Answer:
option (c) n = 201
Step-by-step explanation:
Data provided in the question:
Standard deviation, s = 5.5 ounce
Confidence level = 99%
Length of confidence interval = 2 ounces
Therefore,
margin of error, E = (Length of confidence interval ) ÷ 2
= 2 ÷ 2
= 1 ounce
Now,
E =
here,
z = 2.58 for 99% confidence interval
n = sample size
thus,
1 =
or
n = (2.58 × 5.5)²
or
n = 201.3561 ≈ 201
Hence,
option (c) n = 201
Answer:
A (0,3)
Step-by-step explanation:
The given trapezoid has vertices:
(0,6), (7,12), (7,9) and (0,12).
We want to choose from the given options, a point that is a vertex for the image produced by a dilation about the origin with a scale factor of 1/2.
Note that the mapping for such a dilation is:
This implies that:
Therefore correct choice is (0,3)
Answer:
Step-by-step explanation:
The general equation we use for ballistic motion is ...
where g is the acceleration due to gravity, v₀ is the initial upward velocity, and h₀ is the initial height.
The values of g commonly used are -32 ft/s², or -4.9 m/s². Units are consistent when the former is used with velocity in ft/s and height in feet. The latter is used when velocity is in m/s, and height is in meters.
_____
Dwayne throws a ball with an initial velocity of 73 feet/second. Dwayne holds the ball 5 feet off the ground before throwing it. (h = -16t^2 + 73t + 5)
A watermelon falls from a height of 5 feet to splatter on the ground below. (h = -16t^2 + 5)
Marcella shoots a foam dart at a target. She holds the dart gun 1.5 meters off the ground before firing. The dart leaves the gun traveling 73 meters/second. (h = -4.9t^2 + 73t + 1.5)
Greg drops a life raft off the side of a boat 1.5 meters above the water. (h = -4.9t^2 + 1.5)
_____
<em>Additional comment on these scenarios</em>
The dart and ball are described as being launched at 73 units per second. Generally, we expect launches of these kinds of objects to have a significant horizontal component. However, these equations are only for <em>vertical</em> motion, so we must assume the launches are <em>straight up</em> (or that the up-directed component of motion is 73 units/second).
Answer:
Our equation for the height is:
y(t) = 275 - 16*t^2.
a) To find the average velocity between two times, t1 and t2, (where t2 > t1) the equation is:
Then:
i) t1 = 4s, t2 = 4s + 0.1s = 4.1s
The average velocity is:
And the units will be ft/s, so the average speed is:
-129.6 ft/s
The minus sign is because te pebble is falling down.
ii) t1 = 4s, t2 = 4s + 0.05s = 4.05s
The average velocity is:
So the average speed is -128.9 ft/s
iii) t1 = 4s, t2 = 4s + 0.01s = 4.01s
The average speed is:
The average speed is -128.16 ft/s.
b) The instantaneous velocity of the pebble after 4 seconds can be obtained by looking at the velocity equation, that is the derivative of the height equation.
v(t) = dy(t)/dt.
v(t) = -2*16*t + 0
Then the velocity at t = 4s is:
v(4s) = -32*4 = -128
The instantaneous velocity at t = 4s is -128 ft/s.