Answer:
We have to choose 21 strings to be sure we have chosen at least 6 strings of the same type.
Step-by-step explanation:
Since the string type is determined by the initial and terminal bits as understood from the question, then the value of the bits between the initial and terminal bits is of no concern to us.
Now, to be sure you have atleast 6 of the same type, we select each string five times. By doing this, we have already selected 20 strings because we have 4 strings there. Now if you choose any of the string one more time, we are certain that we must have chosen atleast 6 strings that are the same. This means we have to choose 21 strings to be sure we have chosen at least 6 strings of the same type.
Answer:
l = p - 2w / 2
Step-by-step explanation:
You can make your own problem and plug it in. I substituted 5 for w and 6 for l. That means the perimeter would be 22. Now plug those numbers into all the problems. The one that gives you the correct answer is the top right answer.
6 = 22 - 5 / 2
In the problem 56x31, what are the partial products? To acquire the possible partial products we can just multiply the two numbers to produce the possible numbers at hand. <span><span>
1. </span>56 x 31 = 1736</span><span><span>
2. </span>31 x 56 = 1736</span><span> </span>
<span>Same outcome which is explained by the comutative property of multiplication.</span>
Consecutive integers are numbers that follow each other op in order.
The four integers you're looking for are
-4
-5
-6
-7
This is because they add up to -4 + -5 + -6 + -7 = - 4 - 5 - 6 - 7 = - 22
Answer:
y=-4/1x-13
Step-by-step explanation:
5-1/2-3
5-1 ( 4)
2-3 (-1)
y=-4/1x+b
5=-4/1(2)+b
5=-8+b
b=-13
y=-4/1x-13
Hope this helped, have a Wonderful Day!!