Answer:
Step-by-step explanation:
Given that the observed frequencies for the outcomes as follows:
To check this we can use chi square goodness of fit test.
(Two tailed test at 5% significance level)
Assuming equally likely expected observations are found out and then chi square is calculated as (0-E)^2/E
Df = 6-1 =5
Outcome Frequency Expected frequency (Obs-exp)^2/Exp
1 36 34.83333333 0.03907496
2 30 34.83333333 0.670653907
3 41 34.83333333 1.091706539
4 40 34.83333333 0.766347687
5 23 34.83333333 4.019936204
6 39 34.83333333 0.498405104
209 209 7.086124402
p value =0.214
Since p >alpha, we accept null hypothesis
It appears that the loaded die does not behave differently than a fair die at 5% level of significance
To solve these problems, you usually have a favor the numerator. For example, on #7, label your terms, 1 from m^2 is A, -6m from -6 is B and your constant, 8, is C. Next you have to find out what two numbers can be multiplied to give you for C term but can also add together to give you your B term. For this specific problem, -4 and -2 gives you 8 and also adds to be -6. After that, you put your -4 and -2 into their own separate parentheses along with the M from M^2. It should look this like: (m-4)(m-2)/(m-2). Now you notice that the binomial on the bottom and one of the binomials on top is the exact same, so you would cancel that out and your answer would be (m-4).
Answer:
when 
.
Step-by-step explanation:
Given: 
To find: Value of 
, substituting the known value of 
We have,
Putting in equation 
, we get
Hence, the value of is 
, when the value of is 
.
Answer:
we have, 1953125=5⁹, so it cannot be a perfect square. If the last digit of a given number is 5, then the last three digits must be perfect squares, 025 or 225 or 625. Otherwise, that number cannot be a perfect square. And as 125 is not a perfect square, so no number ending with 125 can be a perfect square
