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.
She run in 28 days = 84 miles
So, 1 day, it will be = 84 / 28 = 3 miles
Now, in 1 year it would be = 365 * 3 = 1095
So, your final answer is 1095 miles
Hope this helps!
Answer:
900 front row seats
Step-by-step explanation:
Tickets for front row seat = 220 dollars (y)
Cost of remaining seats ticket = 50 dollars (z)
Number of front row seats = x
Number of remaining seats = 2050 (w)
Total amount earned when stadium is full = 300,500 dollars (T)
so if we formulate this using our variables y,z,x,w and T problem it becomes:
y*x + z*w = T, which says the
Tickets for front row seat times Number of front row seats plus Cost of remaining seats ticket times Number of remaining seats gives us Total amount earned by the stadium.
220*x + 2050*50 = 300500
220*x = 300500-102500
220*x = 198000
divide both side by 220
x = 900
Answer:
-5
Step-by-step explanation:
an easy way to do it is to flip it so its 7-12 or you can do 12-7 and you get 5 then just add a - sign but thats my way and its weird lol
bearing in mind that perpendicular lines have negative reciprocal slopes, so
![\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D~%5Chspace%7B10em%7D%5Cstackrel%7Bslope%7D%7By%3D%5Cstackrel%7B%5Cdownarrow%20%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7Dx-1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
so we're really looking for a line whose slope is 3 and runs through (1,5)
