Answer:
distributive
Step-by-step explanation:
Answer:
7/40
Step-by-step explanation:
probability of taking 7 vehicles from 10 vehicles = 10C₇
number of ways taking 2 SUVs and 5 trucks = 3C₂* 7C₅ = 63
number of ways taking 3 SUVs and 4 trucks =3C₁*7C₄ = 35
number of ways taking 2 SUVs and 5 trucks or 3 SUVs and 4 trucks = 63 + 35 = 98
The probability that any 7 randomly chosen parking spots have 2 SUVs and 5 trucks or 3 SUVs and 4 trucks is = 98/120 = 49/60
number of ways taking 7 randomly chosen vehicles, exactly 1 is an SUV = 3C₁*7C₆ = 21
The probability that of any 7 randomly chosen vehicles, exactly 1 is an SUV is 21/120 = 7/40
It would be a) 2310
hope this helps
Let us assume for starters that it is true.
The area of the square is s^2, so that is where you start.
(1 + 2*sqrt(2) ) ^2 = 1 + 4*sqrt(2) + 8 Taking 1 away form that does not leave you with 8.
Let's start again.
If all sides are reduced by 1 , but the square is retained, then the length of the side is 2sqrt(2). Now we have
s^2 = (2sqrt(2) )^2 = 4 * 2 = 8
The question is really not worded clearly enough to give an answer. I think it's true, but the language has to be twisted a bit.
Answer: I'd answer true, but don't be surprised if it is not interpreted the way I did on my second try.
Answer:
6a. r^5
6b. 18(4+√2)
Step-by-step explanation:
6a. The first term of the sequence is √r, and the common ratio is √r. Hence the 10th term will be ...
a1·r^(n-1) . . . for a1=√r and n=10.
√r·(√r)^(10-1) = (√r)^10 = r^5
_____
6b. The sum of n terms is given by ...
Sn = a1·(R^n -1)/(R -1)
For a1=√8 and the common ratio R = √8, the sum of 4 terms is ...
Or, you could add up the 4 terms: