Answer:
1 The ratio of the measure of the central angle to the measure of the entire circle is StartFraction 5 Over 2 pi EndFraction
3 The area of the sector is 250 units².
5 The area of the sector is more than half of the circle’s area
Answer:
Step-by-step explanation:
Type I error occurs if the null hypothesis is rejected when in the real sense, it is actually true.
Type II error occurs if the null hypothesis is not rejected when in the real sense, it is actually false.
The null hypothesis is
H0: μ = 9.6 hours
The alternative hypothesis is
Ha: μ > 9.6
Suppose that the results of the sample lead to non rejection of the null hypothesis if in fact the mean running time has increased it means that
a) the conclusion is a type I error because the null hypothesis was of rejected even though it was false
Answer:
Julie weighs 35 lbs, Ted weighs 70 lbs, Mike weighs 105 lbs.
Step-by-step explanation:
Let the weight of Julie be ![x](https://tex.z-dn.net/?f=x)
Given:
Ted weighs twice as much as Julie
Weight of Ted = ![2x](https://tex.z-dn.net/?f=2x)
Mike weighs three times as much as Julie.
Weight of Mike = ![3x](https://tex.z-dn.net/?f=3x)
Together, Ted, Mike, and Julie weigh 210 lbs.
![x+2x+3x=210](https://tex.z-dn.net/?f=x%2B2x%2B3x%3D210)
Solving above we get;
![6x=210\\x=\frac{210}{6}=35\ lbs](https://tex.z-dn.net/?f=6x%3D210%5C%5Cx%3D%5Cfrac%7B210%7D%7B6%7D%3D35%5C%20lbs)
Hence Julie weight is 35 lbs.
Weight of Ted = ![2x=2\times 35 = 70 \ lbs](https://tex.z-dn.net/?f=2x%3D2%5Ctimes%2035%20%3D%2070%20%5C%20lbs)
Hence Ted weight is 70 lbs.
Weight of Mike = ![3x=3\times 35 = 105 \ lbs](https://tex.z-dn.net/?f=3x%3D3%5Ctimes%2035%20%3D%20105%20%5C%20lbs)
Hence Mike weight is 105 lbs.