Solution :
95% confidence interval for the difference between the means of ACT scores between two schools, is given by :
![$[(\overline X_1 - \overline X_2)-ME, (\overline X_1 - \overline X_2)+ME]$](https://tex.z-dn.net/?f=%24%5B%28%5Coverline%20X_1%20-%20%5Coverline%20X_2%29-ME%2C%20%28%5Coverline%20X_1%20-%20%5Coverline%20X_2%29%2BME%5D%24)
M.E. , Margin of error,
= 9.222
s = 3.03
= 1.4
Therefore, 95% CI = [(25.90-27.70) - 1.4 , (25.90-27.70) + 1.4]
= [-3.2, -0.4]
Therefore, the lower bound is -3.2
You use Pythagoras theorem for this:
You add 52 and 20 for AC which is 72
So you do
X^2 = 72^2 + 30^2
X^2 =5184 + 900
X^2= 6084
X = 6084 squared root
X= 78
Short Answers
P(Both) = 0.1792
Givens
The total number of roles = 100
Total number of heads = 56
Number of times rolled a four = 32 [check this for me].
Set up the probabilities.
P(H) = 56/100
P(4) = 32/100
P(H)*P(4) = P(Both)
Substitute and solve
P(Both) = 56/100 * 32/100 = 1792/10000
P(Both) = 112/625 because there are many cancellations. Or you could just express it as 0.1972
I think it should be F=45
Go on the graph look at the y axis and x axis go to the x axis look for 2 and put the point and then go to postive 3 and put a point and that’s your answer.
