<u>Question 1 solution:</u>
You have two unknowns here:
Let the Water current speed = W
Let Rita's average speed = R
We are given <em>two </em>situations, where we can form <em>two equations</em>, and therefore solve for the <em>two unknowns, W, R</em>:
Part 1) W→ , R←(against current, upstream)
If Rita is paddling at 2mi/hr against the current, this means that the current is trying to slow her down. If you look at the direction of the water, it is "opposing" Rita, it is "opposite", therefore, our equation must have a negative sign for water<span>:
</span>R–W=2 - equation 1
Part 2) W→ , R<span>→</span>(with current)
Therefore, R+W=3 - equation 2
From equation 1, W=R-2,
Substitute into equation 2.
R+(R–2)=3
2R=5
R=5/2mi/hr
So when W=0 (still), R=5/2mi/hr
Finding the water speed using the same rearranging and substituting process:
1... R=2+W
2... (2+W)+W=3
2W=1
W=1/2mi/hr
Answer:
-1/2=x
Step-by-step explanation:
-9 + x = 5x - 7
-x -x
-9=4x-7
+7 +7
-2=4x
-2/4 =4x/4
-1/2=x
The correct answer is C. 20 characters
Answer:
Step-by-step explanation:
4a² b³ * (9a⁴b² - 4a² + 3) = 4a²b³ * 9a⁴b² - 4a²b³*4a² + 4a²b³*3
= 36a²⁺⁴ b³⁺² -16a²⁺²b³ + 12a²b³
= 36a⁶b⁵ - 16a⁴b³ + 12a²b³
Answer:
The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]
Step-by-step explanation:
I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.
In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1
If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)
The values of the cumulative distribution function of W, denoted by 
, can be found in the attached file. Lets call 
. We have
Thus
by looking at the table, we find that y = 1.28, therefore
The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.
The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .
