Answer:
Step-by-step explanation:
step 1
Find the measure of the arc DC
we know that
The inscribed angle measures half of the arc comprising
![m\angle DBC=\frac{1}{2}[arc\ DC]](https://tex.z-dn.net/?f=m%5Cangle%20DBC%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20DC%5D)
substitute the values
![60\°=\frac{1}{2}[arc\ DC]](https://tex.z-dn.net/?f=60%5C%C2%B0%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20DC%5D)
step 2
Find the measure of arc BC
we know that
----> because the diameter BD divide the circle into two equal parts
step 3
Find the measure of angle BDC
we know that
The inscribed angle measures half of the arc comprising
![m\angle BDC=\frac{1}{2}[arc\ BC]](https://tex.z-dn.net/?f=m%5Cangle%20BDC%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20BC%5D)
substitute the values
![m\angle BDC=\frac{1}{2}[60\°]](https://tex.z-dn.net/?f=m%5Cangle%20BDC%3D%5Cfrac%7B1%7D%7B2%7D%5B60%5C%C2%B0%5D)
therefore
The triangle DBC is a right triangle ---> 60°-30°-90°
step 4
Find the measure of BC
we know that
In the right triangle DBC
substitute the values
2+42/-2-5(-3) Multiply -5 and -3.
2+42/-2+15 Now, add the non fraction numbers. (2 and 15)
17+42/-2 Here, you want to reduce the fraction if you can, and yes, you can. Find the divisible number that is the lowest one that both can be evenly divided by. For this problem, it’s -2.
17-21. And from here, its easy. Just a simple subtraction problem.
-4 should be the correct answer.
This would be 11.83 I hope you pass your test/quiz :)
1.) <1=25°, <2=155°, <4=155°
Answer:
71.123 mph ≤ μ ≤ 77.277 mph
Step-by-step explanation:
Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:
≤ μ ≤ 
Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and 
is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.
the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.
So, if we replace m by 74.2, s by 5.3083, n by 10 and 
by 1.8331, we get that the 90% confidence interval for the mean speed is:
≤ μ ≤ 
74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077
71.123 ≤ μ ≤ 77.277
