<span>Using the kinematic equations:
(final velocity)^2 = (initial velocity)^2 - 2 * acceleration * distance
Assuming the acceleration/deceleration on the car is constant from a constant force on the brakes. Converting from mph to m/s using 0.447 (so 34 mph is 15.2 m/s)
(0)^2 = (15.2)^2 - 2 * acceleration * 29
acceleration = 4.0 m/s^2
Had the car been going 105.4 mph (47.1 m/s)
(0)^2 = (47.1)^2 - 2 * 4 * distance
distance = 277 meters</span>
I believe that this problem has the following choices:
It must be equal to BQ .<span>
It must be wider than when he constructed the arc centered at
point A.
It must be equal to AB .
It must be the same as when he constructed the arc centered
at point A.</span>
The correct answer is the last one:
It must be the same as when he constructed the arc centered
at point A.
<span> </span>
Answer:
2.865
Step-by-step explanation:
pls rate five stars, thanks :)
⇒ x³y⁴ × x⁵y³
⇒ x³⁺⁵y⁴⁺³
⇒ x⁸y⁷
In short, Your Answer would be x⁸y⁷
Hope this helps!
Answer:
Step-by-step explanation:
Given that the mean incubation time for a type of fertilized egg kept at a certain temperature is 25 days.
Let X be the incubation time for a type of fertilized egg kept at a certain temperature is 25 days.
X is N(25, 1)
a) Normal curve is in the attached file
b) the probability that a randomly selected fertilized egg hatches in less than 23 days
=
we convert x into Z score and use std normal distn table to find probability
i.e. we can say only 2.5% proportion will hatch in less than 23 days.
