Using logarithms property of log(x)+log(y)=log(xy)
so here, you can sum the equation to;
so you can simply say that;
and by multiplying (x+6)*(x-6)
and as you know also that;
is same as
so you can simply state it as;
And you can check your work by substituting with 10 instead of x in the original function.
Hope this helps!
Answer:
1/4: 1s/4
7/8: 7x/s
Step-by-step explanation:
To answer this problem, please subtract 5 1/3 mile from 12 1/3 mile.
That'd be
12 1/3
- 5 1/3
-----------------
7 mi
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:
For z I don't know what to do but thats for x, maybe someone else could find z for you :)
Step-by-step explanation:
5x - 89 = 61
+89 +89
----------------------
5x = 50
---- -----
5 5
x = 10
