Answer:
The coefficients are: 2, 12, 4.
Step-by-step explanation:
<span>y=2\3x-17and 4x-6y=-7 implies - 6y= - 4x -7, and then y= (-4/-6)x +7/6 and so
y=2/3x+7/6
but when we have y=ax+b and y=a'x+b', the two lines are parallels if a=a', in our case 2/3=2/3 the lines are parallel
</span>
Answer:
The required probability is 0.927
Step-by-step explanation:
Consider the provided information.
Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course.
That means 95% of students didn't enrolled in SAT prep course.
Let P(SAT) represents the enrolled in SAT prep course.
P(SAT)=0.05 and P(not SAT) = 0.95
30% of the SAT prep students were admitted to their first choice college, as were 20% of the other students.
P(F) represents the first choice college.
The probability he didn't take an SAT prep course is:
![P[\text{not SAT} |P(F)]=\dfrac{P(\text{not SAT})\cap P(F) }{P(F)}](https://tex.z-dn.net/?f=P%5B%5Ctext%7Bnot%20SAT%7D%20%7CP%28F%29%5D%3D%5Cdfrac%7BP%28%5Ctext%7Bnot%20SAT%7D%29%5Ccap%20P%28F%29%20%7D%7BP%28F%29%7D)
Substitute the respective values.
![P[\text{not SAT} |P(F)]=\dfrac{0.95\times0.20 }{0.05\times0.30+0.95\times0.20}](https://tex.z-dn.net/?f=P%5B%5Ctext%7Bnot%20SAT%7D%20%7CP%28F%29%5D%3D%5Cdfrac%7B0.95%5Ctimes0.20%20%7D%7B0.05%5Ctimes0.30%2B0.95%5Ctimes0.20%7D)
![P[\text{not SAT} |P(F)]\approx0.927](https://tex.z-dn.net/?f=P%5B%5Ctext%7Bnot%20SAT%7D%20%7CP%28F%29%5D%5Capprox0.927)
Hence, the required probability is 0.927
Answer:
f = 100 + 0.133d
Step-by-step explanation:
For each trip, they charge an initial fee of 100 in addition to a constant fee for each vertical meter climbed.
For example, for 3000 meters of climbing, they charged the additional constant fee of 400.
Therefore, per meter climbing they charges
.
Therefore, if f represents the fee in dollars of a trip they climbed d vertical meters then we can model the situation as
f = 100 + 0.133d (Answer)