Let's call their parts w,m and s
if will paid 1/3, then m+s=2w (they paid 2/3, which is twice as much as will did)
Now, we know that:
Micah and Sue paid in the ratio 2:3.
this means the 3m=2s
and m=2/3s
Again:
m+s=2w
and we substitute m:
2/3s+s=2w
5/3s=2w// multiply both sides by 3
5s=6w
we also know that s=w+6 (from the last sentence) so we substitute:
5(w+6)=6w
5w+30=6w
30=w
so, Will paid 30, Sue paid 36 (six more than him), Mike paid 24 (24:36 is the same ratio as 2:3, you can check this by dividing both 24 and 36 by 12: you have 2 and 3)
and the total was 30+36+24=90.
Answer:
It is not possible.
Step-by-step explanation:
How exact is exact? Since the circumference of a circle depends on pi, you have to discuss the nature of pi. C = pi * the diameter.
You are getting to a place where the air is pretty thin when you start dealing with pi. Pi has an infinite number of digits associated with it. Not only that, but if I were to tell you what the 100th digit was, you would have a random chance of figuring out what the 99th digit was or the 101 digit should be, that's only to the hundredth digit. The circumference would go beyond what we can measure with the thousandth digit.
In addition, pi is an irrational number. That means it cannot be represented by any kind of a fraction.
Answer:/?idk
Step-by-step explanation:
Answer:
P=0.147
Step-by-step explanation:
As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2
We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.
We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.
So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7
P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02
P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007
P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12
P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147
Let X be the number of lightning strikes in a year at the top of particular mountain.
X follows Poisson distribution with mean μ = 3.8
We have to find here the probability that in randomly selected year the number of lightning strikes is 0
The Poisson probability is given by,
P(X=k) = 
Here we have X=0, mean =3.8
Hence probability that X=0 is given by
P(X=0) = 
P(X=0) = 
P(X=0) = 0.0224
The probability that in a randomly selected year, the number of lightning strikes is 0 is 0.0224
