The lower and upper limit of the 95% confidence interval is 183.2 and 196.0 unit.
What is a confidence interval ?
When you run your experiment again or resample the population in the same way, you can expect your estimate to fall within a certain range of values a certain percentage of the time. This is known as the confidence interval.
Main Body:
Given in question :
x- bar = 189.6
standard deviation = 10
C.I. = 95% =0.95
N=12
so, degree of freedom (D.F) = N-1 = 11
confidence level (α) = (1- C.I.)/2
= 0.025
according to t- distribution table for d.f = 11 and α= 0.025 , my result is 2.201
Now dividing standard deviation by square root of samples, we get
= 10/√12= 2.887
Lower limit = Mean - 2.201*2.887
= 189.6- 6.354
= 183.246
Upper limit = Mean + 2.201*2.887
= 189.6 + 6.354
= 195.954
Hence the result is 183.2 and 196.
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Not sure but probably it’s b or c
Step-by-step explanation:
To find the equation you must first find the gradient and also look for the points on the graph.
I see point ( 0, 2) and point (4, 1)
Hence first find the gradient before equation
the gradient = y2-y1/x2-x1
y2 = 1
y1 = 2
x2 = 4
X1 =0
if you put in the values it will look like
1-2/4-0
-1/4
the gradient = -1/4
So for the equation the formula is
remember m is the gradient
y - y1 = m ( x - x1 )
so put in the value from only one point so you can choose either (0,2 ) or (4,1 )
I will go with (4, 1)
y - y1 = m ( x - x1 )
y - 1 = m ( x - 4 )
y -1 = -1/4 ( x - 4)
4 ( y-1) = -1(x-4)
4y- 4 = -1x +4
4y - 4 -4 = -1x
4y -8= -1x
4y + 1x -8 = 0
this is the equation.
thanks and good luck.
Answer:
(E) 450
Step-by-step explanation:
Let x represent the number of gallons of solution X that must be added to the mix. Then the amount of sugar in the mix is ...
0.20x +0.40·150 = 0.25(150 +x)
0.20x +60 = 37.5 +0.25x . . . eliminate parentheses
22.5 + 0.20x = 0.25x . . . . . . subtract 37.5
22.5 = 0.05x . . . . . . . . . . . . . subtract 0.20x
450 = x . . . . . . . . . . . . . . . . . . multiply by 20
450 gallons of solution X must be added to make a 25% sugar solution.