Answer:
The 95% confidence interval for the mean change in blood pressure, 
, is 1.744 < < 2.816
Step-by-step explanation:
The parameters given are;
The number of people in the sample, n = 9
The average drop in blood pressure, = 2.28 points
The standard deviation, σ = 0.82
The confidence level = 95%
The formula for confidence interval with a known mean is given as follows;
Where:
= z - score at 95% confidence level = 1.96
We therefore have;
Which gives the 95% confidence interval for the mean change in pressure, , is given as follows;
1.744 < < 2.816
Answer:
x=36
Step-by-step explanation:
5x=180
x=180/5=36
Answer:
3.5p ≤ 40
Step-by-step explanation:
The inequality that best describes this problem is 3.5p ≤ 40;
Given parameters:
Total amount with me = $40
Cost per package of hot dog = $3.5
Unknown;
The inequality that represents the problem = ?
Solution:
p is the number of hotdogs that can be bought.
So, total hotdog that can be bought if each hotdog cost $3.5 per box is;
p x 3.5
Now,
this will equal the total amount at hand;
p x 3.5 ≤ 40
3.5p ≤ 40
The standard formula of circle is x^2 +y^2 = r^2
hope this will help you getting the length of radius sure right easy
r = sqrt8 = 2sqrt2
hope this will help you
well, first off let's check those two points, we know it's centerd at (-26 , 120) and we also know it passes through (0 , 0), so the distance between those two points is its radius
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(~~-26 - 0~~)^2 + (~~120 - 0~~)^2} \implies r=\sqrt{(-26)^2 + (120 )^2} \\\\\\ r=\sqrt{( -26 )^2 + ( 120 )^2} \implies r=\sqrt{ 676 + 14400 } \implies r=\sqrt{ 15076 } \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B0%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-26%7D~%2C~%5Cstackrel%7By_2%7D%7B120%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bradius%7D%7Br%7D%3D%5Csqrt%7B%28~~-26%20-%200~~%29%5E2%20%2B%20%28~~120%20-%200~~%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%28-26%29%5E2%20%2B%20%28120%20%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20r%3D%5Csqrt%7B%28%20-26%20%29%5E2%20%2B%20%28%20120%20%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%20676%20%2B%2014400%20%7D%20%5Cimplies%20r%3D%5Csqrt%7B%2015076%20%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{\sqrt{15076}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(\sqrt{15076})^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill](https://tex.z-dn.net/?f=%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Chspace%7B5em%7D%5Cstackrel%7Bcenter%7D%7B%28%5Cunderset%7B-26%7D%7Bh%7D~~%2C~~%5Cunderset%7B120%7D%7Bk%7D%29%7D%5Cqquad%20%5Cstackrel%7Bradius%7D%7B%5Cunderset%7B%5Csqrt%7B15076%7D%7D%7Br%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%20~~%20x%20-%20%28-26%29%20~~%20%29%5E2%20~~%20%2B%20~~%20%28%20~~%20y-120%20~~%20%29%5E2~~%20%3D%20~~%28%5Csqrt%7B15076%7D%29%5E2%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%28x%2B26%29%5E2%2B%28y-120%29%5E2%20%3D%2015076~%5Chfill)
