Answer:
B. The histogram appears to roughly approximate a normal distribution. The frequencies generally increase to a maximum and then decrease, and the histogram is symmetric.
Step-by-step explanation:
The Graph of Normal Distribution is like a bell-shaped. Here the value of y is less for the lower value of x and then the value of y is increased for a larger value of x, but after some time value of y is again getting decrease as the value of x increases. For the Histogram to appear to a normal distribution, the graph of histogram must have the same nature. Thus option B is only the correct option.
The histogram is made up of columns bar with no gaps between bars with different labels of numeric data of different heights shows the size of the group of different labels.
Answer:
w = w L = 2w h = h
Volume: V = Lwh
10 = (2w)(w)(h)
10 = 2hw^2
h = 5/w^2
Cost: C(w) = 10(Lw) + 2[6(hw)] + 2[6(hL)])
= 10(2w^2) + 2(6(hw)) + 2(6(h)(2w)
= 20w^2 + 2[6w(5/w^2)] + 2[12w(5/w^2)]
= 20w^2 + 60/w + 120/w
= 20 w^2 + 180w^(-1)
C'(w) = 40w - 180w^(-2)
Critical numbers:
(40w^3 - 180)/w^2 = 0
40w^3 -180 = 0
40w^3 = 180
w^3 = 9/2
w = 1.65 m
L = 3.30 m
h = 1.84 m
Cost: C = 10(Lw) + 2[6(hw)] + 2[6(hL)])
= 10(3.30)(1.65) + 2[6(1.84)(1.65)] + 2[6(1.84)(3.30)])
= $165.75 cheapest cost
Step-by-step explanation:
The expected value for the game would be the first one negative three
Answer:
5
Step-by-step explanation:
keep change flip
3 - -2
3 + 2
(a) The maximum height of the ball above the ground is 12.5 m and the time of motion is 1.43 s.
(b) The time taken for the ball to contact the other player at 0.5 m above the ground is 3.0 s.
<h3>Maximum height reached by the ball</h3>
The maximum height reached by the ball is calculated as follows.
At maximum height, the final velocity, v = 0
dh/dt = v = 0
dh/dt = -2(4.9)t + 14
0 = -9.8t + 14
9.8t = 14
t = 1.43 s
H(1.43) = -4.9(1.43)² + 14(1.43) + 2.5
H(1.43) = -10.02 + 20.02 + 2.5
H(1.43) = 12.5 m
<h3>Time to reach maximum height</h3>
12.5 = -4.9t² + 14t + 2.5
4.9t² - 14t + 10 = 0
t = 1.43 s
<h3>Time for the ball to reach 0.5 m above the ground</h3>
0.5 = -4.9t² + 14t + 2.5
4.9t² - 14t + - 2 = 0
t = 3.0 seconds
The complete question is below:
In a volleyball match Hanein serves the volleyball at 14 m/s, from a height of 2.5 m above the court. The height of the ball in flight is estimated using the equation, h = -4.9t² + 14t + 2.5, where t is the time in second and h is height above ground, in metres.
Learn more about time of motion here: brainly.com/question/2364404
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