X=1
To solve this you start by adding like terms on the left side and doing the distributive property and adding like terms on the right. This would get you
4x+7=3x+8
Then you minus 3x on both sides getting
x+7=8
Then subtract 7 from both sides getting x=1.
The first chair can be any one of the 15.
For each of those ...
The second chair can be any one of the remaining 14.
For each of those ...
The third chair can be any one of the remaining 13.
For each of those ...
The fourth chair can be any one of the remaining 12.
Number of ways to fill the 4 chairs = (15 x 14 x 13 x 12) = 32,760 .
But ...
Each set of 4 people can be seated in (4 x 3 x 2 x 1) = 24 orders.
So each group of 4 people is represented 24 times among the 32,760.
If the order doesn't matter, you're really asking how many different
groups of 4 people can occupy the front row.
That's (32,760) / (24) = 1,365 sets of 4 members, in any order.
Answer:
I will be answering numbers 3), 4), and 6) (so that you get most of the concept and that you are able to solve the questions by yourself), and also 9).
Step-by-step explanation:
3)
-15/4b+5/6b
So, first find the common denominator of both fractions. 4 times 6 is 24. -15/4b= -90/24b. 5/6b=20/24b. -90/24b+20/24b=-70/24b=-35/24b=-1 11/24.
4)
60m-15(4-8m)+20
What we need to do first is remove the parentheses.
60m-(15*4-15*8m)+20
=60m-60-120m+20.
So, here, we add the numbers without 'm' and the numbers with 'm.' (Always include the sign in front of the number, like in the equation, we include -60 since that's the sign in front of 60.)
60m-60-120m+20
=-60+20=-40
60m-120m=-60m
=-60m-40
6)
9y-15y+12-6y
Like I said before on number 4, we do about the exact same thing (but there are no parentheses to remove, so we skip that step.)
9y-15y+12-6y
=9y-15y-6y
=-12y+12
9)
It is basically the same thing as we did in the previous problems, just in a different way of saying it.
12x+12x+2x+2x+x+9
=28x+9 yards
Answer:
a) 0.06
b) 0.3
c) 0.44
Step-by-step explanation:
We are given the following in the question:
A and B are two independent events.
P(A) = 0.3
P(B) = 0.2
a) P(A intersect B)
b) P(A|B)
c) P(A union B)
The user below me is correct.
