43 which is very close to the mean. We also have that 7 onservations are lower than 7.4 while 8 are bigger than 7.4; hence, the diagram is rather balanced and not left-skewed. We cannot tell immediately which one is larger since the values are too close. Any such random process can usually be approximated to a greater or smaller degree by a normal curve; the more points, the better. The histogram shows this (it is kind of a discrete normal curve); all points except 4 will be in this interval of bars.
A. 1/6, .16666666, 16.6666666%
B. 4/6, .64, 64%
C. 5/6, .80, 80%
