Domain is all the x values (2,-1,4,-3)and range is the y values (0,4,2,0)
Answer:
21+/-sqrt(253)=x
So one value for x is 21+sqrt(253)
and another is 21-sqrt(253)
Problem:
Given (21,7) and (x,1), find all x such that the distance between these two points is 17.
Step-by-step explanation:
Change in x is x-21
Change in y is 7-1=6
distance^2=(change in x)^2+(change in y)^2
17^2=(x-21)^2+(6)^2
289=(x-21)^2+36
Subtract 36 on both sides:
289-36=(x-21)^2
253=(x-21)^2
Take square root of both sides:
+/-sqrt(253)=x-21
Add 21 on both sides:
21+/-sqrt(253)=x
Answer:

Step-by-step explanation:
If we approximate the binomial distribution with a normal distribution, we have to apply a correction factor for the fact that we are now dealing with a continuous variable instead of a discrete one, as it was with the binomial distribution.
The probability of no more than 35 defective CDs: P(X<35)
In this case, as X=35 is not included in the interval, we start the interval from X=35-0.5=34.5.

being Pb the probability under the binomial distribution and Pn the probability under the normal distribution.
The area for the normal distribution is the one below X=34 (or P(X<34)).
d = distance between the two cities
v₁ = average speed while going from chicago to kansas city = 440 knots
t₁ = time taken to travel distance going from chicago to kansas city
time taken to travel distance going from chicago to kansas city is given as
t₁ = d/v₁
t₁ = d/440 eq-1
v₂ = average speed while going from kansas city to chicago = 110 knots
t₂ = time taken to travel distance going from kansas city to chicago
time taken to travel distance going from kansas city to chicago is given as
t₂ = d/v₂
t₂ = d/110 eq-2
Given that :
t₂ = t₁ + 3
using eq-1 and eq-2
(d/110) = (d/440) + 3
d = 440