Answer: Is there a picture you could show? I don't know unless I can see a picture.
Answer:
a range of values such that the probability is C % that a rndomly selected data value is in that range
Step-by-step explanation:
complete question is:
Select the proper interpretation of a confidence interval for a mean at a confidence level of C % .
a range of values produced by a method such that C % of confidence intervals produced the same way contain the sample mean
a range of values such that the probability is C % that a randomly selected data value is in that range
a range of values that contains C % of the sample data in a very large number of samples of the same size
a range of values constructed using a procedure that will develop a range that contains the population mean C % of the time
a range of values such that the probability is C % that the population mean is in that range
Answer:
See below.
Step-by-step explanation:
The rocket's flight is controlled by its initial velocity and the acceleration due to gravity.
The equation of motion is h(t) = ut + 1.2 g t^2 where u = initial velocity, g = acceleration due to gravity ( = - 32 ft s^-2) and t = the time.
(a) h(t) = 64t - 1/2*32 t^2
h(t) = 64t - 16t^2.
(b) The graph will be a parabola which opens downwards with a maximum at the point (2, 64) and x-intercepts at (0, 0) and (4, 0).
The y-axis is the height of the rocket and the x-axis gives the time.
Maximum height = 64 feet, Time to maximum height = 2 seconds, and time in the air = 4 seconds.
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 
Answer:y=10
Step-by-step explanation: