Y- intercept is a point where any graph crosses the y- axis.
X- intercept is a point where any graph crosses the x- axis.
This means the coordinate of the point of intersection will always have the x point as 0. So any point of the form ( 0, y) is the y- intercept. Any point of the form (x,0) is the x- intercept.
Given point are :
(0,-6) : y intercept
(-2,0) : x intercept
(-6,0): x- intercept
(0,-2): y- intercept
Let us say that the procedure of completely filling up the
pool with water is called “1 job”. Therefore the rates of the equipment in
doing the job would be:
<span>Rate of handheld hose = 1 job / 8 minutes ---> 1</span>
<span>Rate of lawn sprinkler = r --->
2</span>
<span>Rate of the two combines = 1 job/ 5 minutes ---> 3</span>
So we are to find the equation to use, in this case, we
simply have to add equations 1 and 2 to get 3:
1 / 8 + r = 1 / 5
Multiplying both sides by 5:
<span>(5 / 8) + 5 r = 1 --->
ANSWER</span>
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
Answer:
55
Step-by-step explanation:
412.5 / 7.5 =55
Answer:
83.79 kJ
Step-by-step explanation:
The potential energy is the product of weight and height, where weight is in newtons.
PE = Mgh = (190 kg)(9.8 m/s^2)(45 m) = 83,790 J
