Answer:
(a) P(next ball with blue) = 
(b) P(Not to be red) = 
(c) P(drawing 70 times yellow ball) = 
.
Step-by-step explanation:
Given that,
The total outcomes of the color of tile are :-
Red , Red, Blue, Yellow, Blue, Red, Green, Red, Blue, Green.
From the Question,
Total number of outcomes = 10
(a) What is the probability that her next draw will be blue ?
Red = 4,
Blue = 3,
Green = 2,
Yellow = 1.
P(next ball with blue) = .
(b) What is the probability that her next draw will NOT be red?
P(Red ball) = 
But we have to find
P(Not to be red) =
(c) If Amy draws 70 more times with the same tiles from the question above, what probability would you expect a yellow?
P(Yellow ball) = 
drawing it 70 times we get the probability
P(drawing 70 times yellow ball) = .
9/10 = x/13. Solve for x. X = 11.7.
Your answer would be 11.7oz. Hope this helps
Answer:
12
Step-by-step explanation:
Using the pythagorean theorem, we know that a^2+ b^2=c^2
In our case, we know b, which is 5 and c which is 13.
If we plug that into the equation, we get a^2+ 5^2 = 13^2
Next, we simplify the equation. a^2 + 25 = 169
Then we subtract 25 from 169. 169 - 25 = 144.
Lastly, we do the square root of 144 and we get 12 which is the missing side.
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
