9514 1404 393
Answer:
- 7.5 ft
- 32.5 ft, 5 ft
- 10.7 ft
Step-by-step explanation:
a) The starting height is h(0) = 7.5 feet, the constant in the quadratic function.
The irrigation system is positioned <u> 7.5 </u> feet above the ground
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b) The axis of symmetry for quadratic ax^2 +bx +c is x = -b/(2a). For this quadratic, that is x=-10/(2(-1)) = 5. This is the horizontal distance to the point of maximum height. The maximum height is ...
h(5) = (-5 +10)(5) +7.5 = 32.5 . . . feet
The spray reaches a maximum height of <u> 32.5 </u> feet at a horizontal distance of <u> 5 </u> feet from the sprinkler head.
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c) The maximum distance will be √32.5 + 5 ≈ 10.7 ft.
The spray reaches the ground at about <u> 10.7 </u> feet away.
Answer:
The third option.
Step-by-step explanation:
Hence the 3rd option.
Answer:
It is not normally distributed as it has it main concentration in only one side.
Step-by-step explanation:
So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).
So, let us begin the groupings into their different classes, shall we?
Data given:
0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.
(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.
(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31
(3). 0.4 - 0.6 : there are 2 values that falls into this category.
(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.
Class interval frequency.
0.00 - 0.20. 15.
0.20 - 0.40. 2.
0.4 - 0.6. 2.
Answer:
m∠KHL = 43°
Step-by-step explanation:
From the picture attached,
m∠KHL ≅ m∠GHL [Given in the picture]
Now substituting the values of the angles,
(3x + 1) = (5x - 27)
1 + 27 = 5x - 3x
28 = 2x
14 = x
m∠KHL = (3x + 1)° = (3 × 14) + 1
= 42 + 1
= 43°
Therefore, measure of ∠KHL = 43°
That is true. it is part of natural numbers.
