Answer:
Confidence Interval: (21596,46428)
Step-by-step explanation:
We are given the following data set:
10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929
Formula:
where 
are data points, 
is the mean and n is the number of observations.
Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821
Confidence interval:
Putting the values, we get,
Answer:what grade math is this
Step-by-step explanation:
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
The slope is -2. this is because if you look at the graph it goes down 2 units and it goes right to the once so it’ll be -2/1 which equals -2.
