Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Select two options. x < 5 –6x – 5 < 10 – x –
6x + 15 < 10 – 5x A number line from negative 3 to 3 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right. A number line from negative 3 to 3 in increments of 1. An open circle is at negative 5 and a bold line starts at negative 5 and is pointing to the left.
2 answers:
Answer:
The accurate representations are:
- 6x + 15 < 10 - 5x ⇒ 3rd option
A circle is open at 5 with a bold line extending right from 5 ⇒ 4th option (see diagram)
Step-by-step explanation:
Given the inequality is -3(2x - 5) < 5(2 - x)
First, simplify both sides:
∵ -3(2x - 5) = -3(2x) + -3(-5)
Remember that (-)(-) results in (+)
Thus, -3(2x - 5) = - 6x + 15
∵ 5(2 - x) = 5(2) + 5(-x)
Remember that (+)(-) results in (-)
Then, 5(2 - x) = 10 - 5x
Now we have: - 6x + 15 < 10 - 5x
Subtract 15 from both sides
So: - 6x < -5 - 5x
Add 5x to both sides
Thus, - x < - 5
Note that since the coefficient of x is negative, we reverse the inequality sign when dividing both sides by it
∵ The coefficient of x is -1
Therefore, dividing both sides by -1 gives us x > 5
The accurate representations are:
- 6x + 15 < 10 - 5x ⇒ 3rd option
A circle is open at 5 with a bold line extending right from 5 ⇒ 4th option (see diagram)
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Response:
The probability that a student has a pet, given they do not have any siblings is:
Option: D ( 60%)
Step-by-step breakdown:
Let A represent the situation where a student lacks a sibling.
Let B signify the occurrence that a student has a pet.
Consequently, A∩B refers to the event in which a student is without siblings but possesses a pet.
Let P denote the chance of an event happening.
We need to determine:
P(B|A)
From our knowledge:
From the data provided:
P(A)=0.25
and P(A∩B)=0.15
Thus,
which expressed as a percentage is:
Therefore, the probability is:
60%