The length of the route of a bicylist that has covered 5/7 of his route and an additional 40 miles and yet to cover 118 miles less than 0.75 of his route is 168 miles.
<h3>How to form equation and solve for the variable?</h3>
The bicyclist covered 5 / 7 of his route and an addditional 40 miles.
Let
x = distance of his route in miles
Hence,
distance covered = 5 / 7 x + 40
He has yet to cover 118 miles less than 0.75 of his route. Therefore,
distance not covered = 0.75x - 118
Distance of his route = 5 / 7 x + 40 + 0.75x - 118
x = 5 / 7 x + 40 + 0.75x - 118
x = 1.464x - 78
x - 1.464x = -78
-0.464x = -78
x = 78 / 0.464
x = 168.103448276
x = 168 miles
learn more on equation here: brainly.com/question/13887440
You could simplify this work by factoring "3" out of all four terms, as follows:
3(x^2 + 2x - 3) =3(0) = 0
Hold the 3 for later re-insertion. Focus on "completing the square" of x^2 + 2x - 3.
1. Take the coefficient (2) of x and halve it: 2 divided by 2 is 1
2. Square this result: 1^2 = 1
3. Add this result (1) to x^2 + 2x, holding the "-3" for later:
x^2 +2x
4 Subtract (1) from x^2 + 2x + 1: x^2 + 2x + 1 -3 -1 = 0,
or x^2 + 2x + 1 - 4 = 0
5. Simplify, remembering that x^2 + 2x + 1 is a perfect square:
(x+1)^2 - 4 = 0
We have "completed the square." We can stop here. or, we could solve for x: one way would be to factor the left side:
[(x+1)-2][(x+1)+2]=0 The solutions would then be:
x+1-2=0=> x-1=0, or x=1, and
x+1 +2 = 0 => x+3=0, or x=-3. (you were not asked to do this).
Hi!!! So to do this problem, you need to find the mean values for each sample. I will walk you through finding the mean of sample 1: first you want to add all of the values for sample 1, which are 4,5,2,4 and 3. Once you add those values you get 18. Then you must divide that 18 by the number of terms you added. The numbers you added were 4,5,2,4 and 3 like I said earlier which is 5 numbers. You divide 18 by 5 to get your mean, which is 3.6
Answer: (B) Sample 2
sample 1 mean = 3.6
sample 2 mean = 4.2
sample 3 mean = 3.8
sample 4 mean = 4
With what? There is no question
Basically anything greater than zero...k>0
