A is the answer because z=(8-10.5)/1.5= -1.67
Hope this help u
Answer:
Step-by-step explanation:
Suppose that at time
we have x mold cells. We are told that after 12 minutes the amount present is double. That is a
we have 2x cells. Then, at
we have (2x)*2 = 4x. We can continue as follows
we have (4x)*2 = 8x.
we have (8x)*2 = 16x
(one hour later) we have (16x)*2 = 32x.
So after one hour from now we have 32x cells.
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:
Equation is S = 40w + 650
So S = 40(16) + 650
S = 640 + 650
S = $1,290
Step-by-step explanation:
this because each week (W) he adds $40 and the $650 is the initial amount in his savings account. So since 16 weeks went by, just multiply that by 40 and add 650 to that product.