A dog walks at a speed of 3mi/hr for 2 hours. How far (distance) did the dog walk? *
1 answer:
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The coordinates of point F are (15/4 , -5/2)
- The coordinates of point G = (-3/4 , -3)
From the image attached:
AB divided into 4 similar parts, hence
- E is the mid-point of AB
- D is the mid-point of AE
- F is the mid-point of EB
Note that the rule of the mid-point states that:
When M (x , y) is the mid-point of the segment of AB, where A (x1 , y1) and B (x2 , y2), Then x = (x1 + x2)/2 and that of y = (y1 + y2)/2
Then lets solve for points E, F, D
Since A (-3 , -1) and B (6 , -3),
E is the mid-point of AB
Then E =[(-3 + 6)/2, (-1 + -3)/2]
= (3/2 , -2)
Since F is the mid-point of EB,
E (3/2 , -2) , B (6 , -3)
Then F = [(3/2 + 6)/2 , (-2 + -3)/2]
= (15/4 , -5/2)
Since D is the mid-point of AE, A (-3 , -1), E (3/2 , -2)
Then D = [(-3 + 3/2)/2 , (-1 + -2)/2]
= (-3/4 , -3/2)
Note also that:
BC is said to be an horizontal segment due to the fact that B and C have similar y - coordinate and G can be found on BC.
Since the y-coordinate of G is similar to y-coordinate of B and C
Then y-coordinate of G is = -3
Since DG ⊥ BC
Then DG = vertical segment
So G has similar x-coordinate of D
Then The x-coordinate of G = -3/4
The G = (-3/4 , -3)
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Answer:
a) 0.283 or 28.3%
b) 0.130 or 13%
c) 0.4 or 40%
d) 30.6 mm
Step-by-step explanation:
z-score of a single left atrial diameter value of healthy children can be calculated as:
z= where
- X is the left atrial diameter value we are looking for its z-score
- M is the mean left atrial diameter of healthy children (26.7 mm)
- s is the standard deviation (4.7 mm)
Then
a) proportion of healthy children who have left atrial diameters less than 24 mm
=P(z<z*) where z* is the z-score of 24 mm
z*= ≈ −0.574
And P(z<−0.574)=0.283
b) proportion of healthy children who have left atrial diameters greater than 32 mm
= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm
z*= ≈ 1.128
1-P(z<1.128)=0.8703=0.130
c) proportion of healthy children have left atrial diameters between 25 and 30 mm
=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm
z(30)= ≈ 0.702
z(25)= ≈ −0.362
P(z<0.702)=0.7587
P(z<−0.362)=0.3587
Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4
d) to find the value for which only about 20% have a larger left atrial diameter, we assume
P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.
Then P(z<z*)=0.8 and z*=0.84. That is
0.84= solving this equation for X we get X=30.648