Twenty two three's go into 70.
The confidence interval formula is computed by:
Xbar ± Z s/ sqrt (n)
Where:
Xbar is the mean
Z is the z value
S is the standard deviation
N is the number of samples
So our given are:
90% confidence interval with a z value of 1.645
Sample size 40, 45
Mean 180, 179
Standard deviation 2, 4
So plugging that information in the data will give us a
confidence interval:
For 1:
Xbar ± Z s/ sqrt (n)
= 180 ± 1.645 (2 / sqrt (40))
= 180 ± 1.645 (0.316227766)
= 180 ± 0.520194675
= 179.48, 180.52
For 2:
Xbar ± Z s/ sqrt (n)
= 179 ± 1.645 (4 / sqrt (45))
<span>= 179 ± 1.645 (0.596284794)</span>
therefore, the answer is letter b
Answer:
92 attendees had activity cards
Step-by-step explanation:
Let x be the number of students with activity cards. Then 130-x is the number without, and the total revenue is ...
7x +10(130 -x) = 1024
7x +1300 -10x = 1024 . . . . eliminate parentheses
-3x = -276 . . . . . . . . . . . . . collect terms; subtract 1300
x = 92 . . . . . . divide by 3
92 students with activity cards attended the dance.
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<em>Comment on the solution</em>
Often, you will see such a problem solved using two equations. For example, they might be ...
Let 'a' represent the number with an activity card; 'w' the number without. Then ...
- a+w = 130 . . . . the total number of students
- 7a +10w = 1024 . . . . the revenue from ticket sales
The problem statement asks for the value of 'a', so you want to eliminate w from these equations. You can do that using substitution. Using the first equation to write an expression for w, you have ...
w = 130-a
and making the substitution into the second equation gives ...
7a +10(130 -a) = 1024
This should look a lot like the equation we used above. There, we skipped the extra variable and went straight to the single equation we needed to solve.
Step-by-step explanation:
multiply 2 with the bracket numbers
6x+16=0
6x= - 16
x=-16/6
x= - 8/3
