Andrew walks along a road which can be modeled by the equation y=3x, where (0,0) represents his starting point. When he reaches
1 answer:
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The total number of gifts = x+y.
The inequality is:
Key chains cost $1, Magnets $0.50
Total Cost = x + 0.5y
Inequality is:
Without graphing you can solve system by using substitution:
This is one solution where the maximum x value is given.
So the most keychains that can be purchased is 16. However, because magnets are cheaper, more can be purchased as long as cost remains under 20.
If you solve both inequalities for "y", you get the upper and lower bounds for how many magnets can be purchased given a quantity of keychains.
This is complete solution which gives all possible combinations.
(Graph is Attached)
The inequality is:
Key chains cost $1, Magnets $0.50
Total Cost = x + 0.5y
Inequality is:
Without graphing you can solve system by using substitution:
This is one solution where the maximum x value is given.
So the most keychains that can be purchased is 16. However, because magnets are cheaper, more can be purchased as long as cost remains under 20.
If you solve both inequalities for "y", you get the upper and lower bounds for how many magnets can be purchased given a quantity of keychains.
This is complete solution which gives all possible combinations.
(Graph is Attached)
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