Answer:
I'm not sure if i'm correct but i'm guessing no
Step-by-step explanation:
<h3>Answer: The lines are perpendicular</h3>
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Work Shown:
Use the slope formula to find the slope of the line through the two given points.
m = (y2 - y1)/(x2 - x1)
m = (-4 - (-5))/(6 - (-7))
m = (-4 + 5)/(6 + 7)
m = 1/13
The slope of the line through the two given points is 1/13.
This is not equal to -13, which was the slope of the first line, so the two lines are not parallel.
However, the two lines are perpendicular because multiplying the two slope values leads to -1
(slope1)*(slope2) = (-13)*(1/13) = -1
note: It might help to think of -13 as -13/1 to help multiply the fractions.
Answer:
Q1
cos 59° = x/16
x = 16 cos 59°
x = 8.24
Q2
BC is given 23 mi
Maybe AB is needed
AB = √34² + 23² = 41 (rounded)
Q3
BC² = AB² - AC²
BC = √(37² - 12²) = 35
Q4
Let the angle is x
cos x = 19/20
x = arccos (19/20)
x = 18.2° (rounded)
Q5
See attached
Added point D and segments AD and DC to help with calculation
BC² = BD² + DC² = (AB + AD)² + DC²
Find the length of added red segments
AD = AC cos 65° = 14 cos 65° = 5.9
DC = AC sin 65° = 14 sin 65° = 12.7
Now we can find the value of BC
BC² = (19 + 5.9)² + 12.7²
BC = √781.3
BC = 28.0 yd
All calculations are rounded
Answer:

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795
Step-by-step explanation:
Given
See attachment for class
Solving (a): Fill the midpoint of each class.
Midpoint (M) is calculated as:

Where
Lower class interval
Upper class interval
So, we have:
Class 63-65:

Class 66 - 68:

When the computation is completed, the frequency distribution will be:

Solving (b): Mean and standard deviation using 1-VarStats
Using 1-VarStats, the solution is:


<em>See attachment for result of 1-VarStats</em>