Solution :
The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.
Null hypothesis, 
That is the loaded die behaves as a fair die.
Alternative hypothesis, 
: loaded die behave differently than the fair die.
Number of attempts , n = 200
Expected frequency, 
Test statistics, 
≈ 5.8
Degrees of freedom, df = n - 1
= 6 - 1
= 5
Level of significance, α = 0.10
At α = 0.10 with df = 5, the critical value from the chi square table
= 9.236
Thus the critical value is 
![$P \text{ value} = P[x^2_{df} \geq x^2]$](https://tex.z-dn.net/?f=%24P%20%5Ctext%7B%20value%7D%20%3D%20P%5Bx%5E2_%7Bdf%7D%20%5Cgeq%20x%5E2%5D%24)
![$=P[x^2_5\geq 5.80]$](https://tex.z-dn.net/?f=%24%3DP%5Bx%5E2_5%5Cgeq%205.80%5D%24)
= chi dist (5.80, 5)
= 0.3262
Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject 
at 10% LOS.
Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.
Answer:
Step-by-step explanation:
The given expression is:
-3(6f - 12) = 36 - 18f
To prove that L.H.S=R.H.S:
Multiply -3 (6f-12)
-3(6f)-3(-12)
=-18f+36
=36-18f
Hence it is proved that L.H.S = R.H.S....
You would set it up as a fraction putting 35 over 100 and 98 over x then cross multiply to get 280
Answer:
<em>Solution: x=5, y=4</em>
Step-by-step explanation:
<u>System of Equations</u>
We need to solve:
- 5x + 9y = 11
x - 3y = - 7
There are several ways to solve a system of linear equations. We'll take advantage of the second equation since it has the x with coefficient 1 and solve it for x:
x = - 7 + 3y
Now replace it into the first equation:
- 5(- 7 + 3y) + 9y = 11
Operate:
35 - 15y + 9y = 11
Simplify:
35 - 6y = 11
Rearrange:
- 6y = 11 - 35 = -24
Solve:
y = -24 / -6
y = 4
Finally, since
x = - 7 + 3y
x = - 7 + 3*4
x = - 7 + 12
x = 5
Solution: x=5, y=4
Answer:
−7x^2+4x+6
Step-by-step explanation:
HOPE THIS HELPS!!!!
