2002=25,160
2001=22,644
2000=19,926.72 (need to round to 19,926)
2001:
25,160/10=2516
25,160-2516
2000:
22,644/10=2,264.4 (10%)
2,264.4/5=452.88 (2%)
2,264.4+452.88=2717.28 (12%)
22,644-2717.28=19.926.72
What does this mean buddy?
Answer:
a=1 or a=5/6
Step-by-step explanation:
I'm going to attempt to factor 6a^2-a-5
a=6
b=-1
c=-5
Find two numbers that multiply to be a*c and add to be b.
a*c=-30 =-6(5)
b=-1 =-6+5
So replace -a with -6a+5a in the expression we started with
6a^2-6a+5a-5
now we factor by grouping
6a(a-1)+5(a-1)
(a-1)(6a-5)
Now let's solve the equation:
(a-1)(6a-5)=0
So a=1 or a=5/6
Since these 2 events are independent , the probability of them both occurring is obtained by multiplying them
P(rain on Wed and Sat) =0.30* 0.60 = 0.18
which is 18%
Answer:
v_top = 2400 mi/hr
v_w = 400 mi/h
Step-by-step explanation:
Given:
- Total distance D = 4800 mi
- Headwind journey time taken t_up= 3 hr
- Tailwind journey time taken t_down = 2 hr
Find:
Find the top speed of Luke's snow speeder and the speed of the wind.
Solution:
- The speed of Luke v_l is in stationary frame is given by:
v_l = v_w + v_l/w
Where,
v_w: Wind speed
v_l/w: Luke speed relative to wind.
- The top speed is attained on his returned journey with tail wind. We will use distance time relationship to calculate as follows:
v_top = D / t_down
v_top = 4800 / 2
v_top = v_down = 2400 mi/hr
- Similarly his speed on his journey up with head wind was v_up:
v_up = D / t_up
v_up = 4800 / 3
v_up = 1600 mi/hr
- Now use the frame relations to find the wind speed v_w:
v_down = v_w + v_l/w
v_up = -v_w + v_l/w
- Solve equations simultaneously:
2400 = v_w + v_l/w
1600 = -v_w + v_l/w
4000 = 2*v_l/w
v_l/w = 2000 mi/h
v_w = 400 mi/h
