The best correct answer is option E. If two or more polygons are congruent, then rigid transformations can be used to map one polygon onto the other. <span>Two figures are said to be congruent when they have the same shape and
size or if one object is a mirror image of the other object.</span>
Answer:
The answer to your question is: Japan
Step-by-step explanation:
Mobile cost in Spain = € 352.5
Mobile cost in Japan = ¥39856
Exchange rate = £1 = €1.41
£1 = ¥188
Cost of mobile in pounds
£ 1 ------------------ € 1.41
x ----------------- €352.5
x = (352.5 x 1) / 1.41
x= £250
£1 ------------------ ¥188
x ----------------- -¥ 39856
x = (39856 x 1) / 188
x = £212
Then, it was cheaper in Japan
To figure out this question, I first divided 24 into 3, which gives me an answer of 8. then I divided 52 into 7 , which also is 8 . because the two numbers are the same, I know the ratios are equivalent.
Answer:1.8x – 10 = –4; x = 1.8 x minus 10 equals negative 4; x equals StartFraction 10 Over 2 EndFraction.
1.8 - 10= -4 x= -10/9
Step-by-step explanation:
(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.
