Answer:
Test statistic = -2.25
P-value = 0.0199
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 450 gram
Sample mean, 
= 441 grams
Sample size, n = 16
Alpha, α = 0.05
Sample variance = 256
First, we design the null and the alternate hypothesis
We use one-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have
Now,
Degree of freedom =
We can calculate the p-value from the table as:
P-value = 0.0199
Conclusion:
Since the p-value is smaller than the significance level we fail to accept the null hypothesis and reject it.
Thus, there is enough evidence to support the claim that the machine is under filling the bags .
Answer:
answer 60,40 i think
Step-by-step explanation:
the perimiter is the outside of an object, when finding the perimiter you add the suroundding together, when it comes to area you multiply
Answer:
g = 12.7 cm
Step-by-step explanation:
Here, we want to find the length of the side g
To do this, we shall employ the use of the cosine rule
Mathematically, by using the general form of the rule, we have it that ;
g^2 = e^2 + f^2 -2efCosG
So we have;
g^2 = (9.8)^2 + (3)^2 - 2(9.8)(3) cos 167
g^2 = 162.33
g = √162.33
g = 12.7
I think it’s D. Hope this helps
Check the picture below.
so the lunchbox is really a semi-circle with a radius of 3, sitting on top of a square of 6x6, with a lenght of 10.
the volume of the lunchbox then will be the area of the semi-circle and square times the length.
![\bf \left[ \stackrel{\textit{semi-circle's area}}{\frac{\pi 3^2}{2}}+\stackrel{\textit{square's area}}{(6\cdot 6)} \right]\stackrel{length}{10}\implies \left( \frac{9\pi }{2}+36\right)(10)](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Bsemi-circle%27s%20area%7D%7D%7B%5Cfrac%7B%5Cpi%203%5E2%7D%7B2%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bsquare%27s%20area%7D%7D%7B%286%5Ccdot%206%29%7D%20%5Cright%5D%5Cstackrel%7Blength%7D%7B10%7D%5Cimplies%20%5Cleft%28%20%20%5Cfrac%7B9%5Cpi%20%7D%7B2%7D%2B36%5Cright%29%2810%29)
