Solve the following inequality. -3.55g<-28.4 which graph shows the correct solution?
2 answers:
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Answer:
Data that we know:
He drove 504km in each direction.
Going, the average speed was 14km/h more than returning.
The total trip lasted 21 hours.
Ok, to solve this first:
Let's define the variables:
Sg = Average speed when going.
Sr = Average speed when returning.
Then we can write one of our relationships as:
Sg = Sr + 14km/h.
Now, you can recall the relation:
Time = Distance/speed.
Then we can write the equation that represents the total time of the travel as:
504km/Sg + 504km/Sr = 21h
Now we can replace the Sg by Sr + 14km/h, and get:
504km/(Sr + 14km/h) + 504km/Sr = 21h.
Now we must multiplacate by each denominator, in order to remove them:
504km*Sr + 504km*(Sr + 14km/h) = 21h*Sr*(Sr + 14km/h)
Sr*1008km + 7056km^2/h = 21h*Sr^2 + 294km*Sr
Now we can write a cuadratic equation, where i will ignore the units so it is easier to read, as:
21*Sr^2 -714*Sr - 7056 = 0
The solutions are given by the Bhaskara formula:
We have two solutions, one negative and one positive, and because Sr represents an average speed, it must be a positive number, then we choose the positive solution:
Sr = (714 + 1050)/42 km/h = 42km/h
Now we have the average speed for the returning, with this we can find Sg.
Sg = Sr + 14km/h = 42km/h + 14km/h = 56km/h.
Answer:
Probability that Adam and Ella will be selected:
Step-by-step explanation:
Probabilities
The probability of a random event E to occur is a real number between 0 and 1, both inclusive, where 0 indicates an impossible event and 1 a sure event. There are many techniques to compute probabilities depending on the particular situation and distribution.
This question will be solved by simple calculations and logic, given its simplicity. We know the middle school chess club has 5 members: Adam, Bradley, Carol, Dave, and Ella. Two of them are going to be selected at random to participate in the county chess tournament. We can calculate the number of different ways it can be done without any restriction. It's called the sample space.
The sample space of this event is the combination of 5 members regardless of their position. If {a,b,c,d,e} are the five members, then the possible combinations are {ab,ac,ad,ae,bc,bd,be,cd,ce,de}. Notice that there are only 10 possibilities because the combination ab is the same as ba since it's the same team for the tournament.
We can see there is only one combination of two specific letters out of 10. If a=Adam and e=Ella, only one combination is ae, the other 9 don't include both members, so the probability is