Here we have a situation where the probability of a driver wearing seat belts is known and remains constant throughout the experiment of stopping 20 drivers.
The drivers stopped are assumed to be random and independent.
These conditions are suitable for modelling using he binomial distribution, where
where n=number of drivers stopped (sample size = 20)
x=number of drivers wearing seatbelts (4)
p=probability that a driver wears seatbelts (0.35), and
C(n,x)=binomial coefficient of x objects chosen from n = n!/(x!(n-x)!)
So the probability of finding 4 drivers wearing seatbelts out of a sample of 20
P(4;20;0.35)
=C(20,4)*(0.35)^4*(0.65)^16
= 4845*0.0150061*0.0010153
= 0.07382
The velocity of the bus is:
velocity (bus) = 36 miles / 40 min
velocity (bus) = 0.9 miles / min
Since the car is 1.5 times faster, so the velocity of the
car is:
velocity (car) = 1.5 * 0.9 miles / min
velocity (car) = 1.35 miles / min
At the meeting point, the sum of the distance is equal to
36 miles. Therefore:
1.35 t + 0.9 t = 36
2.25 t = 36
t = 16 min
So they will meet after 16 minutes.
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
I'm pretty sure simplest form would still be 865:2678
I could be wrong tho
Step-by-step explanation:
Transcribed image text: Spectra Analysis 7 100 80 60 40 20 150 50 STRUCTURE 60 40 20 0 180 160 140 120 100 8O
