Answer:
The claim that the scores of UT students are less than the US average is wrong
Step-by-step explanation:
Given : Sample size = 64
Standard deviation = 112
Mean = 505
Average score = 477
To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.
Solution:
Sample size = 64
n > 30
So we will use z test
Formula :
Refer the z table for p value
p value = 0.9772
α=0.05
p value > α
So, we accept the null hypothesis
Hence The claim that the scores of UT students are less than the US average is wrong
Given =
n=6,
r=1/4
a=2
Sn = a+ar+ar²+ar³+ar⁴+ar⁵= 2+0.5+0.125+0.03125+0.0078125+0.00195~ 2.67
There are 2 possibilities for where A can be: one where C is 30° (and A is 60°) and another where C is 60° (and A is 30°). Since it's not specified, we can find both.
30°:
drawing the triangle on a graph, you can see that point C is 4 units above point B, so we know that one side of the triangle is 4. Once we find the other "leg" of the triangle (the one that's parallel to the x-axis), we can just add that value to B to find the x coordinate of A.
If angle C is 30°, using the side ratios of a 30-60-90 triangle, that side is "a√3", and the side we're looking for is a. So, to find a, we just divide 4 by √3. In that case, point A is 4/(√3) units to the right of -2√3. We can rationalize 4/(√3) like this:
(4√3)/3
and then add that to 2√3:
(4√3)/3 + -2√3
(4√3)/3 + (-6√3)/3 = (-2√3)/3
We know that the x-coordinate of A is (-2√3)/3, and the y-coordinate is -1 because B is a right angle and we're just moving horizontally. So, if C is 30° and A is 60°, point A is at ((-2√3)/3, -1).
60°:
in this case, the leg we know is "a" and the leg we're looking for is "a√3". So, we can multiply 4 by √3 to get the distance from B:
4 x √3 = 4√3
4√3 + -2√3 = 2√3
So the x-coordinate of A here is 2√3, and the y-coordinate is still -1: (2√3, -1).
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Answer:
A. drag it to the right
b. drag it to the left
Step-by-step explanation: