<span>There is 1 way to win 6 games in a row
wwwwww
There are 2 ways to win 5 games in a row
wwwwwl, lwwwww
There are 5 ways to win 4 games in a row
llwwww, wwwwll, lwwwwl, wlwwww, and wwwwlw.
There are 2^n to arrange 6 wins and losses which adds up to 2^(6) = 64
So the Probability they win all four games in a stretch is 5 + 2 + 1 ways = 8 ways.
So the probability = 8/64 = 1/8</span>
The answer is 0 < x <span>≤ 7
</span>
First, we know that width = x
Which means that length = x +18
So, the possible equation for the Table's area is
X (X + 18) ≤ 175
X^2 + 18x - 175 <span>≤ </span>0
Next, we need to calculate is by using complete square method
x^2 + 18x + 81 <span>≤ 175 + 81
(x + 9)^2 </span><span>≤ 256
|x + 9| </span><span>≤ sqrt(256)
|x + 9| </span><span>≤ +-16
-16 </span>≤ x + 9 <span>≤ 16
</span>-16 - 9 ≤ x <span>≤ 16 - 9
</span>-25 ≤ x <span>≤ 7
Since the width couldn't be negative, we can change -25 with 0,
so it become
</span> 0 < x ≤ 7
The ribbon would be 9 yards long
Answer:
y = x + 1 y = -x + 21
(0, 1) (0, 21)
(1, 2) (1, 20)
(2, 3) (2, 19)
(3, 4) (3, 18)
Step-by-step explanation:
Answer:
The data provide strong evidence that young men weigh more on average than old men in the U.S
Step-by-step explanation:
Given :
The null hypothesis ; H0 : μ1 = μ2
The alternative hypothesis ; H1 : μ1 > μ2
T score = 5.3 ; Pvalue = < 0.0001
The decision region :
If Pvalue < α ; We reject the Null
If Pvalue > α ; We fail to reject the Null
When the α - level isn't stated, we usually assume a α - level of 5%
However, even at lower alpha level of 1% = 0.01 ;
The Pvalue < α
Hence, we can conclude that there is significant evidence that there is difference in the mean weight of young men and old men in the U.S
