Answer:
i got 70
Step-by-step explanation:
Brain 4 Brain gets 42 more sweets Sara
Pip 5 How many sweets does Pip get?
Sara 1
when b = s + 42 find p b = 4s p = 5s
4s = s + 42 b/4 = s p = 5b/4
3s = 42
s = 14 b = 56 p = 70
I initially found the answer by trail and error, then I could figure out the equation.
s b p
1 4 5
10 40 50
16 64 90
14 56 70
Answer:
are you trying to find the place on the number line?
if so, its a open circle starting at -3.0 then going to the right it stops at 1.0
PLEASE BE MORE DEFINED CUZ IM SUPER CONFUSED ILL EDIT MY ANSWER IF THIS IS WRONG BUT JUST TELL ME EXACTLY WHAT I NEEDA DO :)
<em>also i stan your profile picture uwu</em>
Answer:
Graph A
Step-by-step explanation
I look at the y-intercept first, and the y-intercept is -4, so looking at each graph you can see both A and D have y-intercepts of -4, since the line intercepts, or goes through the y-axis (the vertical line of the graph) at -4.
Then, you look at the variable that's changing, which is -5x. Since it's a negative, A is the graph that is negatively decreasing, because of its position on the graph. The line is going down, starting in the negative side.
Answer:
45% probability that a randomly selected customer saw the advertisement on the internet or on television
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a customer saw the advertisement on the internet.
B is the probability that a customer saw the advertisement on television.
We have that:
In which a is the probability that a customer saw the advertisement on the internet but not on television and 
is the probability that the customers saw the advertisement in both the internet and on television.
By the same logic, we have that:
12% saw it on both the internet and on television.
This means that 
20% saw it on television
This means that 
37% of customers saw the advertisement on the internet
This means that 
What is the probability that a randomly selected customer saw the advertisement on the internet or on television
45% probability that a randomly selected customer saw the advertisement on the internet or on television
