A number used to multiply a variable is a
2 answers:
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<span>Question 2.2. Which ordered pairs make the inequality true?</span><span>2x + y > –4</span>The solutions are (-1, 2) and (1, -5), look at the graph in the attachment.
Question 3.3. What is the slope of the line represented by the equation?
There is no equation
Question 4.4. What is the slope of the line represented by the equation 6x - 3y = 4?
Convert to slope-intercept form:
6x - 3y = 4
Subtract 6x to both sides:
-3y = -6x + 4
Divide -3 to both sides:
y = -6/-3x + 4/-3
Simplify:
y = 2x - 4/3
Now it's in slope intercept form, y = mx + b, where 'm' is the slope. So the slope here is 2.
Question 5.5. What is the simplified form of the expression?
15y - 3(4y + 10)
Distribute -3 into the parenthesis:
15y - 12y - 30
Combine like terms:
3y - 30
Question 3.3. What is the slope of the line represented by the equation?
There is no equation
Question 4.4. What is the slope of the line represented by the equation 6x - 3y = 4?
Convert to slope-intercept form:
6x - 3y = 4
Subtract 6x to both sides:
-3y = -6x + 4
Divide -3 to both sides:
y = -6/-3x + 4/-3
Simplify:
y = 2x - 4/3
Now it's in slope intercept form, y = mx + b, where 'm' is the slope. So the slope here is 2.
Question 5.5. What is the simplified form of the expression?
15y - 3(4y + 10)
Distribute -3 into the parenthesis:
15y - 12y - 30
Combine like terms:
3y - 30
Answer:
Choice number one:
.
Step-by-step explanation:
- Let
be the event that the number on the first card is even.
- Let
be the event that the number on the second card is even.
The question is asking for the possibility that event and
happen at the same time. However, whether
occurs or not will influence the probability of
. In other words,
and
are not independent. The probability that both
and
occur needs to be found as the product of
- the probability that event
- the probability that event
5 out of the ten numbers are even. The probability that event occurs is:
.
In case A occurs, there will only be four cards with even numbers out of the nine cards that are still in the bag. The conditional probability of getting a second card with an even number on it, given that the first card is even, will be:
.
The probability that both and
occurs will be:
.