Can you use a calculator? If yes, then isolate 7) cot 8) tan 2 theta, etc. Then, input the inverse of that trig into the calculator.
Ex. #7 cot theta= -6/5
Then, input arccot (-6/5) (aka inverse cot (-6/5)) into your calculator.
The histogram is especially useful in comparing mean and median values of a variable. We have that 5.5+6+7+10+7.5+8+9.5+9+8.5+8+7+7.5+6+6.5+5.5=111.5 Since there are 15 values, their mean is 111.5/15=7.43 which is very close to the mean. We also have that 7 onservations are lower than 7.4 while 8 are bigger than 7.4; hence, the diagram is rather balanced and not left-skewed. We cannot tell immediately which one is larger since the values are too close. Any such random process can usually be approximated to a greater or smaller degree by a normal curve; the more points, the better. The histogram shows this (it is kind of a discrete normal curve); all points except 4 will be in this interval of bars.
Answer:
-5 and +5
Step-by-step explanation:
add twelve to both sides
New equations: 2x= -+10
Divide by 2 on both sides
x=-5, +5
Answer:
D.
Step-by-step explanation: