If 1 yard = 3 feet and 1 mile = 5,280 feet, how many yards are there in 4 miles?
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Answer:
The number of times coin thrown was <u>500</u>.
Step-by-step explanation:
Given:
A coin lands on heads 200 times. the relative frequency of heads is 0.4.
Now, to find the times coin was thrown.
Let the number of times coin thrown was be
Relative frequency = 0.4.
Number of lands on heads = 200.
So, to get the number of times coin was thrown we put formula:
<em>By cross multiplying we get:</em>
<em>Dividing both sides by 0.4 we get:</em>
Therefore, the number of times coin thrown was 500.
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