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
The answer is 3 3x3x3=9
Hope this helped:)
Answer:
Given the expression: log 2 -log 6
Use the logarithmic rule :


Use the above rule to solve the expression:
[∴
]
or
[∴
]
or
value of : log 2 -log 6 = 0.30102999566 - 0.77815125038 = -0.47712125472
<h3>
Answer: 81</h3>
Work Shown:
x = number of cats
(4 cats)/(5 dogs) = (x cats)/(45 dogs)
4/5 = x/45
4*45 = 5x
180 = 5x
5x = 180
x = 180/5
x = 36
There are 36 cats and x+45 = 36+45 = 81 cats and dogs total.