Required information An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiv
1 answer:
Answer:
So on this case the 95% confidence interval would be given by (563.997;741.203)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean for the sample
population mean (variable of interest)
s=311.7 represent the sample standard deviation
n=50 represent the sample size
Soltuion to the problem
The confidence interval for the mean is given by the following formula:
(1)
In order to calculate the critical value we need to find first the degrees of freedom, given by:
Since the Confidence is 0.95 or 95%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 95% confidence interval would be given by (563.997;741.203)
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Three important properties of the diagonals of a rhombus that we need for this problem are:
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus
First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°.
Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.
For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.
Similarly, we have
Now, to find the lengths of the diagonals,
So, the lengths of the diagonals are 10 and 10√3.
Answer: 10 and 10√3 units
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus
First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°.
Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.
For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.
Similarly, we have
Now, to find the lengths of the diagonals,
So, the lengths of the diagonals are 10 and 10√3.
Answer: 10 and 10√3 units
Answer:
the probability the die chosen was green is 0.9
Step-by-step explanation:
Given that:
A bag contains two six-sided dice: one red, one green.
The red die has faces numbered 1, 2, 3, 4, 5, and 6.
The green die has faces numbered 1, 2, 3, 4, 4, and 4.
From above, the probability of obtaining 4 in a single throw of a fair die is:
P (4 | red dice) =
P (4 | green dice) = =
A die is selected at random and rolled four times.
As the die is selected randomly; the probability of the first die must be equal to the probability of the second die =
The probability of two 1's and two 4's in the first dice can be calculated as:
=
=
=
=
=
=
The probability of two 1's and two 4's in the second dice can be calculated as:
=
=
=
=
=
∴
The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )
The probability of two 1's and two 4's in both die =
The probability of two 1's and two 4's in both die =
The probability of two 1's and two 4's in both die =
By applying Bayes Theorem; the probability that the die was green can be calculated as:
P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)
P(second die (green) | two 1's and two 4's ) =
P(second die (green) | two 1's and two 4's ) =
P(second die (green) | two 1's and two 4's ) = 0.9
Thus; the probability the die chosen was green is 0.9