The best estimate for this correlation would be B) 0.9.
We can see that the number is constantly going up, which would throw out the D answer.
We can also see that for every time the x goes up 1, the y goes up a little less than one. We can see that in the ordered pairs that exist on the graph such as (3, 2), (8, 6) and (2.1, 1.9).
Since the y values are just lower than the x, the correlation would be just under one. Therefore, 0.9 is an accurate estimation.
I don't know if you meant profit or markup but,
take the difference of $80 & $50 = $30
What percent of 50 is 30? It's 3/5 or 60%. So there is a 60% markup on the original price of 50$.
Let a and b be store A and store B, respectively. Then:
3a+b=63
5a+4b=140
So
12a+4b=252
5a+4b=140
7a=112
a=16
b=15
☺☺☺☺
Answer:
2 one.
Step-by-step explanation:
<span>et us assume that the origin is the floor right below the 30 ft. fence
To work this one out, we'll start with acceleration and integrate our way up to position.
At the time that the player hits the ball, the only force in action is gravity where: a = g (vector)
ax = 0
ay = -g (let's assume that g = 32.8 ft/s^2. If you use a different value for gravity, change the numbers.
To get the velocity of the ball, we integrate the acceleration
vx = v0x = v0cos30 = 103.92
vy = -gt + v0y = -32.8t + v0sin40 = -32.8t + 60
To get the positioning, we integrate the speed.
x = v0cos30t + x0 = 103.92t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + 60t + 4
If the ball clears the fence, it means x = 0, y > 30
x = 0 -> 103.92 t - 350 = 0 -> t = 3.36 seconds
for t = 3.36s,
y = -16.4(3.36)^2 + 60*(3.36) + 4
= 20.45 ft
which is less than 30ft, so it means that the ball will NOT clear the fence.
Just for fun, let's check what the speed should have been :)
x = v0cos30t + x0 = v0cos30t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + v0sin30t + 4
x = 0 -> v0t = 350/cos30
y = 30 ->
-16.4t^2 + v0t(sin30) + 4 = 30
-16.4t^2 + 350sin30/cos30 = 26
t^2 = (26 - 350tan30)/-16.4
t = 3.2s
v0t = 350/cos30 -> v0 = 350/tcos30 = 123.34 ft/s
So he needed to hit the ball at at least 123.34 ft/s to clear the fence.
You're welcome, Thanks please :)
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