Let's assume
the price of each train ticket is x
we are given
total number of family members =5
so, total cost of ticket = total number of family members*the price of each ticket
so, total cost of ticket =5*x
so, total cost of ticket =5x
now, we have
each family member buys a boxed lunch for $6.50
so, total cost of lunch box = total number of family members*price of each lunch box
so, total cost of lunch box = 5*6.50
total cost of trip = total cost of ticket+total cost of lunch box
we are given
total cost of trip is $248.50
now, we can plug values
and we get

now, we can solve for x





so,
the price of each train ticket is $43.2..........Answer
X^m means x times itself m times
(-6)^5 means -6 time itself 5 times
simplify it says
remember that
(mn)^x=(m^x)(n^x)
undistribute the -1
(-6)^5=(-1)^5 times (6^5)
(-1)^5=-1 since it is odd power
can be simplified to
-6^5
(note, -6^5 vs (-6)^5 PEMDAS)
Answer:
sss
Step-by-step explanation:
i think because 3 sides are given
Answer: Scientific knowledge?
I’m not really sure what you’re asking, but I hope this helped!
Step-by-step explanation:
Solution:
Population in the city= 10,000
As genetic condition affects 8% of the population.
8 % of 10,000

As, it is also given that, there is an error rate of 1% for condition (i.e., 1% false negatives and 1% false positives).
So, 1% false negatives means out of 800 tested who are found affected , means there are chances that 1% who was found affected are not affected at all.
So, 1% of 800 
Also, 1% false positives means out of 10,000 tested,[10,000-800= 9200] who are found not affected , means there are chances that 1% who was found not affected can be affected also.
So, 1% of 9200 
1. Has condition Does not have condition totals = 800
2. Test positive =92
3. Test negative =8
4. Total =800 +92 +8=900
5. Probability (as a percentage) that a person has the condition if he or she tests positive= As 8% are found positive among 10,000 means 9200 are not found affected.But there are chances that out of 9200 , 1% may be affected

that is Probability equal to 0.01 or 1%.