You start driving north for 16 miles, turn right, and drive east for another 30 miles. At the end of driving, what is your strai
1 answer:
The distance between starting and ending point is 34 miles.
Step-by-step explanation:
Given,
Car moves 16 miles to north then 30 mile to east.
It forms a right angle triangle.
The straight line distance from starting to ending point represents hypotenuse.
To find the distance between starting and ending point.
Formula
By <em>Pythagoras theorem,</em>
h² = b²+l² where h is the hypotenuse, b is base and l is the another side.
Taking, b=16 and l=30 we get,
h² = 16²+30²
or, h =
or, h = = 34
Hence,
The distance between starting and ending point is 34 miles.
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Answer:
(a) The probability that the second part is the one with excessive shrinkage is 0.2000.
(b) The probability that the third part is the one with excessive shrinkage is 0.2000.
Step-by-step explanation:
Let the variable <em>X </em>ₙ denote the <em>n</em>th part that has suffered excessive shrinkage.
(a)
It is provided that two parts are selected at random.
The parts are selected without replacement.
Now, for the two parts selected it is possible that either both have excessive shrinkage or only the second part has excessive shrinkage.
The probability that the second part is the one with excessive shrinkage is:
P (X₂) = P (X₂ ∩ X₁) + P (X₂ ∩ X₁')
(without replacement)
Thus, the probability that the second part is the one with excessive shrinkage is 0.2000.
(b)
It is provided that three parts are selected at random.
The parts are selected without replacement.
Now, for the three parts selected it is possible that either all three have excessive shrinkage or any two of the three has excessive shrinkage or only the third part has excessive shrinkage.
The probability that the third part is the one with excessive shrinkage is:
P (X₃) = P (X₃ ∩ X₂ ∩ X₁) + P (X₃ ∩ X₂ ∩ X₁')
+ P (X₃ ∩ X₂' ∩ X₁) + P (X₃ ∩ X₂' ∩ X₁')
Thus, the probability that the third part is the one with excessive shrinkage is 0.2000.