I think that the correct function in your question should be y = 60x⁻¹ + 0.625x.
With that said, the answer is A. <span>0 < x < ∞
By the process of trial and error of the choices, we can see that this choice satisfies the equation. Take note that y models moose population, so y should always be positive. </span>
Answer:
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Step-by-step explanation:
Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.
Initial velocity u with which the ball is projected upwards = + 120 ft/s
Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Using the formula:
v² - u² = 2 a h,
where
u = initial velocity of the ball = +120 ft/s
v = final velocity of the ball at the highest point = 0 ft/s
a = uniform acceleration acting on the ball = -32 ft/s²
h = height attained
Substituting the values we get;
0² - 120² = 2 × (- 32) h
=> h = 120²/2 × 32 = 225 feet
The height of the ball from the ground at its highest point = 225 feet + 12 feet = 237 feet.
Answer:
1/5
Step-by-step explanation:
-8-(-10) 2 1
---------- = ------ = --------
1-(-9) 10 5
We know that 12 equals one dozen. It is not asking for a dozen, but for 9 dozen. So, we need to multiply 9 by 12. 9*12 = 108.
It says the rolls of first aid tape were divided equally into 4 boxes. We need to divide these 108 rolls into 4 equal groups, since that is what took place in the problem. So, we divide 108 by 4.
108/4 = 27. There are 27 rolls in each box.
I re-orders as 4,5,5,7,8,8,8,10,10.
Mean 7.2222222222222
Median 8
Mode 8
Range 6
Minimum 4
Maximum 10
Count n 9
Sum 65
Quartiles Quartiles:
Q1 --> 5
Q2 --> 8
Q3 --> 9
Interquartile
Range IQR 4
Outliers none
