Average velocity is equal to the change in distance / change in time :
v(avg) = s / t
that's the average distance, but at some point the object could have moved faster / slower and we want to see the instantaneous velocity. The principal is the same. You take a a little piece of the road for a little piece if time and get the avg velocity of the tiny part of the road. But this way it will be not the exactly speed at that moment, it will be just the avg speed in the little fragment. So we must add limits. What will be the velocity for so tiny part of time that it approaches zero?
We get :
lim s/t = ds/dt
t->0
Answer:
- $52
- this estimate is a little low
Explanation:
a) I might choose 6 1/2 hours and $8 per hour, so the estimated earnings would be ...
6(8) +(1/2)(8) = 52 dollars . . . . estimated earnings
These values are purposely chosen so that one is a little high and the other is a little low, hopefully producing a closer estimate than if both numbers were high (as 7 hours, $8, for example).
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b) Using 800 instead of 790 means the pay estimate is about 1/80 = 1.25% high.
Using 6 1/2 instead of 6 3/4 means the hours estimate is about 1/27 ≈ 3.7% low.
So, we expect the product of these values to be slightly low, perhaps by about 2.5%.
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<em>Comment on estimating</em>
If you can estimate the error in the estimate with some reasonable accuracy, you can use that to adjust the estimate to a pretty close value.
The mistake is n third line - adding 3w
96 + 2w =2(80 - 3w)
96 + 2w = 160 - 6w
+ 6w + 6w
96 + 8w = 160
- 96 -96
8w = 64
w = 8 weeks answer
Answer:
Probability of being senior: P(A) = 0.22
Probability of being female: P(B) = 0.46
=> P(A) x P(B) = 0.22 x 0.46 = 0.10
Otherwise, probability of being senior and female: P(A⋂B) = 0.12
=> P(A) x P(B) is not equal to P(A⋂B)
=> A and B are not independent
Hope this helps!
:)