Answer:
Hi There!!
Step-by-step explanation:
Your answers are:
<h2>1. C</h2><h2>2. A</h2><h2 /><h2 />
I hope this helps!!
- abakugosimp
Answer:
11.666666-repeated on and on
Step-by-step explanation:
but you can probably round to 11.7
Weird way to write it but alright! (Sideways)
19pq^-2 x 5pq^6 = ?
These problems are pretty much single operations between each of the variables / constants.
So it's like this:
(19*5)(p*p)(q^-2*q^6) = ?
19*5 is 95.
For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2.
p*p is p^2
For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this:
-2 + 6 = 4
(The two is negative, because the power is negative 2)
So, q^4.
Our final answer is all of the combined.... like a so:
95p^2q^4
Answer:
Probability at least one car will get punctured: 0.39347
Step-by-step explanation:
B(10,000 , 0.00005)
P(X ≥ 1) = 1 - P(X = 0)
= 1 - (1 - 0.00005)^10,000
= 1 - (0.99995)^10,000
= 1 - 0.60652...
= 0.39347 (probability that at least one car will get punctured)
As you can tell P(X ≥ 1) as we have to solve for the probability that at least one car will get punctured. That is of course 1 - [ P(X = 0) ].
Answer:
the 3rd one from top
Step-by-step explanation:
because the coefficient is largest
