To find the scale factor, locate two corresponding sides, one on each figure. Write the ratio of one length to the other to find the scale factor from one figure to the other. In this example, the scale factor from the blue figure to the red figure is 1.6 : 3.2, or 1 : 2.
Answer: 112
Step-by-step explanation: can i get brainliest
Answer:
Step-by-step explanation:
false,it is of degree 2 exactly
Answer:
∠ 1 = 120°
Step-by-step explanation:
5x + 15 and 3x - 3 are same- side interior angles and sum to 180° , then
5x + 15 + 3x - 3 = 180 , that is
8x + 12 = 180 ( subtract 12 from both sides )
8x = 168 ( divide both sides by 8 )
x = 21
Then
5x + 15 = 5(21) + 15 = 105 + 15 = 120
∠ 1 and 5x + 15 are corresponding angles and are congruent , so
∠ 1 = 120°
Answer:
The null hypothesis ![H_o : \mu = 3.2 \ mg/dl](https://tex.z-dn.net/?f=H_o%20%3A%20%20%5Cmu%20%3D%203.2%20%5C%20mg%2Fdl)
The alternative hypothesis ![H_a : \mu > 3.2 \ mg/dl](https://tex.z-dn.net/?f=H_a%20%3A%20%20%5Cmu%20%3E%203.2%20%5C%20mg%2Fdl)
Step-by-step explanation:
From the question we are told that
The mean is ![\mu = 3.2 \ mg/dl](https://tex.z-dn.net/?f=%5Cmu%20%3D%203.2%20%5C%20%20mg%2Fdl)
The standard deviation is ![\sigma = 1.5 mg/dl](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%201.5%20mg%2Fdl)
The sample size is n = 25
The sample mean is ![\= x = 3.75 \ mg /dl](https://tex.z-dn.net/?f=%5C%3D%20x%20%3D%203.75%20%5C%20%20mg%20%2Fdl)
Generally
The null hypothesis is : That the mean of the Arsenic blood concentrations in healthy individuals is ![\mu = 3.2 \ mg/dl](https://tex.z-dn.net/?f=%5Cmu%20%3D%203.2%20%5C%20%20mg%2Fdl)
i.e
The null hypothesis ![H_o : \mu = 3.2 \ mg/dl](https://tex.z-dn.net/?f=H_o%20%3A%20%20%5Cmu%20%3D%203.2%20%5C%20mg%2Fdl)
The alternative hypothesis is : That the mean of the Arsenic blood concentrations in healthy individuals is greater than ![\mu = 3.2 \ mg/dl](https://tex.z-dn.net/?f=%5Cmu%20%3D%203.2%20%5C%20%20mg%2Fdl)
The alternative hypothesis ![H_a : \mu > 3.2 \ mg/dl](https://tex.z-dn.net/?f=H_a%20%3A%20%20%5Cmu%20%3E%203.2%20%5C%20mg%2Fdl)