Answer:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
Step-by-step explanation:
Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.
We find mean = 11.015
Sample std deviation = 3.157
a)
(Right tailed test)
Mean difference /std error = test statistic
p value =0.174
Since p >0.01, our alpha, fail to reject H0
Conclusion:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
0.53(3) + 5 = <span>6.59
</span><span>6.59 is the answer.
</span>
8x-4=-32 so thats an equation that you can use
I think that the correct function in your question should be y = 60x⁻¹ + 0.625x.
With that said, the answer is A. <span>0 < x < ∞
By the process of trial and error of the choices, we can see that this choice satisfies the equation. Take note that y models moose population, so y should always be positive. </span>
Point slope form is y-(-4) = 1/2(x-(-2)
y+4 =1/2 (x+2)
now the standard form 2y+8=x+2
so -x+2y=-6 is the equation for line p in standard form