In rectangle ABCD the diagonals intersect each other at point O and m∠ABD=30°. Find BC if AC = 16 in....
1 answer:
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Draw ΔABC as shown in the figure below.
m∠B = 137°, b = 14, c = 9.
From the Law of Sines,
sin(C)/c = sin(B)/b
or
sin(C)/9 = sin(137°)/14
sin(C) = (9/14)*sin(137°) = 0.4384
m∠C = arcsin(0.4384) = 26°
Because the sum of angles in a triangle is 180°, therefore
m∠A = 180 -137 - 26 = 17°
From the Law of Sines,
a/sin(A) = b/sin(B)
or
a/sin(17°) = 14/sin(137°)
a = (sin(17°)/sin(127°))*14 = 6
Answer:
The solution for the triangle is
a = 6, b = 14, c = 9
m∠A = 17°, m∠B = 137°, m∠C = 26°
m∠B = 137°, b = 14, c = 9.
From the Law of Sines,
sin(C)/c = sin(B)/b
or
sin(C)/9 = sin(137°)/14
sin(C) = (9/14)*sin(137°) = 0.4384
m∠C = arcsin(0.4384) = 26°
Because the sum of angles in a triangle is 180°, therefore
m∠A = 180 -137 - 26 = 17°
From the Law of Sines,
a/sin(A) = b/sin(B)
or
a/sin(17°) = 14/sin(137°)
a = (sin(17°)/sin(127°))*14 = 6
Answer:
The solution for the triangle is
a = 6, b = 14, c = 9
m∠A = 17°, m∠B = 137°, m∠C = 26°
Answer:
For this case we want this probability:
And we can use the z score formula given by:
So we want the probability for the values within two deviations from the mean and from the empirical rule we know that we have approximately 0.95 or 95% of the values within two deviations from the true mean.
And since we have a total of 120 students we expect:
120*0.95 = 114 between 75g and 87 g
Step-by-step explanation:
Previous concepts
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
Let X the random variable who represent the courtship time (minutes).
From the problem we have the mean and the standard deviation for the random variable X.
So we can assume
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
Solution to the problem
For this case we want this probability:
And we can use the z score formula given by:
So we want the probability for the values within two deviations from the mean and from the empirical rule we know that we have approximately 0.95 or 95% of the values within two deviations from the true mean.
And since we have a total of 120 students we expect:
120*0.95 = 114 between 75g and 87 g