Guys please help me out these
2 answers:
Answer:
l (PM): y = -2/3x + 2
y-coordinate: -1
Step-by-step explanation:
B (x, y): (x+1)/2 = 3 x = 5
(y + -2) / 2 = 1 y = 4 ... B (5,4)
slope of AB = (4- -2)/(5-1) = 3/2
PM bisect AB, PM⊥ AB
slope of AB = - 2/3
equation of PM : (y-1)/(x-3) = -2/3
y-1=-2/3x+2
<u>y = -2/3x+3</u>
PA = PB = radius P (6,k)
PA² = PB² ... (6-1)²+(k- -2)² = (6-5)²+(k-4)²
25+k²+4k+4 = 1+k²-8k+16 ... -K² both side
12k = -12 <u>k = -1</u>
P (6 , -1) r = √(6-1)²+(k- -2)² = √(6-1)²+(-1- -2)² = √26
equation of circle: (x-h)² + (y-k)² = r² .. h=6, k=-1, r=√26
I'm sure you can figure out the equation by yourself... :)
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The result of the respective questions are:
- This chi-square test only takes into consideration one variable.
- The type of chi-square test this is is a Goodness of Fit
- df= 3
- NO
a)
This chi-square test only takes into consideration one variable.
b)
The type of chi-square test this is, is a Goodness of Fit
To test the hypothesis, we must determine whether the actual data conform to the assumed distribution.
The "Goodness-of-Fit" test is a statistical hypothesis test that determines how well the data that was seen resembles the data that was predicted.
c)
Parameter
n = 4
Therefore
Degrees of freedom
df= n - 1
df= 4 - 1
df= 3
d)
In conclusion
Parameters
df = 3
Hence
Critical value = 7.814728
Test statistic = 6.6
Test statistic < Critical value, .
NO, the result of this test is not statistically significant.
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