See attachment for the drawing of the perpendicular at a point c on line ab such the ac = 3.5 cm
<h3>How to draw the perpendicular at a point c on line ab such the ac = 3.5 cm?</h3>
The given parameters are:
Length of segment AB = 7 cm
Also, we have:
Point C is located on line segment AB
This means that:
Length of segment AC = Length of segment BC = 3.5 cm
When represented on a line segment, we have:
A C B
See attachment for the drawing of the perpendicular at a point c on line ab such the ac = 3.5 cm
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Hello :
an equation of the circle <span>Center at the A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a = -2 and b =-3 (</span><span>Center A(-2,-3))
r = AP......(P(-2,0))
r² = (AP)²
r² = (-2+2)² +(-3-0)² = 9
</span><span>an equation of the circle that satisfies the stated conditions.
Center </span> A(-2,-3) , passing through P(-2, 0) is : (x+2)² +(y+3)² = 9
Answer:
true
Step-by-step explanation:
hope this helps
Answer:
Mean = 35
Variance = 291.7
Step-by-step explanation:
Data provided in the question:
X : 1, 2, 3, 4, 5, 6
All the data are independent
Thus,
The mean for this case will be given as:
Mean, E[X] = 
or
E[X] = 
or
E[X] = 3.5
For 10 days, Mean = 3.5 × 10 = 35
And,
variance = E[X²] - ( E[X] )²
Now, for this case of independent value,
E[X²] = 
or
E[X²] = 
or
E[X²] = 
or
E[X²] = 15.167
Therefore,
variance = E[X²] - ( E[X] )²
or
variance = 15.167 - 3.5²
or
Variance = 2.917
For 10 days = Variance × Days²
= 2.917 × 10²
= 291.7
B=29.4 !!!!!!!! this how i do it
