Complete question :
On a flight New York to London an airplane travels at a constant speed. An equation relating the distance traveled in miles d to the number of hours flying t is t= 1/500d. How long will it take the airplane to travel 800 miles?
Answer:
1.6 hours
Step-by-step explanation:
Given the function :
Time t taken for flying distance d is given by:
t = 1/500 * d
Tine taken to fly 800 miles;
d = 800 miles
t = 1/500 * (800)
t = 800 / 500
t = 1.6
Hence, time t = 1.6 hours
Answer:
20%
Step-by-step explanation:
- Loss=C.P-S.P
- Loss=10000-8000
- Loss=2000
- Loss%=loss/cp*100
- Loss%=2000/10000*100
- Loss%=20%
-x-10>14+2x
Add x for both side
-x+x-10>14+2x+x
-10>14+3x
Subtract 14 for both side
-10-14>14+3x-14
-24>3x
Divided 3 for both side
-24/3>3x/3
-8>x
x<-8
Or
-x-10>14-2x
-3x-10>14
-3x>24
x<-8.
Um, we see that there is one step that are missing which is:
-3x>24. Hope it help!
Answer:
(a) 283 days
(b) 248 days
Step-by-step explanation:
The complete question is:
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. (a) What is the minimum pregnancy length that can be in the top 11% of pregnancy lengths? (b) What is the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths?
Solution:
The random variable <em>X</em> can be defined as the pregnancy length in days.
Then, from the provided information
.
(a)
The minimum pregnancy length that can be in the top 11% of pregnancy lengths implies that:
P (X > x) = 0.11
⇒ P (Z > z) = 0.11
⇒ <em>z</em> = 1.23
Compute the value of <em>x</em> as follows:

Thus, the minimum pregnancy length that can be in the top 11% of pregnancy lengths is 283 days.
(b)
The maximum pregnancy length that can be in the bottom 5% of pregnancy lengths implies that:
P (X < x) = 0.05
⇒ P (Z < z) = 0.05
⇒ <em>z</em> = -1.645
Compute the value of <em>x</em> as follows:

Thus, the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths is 248 days.