Answer:
The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).
Step-by-step explanation:
Intern No. of Breast
Number Exams Performed X²
1 30 900
2 40 1600
3 8 64
4 20 400
5 26 676
6 35 1225
7 35 1225
8 20 400
9 25 625
<u>10 20 400 </u>
<u> </u><u> ∑ 259 ∑ 7515</u>
Mean= X`= ∑x/n= 259/10= 25.9
Variance = s²= 1/n-1[∑X²- (∑x)²/n]
= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]
= 806.9/9=89.655= 89.66
Standard Deviation= √89.655= 9.4687
Hence
The value of t with significance level alpha= 0.05 and 9 degrees of freedom is t(0.025,9)= 2.262
The 95 % Confidence interval is given by
x`±t(∝,n-1) s/√n
So Putting the values
25.9± 2.262( 9.4687/√10)
= 25.9 ±2.262 (2.9943)
= 25.9 ± 6.7730
= 25.9 +6.7730=32.6730
25.9 -6.7730= 19.1269
= 19.1269, 32.6730
The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).
Answer:
<u>The correct answer is A. 16.5%</u>
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly and to calculate the trend:
2006 2007 Trend
Top Coats 297 223 -24.92%
Parkas 210 210 + 0%
Jackets 213 285 +33.80%
Raincoats 137 259 +89.05
Trench coats 103 127 +23.30%
Total 960 1,104 +15%
2. If the trend shown in these graphs stays constant, what percent of the market will parkas occupy in 2008?
Let's calculate the percent of the market occupied by parkas.
In 2006 = 210/960 = 21.88%
In 2007 = 210/1,104 = 19.02%
In 2008 = 210/1,270 = 16.54% (210 + 0 = 210; 1,104 + 15% = 1,270)
<u>The correct answer is A. 16.5%</u>
Answer:
do you have a photo of the figure?
Answer: 999 games
Step-by-step explanation:
There are many ways to illustrate the rooted tree model to calculate the number of games that must be played until only one player is left who has not lost.
We could go about this manually. Though this would be somewhat tedious, I have done it and attached it to this answer. Note that when the number of players is odd, an extra game has to be played to ensure that all entrants at that round of the tournament have played at least one game at that round. Note that there is no limit on the number of games a player can play; the only condition is that a player is eliminated once the player loses.
The sum of the figures in the third column is 999.
We could also use the formula for rooted trees to calculate the number of games that would be played.

where i is the number of "internal nodes," which represents the number of games played for an "<em>m</em>-ary" tree, which is the number of players involved in each game and l is known as "the number of leaves," in this case, the number of players.
The number of players is 1000 and each game involves 2 players. Therefore, the number of games played, i, is given by
