Consider WXY and BCD with X= C, WX = BC, and WY = BD. Can it be concluded that WXY = BCD by SAS?
2 answers:
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Answer:
The 34 week gestation's period baby weighs 0.11 standard deviations below the mean.
The 41 week gestation's period baby weighs 0.24 standard deviations below the mean.
A. The baby born in week 40 does since its z-score is smaller.
Step-by-step explanation:
Normal model problems can be solved by the zscore formula.
On a normaly distributed set with mean and standard deviation
, the z-score of a value X is given by:
The zscore represents how many standard deviations the value of X is above or below the mean[/tex]\mu[/tex]. Whoever has the lower z-score weighs relatively less.
Find the corresponding z-scores.
Babies born after a gestation period of 32 to 35 weeks have a mean weight of 2800 grams and a standard deviation of 900 grams. A 34-week gestation period baby weighs 2700 grams.
Here, we have .
So
A negative z-score indicates that the value is below the mean.
So, the 34 week gestation's period baby weighs 0.11 standard deviations below the mean.
Babies born after a gestations period of 40 weeks have a mean weight of 3400 grams and a standard deviation of 425 grams. A 40-week gestation period baby weighs 3300 grams.
Here, we have
So, the 41 week gestation's period baby weighs 0.24 standard deviations below the mean.
Which baby weighs relatively less?
A. The baby born in week 40 does since its z-score is smaller.
Problem 1)
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF
-------------------------------------------------------------
Problem 2)
a) The center is (2,-3)
The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b) The radius is 3 and the diameter is 6
From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF
-------------------------------------------------------------
Problem 2)
a) The center is (2,-3)
The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b) The radius is 3 and the diameter is 6
From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.