Given:
t1 = 3.6 h
t2 = 4.5 h
x = speed of boat
y = speed of water
Required:
a) Expression of distance traveled with moving water with 3.6h
Expression of distance traveled with moving water with 4.5h
b) Solve for y
c) Percent of boat's speed is the water current
Solution:
Working formula: distance = velocity*time
a) For travelling downstream, we get the equation
d = (x +y)*3.6
For travelling upstream, we get the equation
d = (x-y)*4.5
b) Setting the distance as equal for travelling upstream or downstream, we arrive at the equation of
(x+y)*3.6 = (x-y)*4.5
3.6x + 3.6y = 4.5x - 4.5y
8.1y =0.9x
y = x/9
c) percentage = 1/9*100% = 11.1%
<em>ANSWERS: a) d = (x+y)*36; d = (x-y)*4.5
</em> <em>b) y = x/9
</em> <em>c) 11.1%</em>
Answer:
Step-by-step explanation:
2X + 1 + 3X - 2 + 4X - 5 = 180
9X - 6 = 180
9X = 174
9 = 174
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9 9
X = 19.333
The range is: 35-60
The IQR is: 11
This value is the first 25%
IQR= interquartile range
Hope this helps ;)
2: c
Step-by-step explanation:
3: a
by Supplementing each answer Into the equation