The percentage change is 20%
Answer:
The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 67 - 4.54 = 62.46 ounches.
The upper end of the interval is the sample mean added to M. So it is 67 + 4.54 = 71.54 ounces.
The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.
Answer:
490 or 500
Step-by-step explanation:
If we round to the 10th, Julie ran 490 feet.
I we round to the 100th, Julie ran 500 sq. ft.
X=2
subtract 5x on both sides
-2x=-4
divide by -1 on both sides
2x=4
divide by 2 on both sides
x= 2
Given :
On the first day of ticket sales the school sold 10 senior tickets and 1 child ticket for a total of $85 .
The school took in $75 on the second day by selling 5 senior citizens tickets and 7 child tickets.
To Find :
The price of a senior ticket and the price of a child ticket.
Solution :
Let, price of senior ticket and child ticket is x and y respectively.
Mathematical equation of condition 1 :
10x + y = 85 ...1)
Mathematical equation of condition 2 :
5x + 7y = 75 ...2)
Solving equation 1 and 2, we get :
2(2) - (1) :
2( 5x + 7y - 75 ) - ( 10x +y - 85 ) = 0
10x + 14y - 150 - 10x - y + 85 = 0
13y = 65
y = 5
10x - 5 = 85
x = 8
Therefore, price of a senior ticket and the price of a child ticket $8 and $5.
Hence, this is the required solution.
