Solution :
In this case we have to use 
for the goodness of fit.
Null hypothesis is : 
Data follows the given distribution
Alternate hypothesis is : 
Data do not follow the given distribution
The level of significance, 
The test statistics is given by :
Chi square = 
Here O is the observed frequencies
E is the expected frequencies
So we have, N = number of the categories = 4
df = degrees o freedom = N - 1
= 4 - 1 = 3
Critical value = 
Calculating the table for the test statistics are given as :
Category prop. O E 
A 0.2 14 20 1.80
C 0.3 28 30 0.13
M 0.1 6 10 1.60
P 0.4 52 40 3.60
Total 1 100 100 7.13
The test statics = chi square = = 7.13
statistics = 7.13
P-value = 0.067767248
P -value >
Therefore, we do not reject the 
. There is sufficient evidence to conclude that the data follows the given distribution. So there is sufficient evidence to conclude the given claim is true.
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Call the number : x
(1/3)x + 5 = x-6
=> (1/3)x + 5 - (x-6) = 0
=> (1/3)x + 5 - x + 6 = 0
Group (1/3)x with -x, 5 with 6
=> [(1/3)x - x] + (5+6) = 0
=> (-2/3)x + 11 = 0
=> (-2/3)x = -11
=> x = -11 : (-2/3) = 33/2.
Recheck : 1/3 x 33/2 + 5 = 33/6 + 5 = 63/6 = 21/2
33/2 - 21/2 = 12/2 = 6 (21/2 is 6 less than 33/2, satisfied.)
Y = kx
15 = -3k
k = -5
so if x = 4
y = kx
y = (-5) (4) = - 20
answer
<span>y = -20 when the x = 4.</span>
