Answer:
14.04 miles per hour
Step-by-step explanation:
The problem is asking for Izumi's speed. The formula of speed is:
- s =
, where "s" means speed, "d" means distance and "t" means time.
The problem is also asking for the unit<u> miles per hou</u>r (
) so, this means that we have to know how many miles Izumi ran, given that the problem only mentioned<u> yards (110 yards).</u>
Let's convert 110 yards to miles, provided that he Izumi ran 1,760 yards in a mile.
- 110 yards ÷ 1760
= 0.0625 miles (this is the distance covered by Izumi in miles)
Let's go back again to the formula: s = 
s =
= 0.0039 
Since, the we arrived at a miles per second unit, we have to convert it to miles per hour.
So, if a minute has 60 seconds, then an hour has 3,600 seconds.
Thus, 0.0039
× 3,600
= 14.04 miles per hour (the answer)
Answer:
15 more
Step-by-step explanation:
John has 36 sweets and he shares them in the ratio 2 : 7.
How many sweets is the larger share?
Answer:
25,722 yen
Step-by-step explanation:
300x85.74 is the simplest way to do this and I am just adding words to compete this
(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.