Answer:x = 75
Step-by-step explanation:y=105 - 5 = 100 ft
z=125 ft
dz/dt=2 ft/sec
by using Pythagorean Theorem, we find x
x^2 + 100^2 = 125^2
x = 75
Sense -4 and +2 would be 6 digits away from each other, we would then have to figure out, what would 4 times x - 2 would equal "x".
We would do, 4×2 = 8.
8 - 2 = 6.
Sense 6 is the amount of how much -4 and +2 would be from each other, this would then mean that x would then equal 8.
Are you asking for the expression?
If so, you plug into the format (x - h)² + (y - k)² = r² where (h, k) is the center and r is the radius.
(x + 2.5)² + (y + 4.4)² = (7/4)²
Simplify to get your equation and answer:
(x + 2.5)² + (y + 4.4)² = 49/16
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, ![$H_0:p_1=p_1=p_3=p_4=p_5=p_6=\frac{1}{6}$](https://tex.z-dn.net/?f=%24H_0%3Ap_1%3Dp_1%3Dp_3%3Dp_4%3Dp_5%3Dp_6%3D%5Cfrac%7B1%7D%7B6%7D%24)
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, ![$E_i=np_i$](https://tex.z-dn.net/?f=%24E_i%3Dnp_i%24)
![$=200 \times \frac{1}{6} = 33.333$](https://tex.z-dn.net/?f=%24%3D200%20%5Ctimes%20%5Cfrac%7B1%7D%7B6%7D%20%3D%2033.333%24)
Test statistics, ![$x^2= \sum^6_{i=1} \frac{(O_i-E_i)^2}{E_i} $](https://tex.z-dn.net/?f=%24x%5E2%3D%20%5Csum%5E6_%7Bi%3D1%7D%20%5Cfrac%7B%28O_i-E_i%29%5E2%7D%7BE_i%7D%20%24)
![$=\frac{(28-33.333)^2}{33.333}+\frac{(29-33.333)^2}{33.333}+\frac{(40-33.333)^2}{33.333}+\frac{(41-33.333)^2}{33.333}+\frac{(28-33.333)^2}{33.333}+$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B%2828-33.333%29%5E2%7D%7B33.333%7D%2B%5Cfrac%7B%2829-33.333%29%5E2%7D%7B33.333%7D%2B%5Cfrac%7B%2840-33.333%29%5E2%7D%7B33.333%7D%2B%5Cfrac%7B%2841-33.333%29%5E2%7D%7B33.333%7D%2B%5Cfrac%7B%2828-33.333%29%5E2%7D%7B33.333%7D%2B%24)
![$\frac{(34-33.333)^2}{33.333}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%2834-33.333%29%5E2%7D%7B33.333%7D%24)
≈ 5.8
Degrees of freedom, df = n - 1
= 6 - 1
= 5
Level of significance, α = 0.10
At α = 0.10 with df = 5, the critical value from the chi square table
![$x^2_{\alpha}= \text{chi inv}(0.10,5)$](https://tex.z-dn.net/?f=%24x%5E2_%7B%5Calpha%7D%3D%20%5Ctext%7Bchi%20inv%7D%280.10%2C5%29%24)
= 9.236
Thus the critical value is ![$x_{\alpha}^2=9.236$](https://tex.z-dn.net/?f=%24x_%7B%5Calpha%7D%5E2%3D9.236%24)
![$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.