D midpoint of EC -----------------> FD parallel to AC and FD=AC/2=14/2=7
<span>2-EB=EA E midpoint of AB </span>
<span>DB=DC D midpoint BC ...............> ED=AC/2=2 </span>
<span>3-T midpoint of SR </span>
<span>U midpoint of QR ---------> TU = QS/2 </span>
<span>QS=2 TU = 4.4 </span>
<span>4- The same steps SR=2 UV=9 </span>
<span>5-N midpoint of KM </span>
<span>O midpoint of ML </span>
<span>* NO parallel to Kl</span>
12001, 12002, 12003 so lol
Answer:
9
Step-by-step explanation:
Answer:
a) Dependent
b) H0: µd = 0
Ha: µd > 0
c) Stat --> Basic Statistics ----> Paired t. then select samples 1 , 2 to get the required output
d) Not enough data
Step-by-step explanation:
<u>Using values found in MINITAB 19 </u>
a) The samples are dependent and this is because the title of the test is the same ( i.e. A group of students given the LSAT )
<u>b) Appropriate hypothesis</u>
H0: µd = 0
Ha: µd > 0
<u>c) The Minitab procedure to be used to test the hypothesis is </u>
click on Stat --> Basic Statistics ----> Paired t. then select samples 1 , 2 to get the required output
d) Not enough data to create a probability plot
