recall your d = rt, distance = rate * time.
let's say we have two trains, A and B, A is going at 85 mph and B at 65 mph.
they are 210 miles apart and moving toward each other, at some point they will meet, when that happens, the faster train A has covered say d miles, and the slower B has covered then the slack from 210 and d, namely 210 - d.
When both trains meet, A has covered more miles than B because A is faster, however the time both have been moving, is the same, say t hours.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{Train A}&d&85&t\\ \textit{Train B}&210-d&65&t \end{array}\\\\ \dotfill\\\\ \begin{cases} \boxed{d}=85t\\ 210-d=65t\\[-0.5em] \hrulefill\\ 210-\boxed{85t}=65t \end{cases} \\\\\\ 210=150t\implies \cfrac{210}{150}=t\implies \cfrac{7}{5}=t\implies \stackrel{\textit{one hour and 24 minutes}}{1\frac{2}{5}=t}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ctextit%7BTrain%20A%7D%26d%2685%26t%5C%5C%20%5Ctextit%7BTrain%20B%7D%26210-d%2665%26t%20%5Cend%7Barray%7D%5C%5C%5C%5C%20%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D85t%5C%5C%20210-d%3D65t%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20210-%5Cboxed%7B85t%7D%3D65t%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20210%3D150t%5Cimplies%20%5Ccfrac%7B210%7D%7B150%7D%3Dt%5Cimplies%20%5Ccfrac%7B7%7D%7B5%7D%3Dt%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bone%20hour%20and%2024%20minutes%7D%7D%7B1%5Cfrac%7B2%7D%7B5%7D%3Dt%7D%20)
Since there are two events happening simultaneously (windy and no sun), we can apply the concept of conditional probability here.
P(A|B) = P(A∩B)/P(B)
where it means that given B is happening, the probability that A is happening as well is the ratio of the chance for A and B to happen simultaneously over the chance of B to happen.
For our case, this can be interpreted as
P(A|B): it is the probability that it is windy (A) GIVEN that there is no sun (B)
P(A∩B) : chance of wind and no sun
P(B) : chance that there is no sun tomorrow
The chance of P(A∩B) is already given as 20% or 0.20. Since there is 10% or 0.10 chance of sun, then chances of having no sun tomorrow is (1-0.10) = 0.90.
Thus, we have P(A|B) = 0.2/0.9 ≈ 0.22 or 22%.
So, answer is B: 22%<span>.</span>
Answer:
Weakest value = 0.0196
Strongest value = 0.5929
Step-by-step explanation:
Given:
Choices
0.691
-0.14
-0.24
-0.77
Computation:
0.691² = 0.477481
-0.14² = 0.0196
-0.24² = 0.0576
-0.77² = 0.5929
Weakest value = 0.0196
Strongest value = 0.5929
1.
Calculate the sum
5x - 10 + 7 = 65 - 20x + 32
Move terms
5x - 3 = 97 - 20x
Collect the like terms and calculate
5x + 20x = 97 +3
Divide both sides by 25
25x = 100
X= 4 ANSWER
I skipped some steps because it would be too long :/
2.
Multiply parenthesis by 8
20x>8(4x - 5) -20
Calculate
20x>32x - 40 - 20
Move variable to the left
20x>32x-60
Collect like terms
20x - 32x > -60
Divide both sides by -12
-12x>-60
X<5 ANSWER
Answer:
What about
Step-by-step explanation:
SUS :0000
jp I'm pretty sure its asa. forgive me if im wrong.
