M is a letter your welcome
Okay, first we need to figure out what one would make. We need to figure out the area of the base x height. The diameter is 8 so we will cut that to 4 so we can do piR2! 3.14x16 would be 50.24. Then we mutiply that by 10 to get 502.4cm3! Okay, now we would divide 2000/502.4 to find out how many times we would have to make the containers to get to 2000. The answer is 3.98, which we will have to round up since its asks for the minimum and you cant make a container 3.98 times. So the answer is 4!
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p = 
Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]](https://tex.z-dn.net/?f=P%28p%5Cleq%200.634%29%20%3D%20%5B%5Cfrac%7Bp%20-%20P%27%7D%7B%5Csqrt%7B%5Cfrac%7BP%27q%27%7D%7Bn%7D%7D%7D%5D%5Cleq%20%5B%5Cfrac%7B0.634%20-%200.74%7D%7B%5Csqrt%7B%5Cfrac%7B0.74%5Ctimes%200.26%7D%7B60%7D%7D%7D%5D)
P![(p\leq 0.634) = P[z\leq -1.87188]](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87188%5D)
P![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87%5D%20%3D%200.0298)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.
For the answer to the question above asking Which test point holds true for 3/2y - 2x>1?
There exists a question that instead of >, the symbol used is ≥. Substitute the value of abscissas and ordinates of the points to x and y, respectively.
The answer to the question above is the first one among the given choices which is <span>A. (1/4, 1)</span>
Answer:
Step-by-step explanation:
Simplified the following system of equations into a linear equation.
Graphed (plot the points) the linear equation onto the graph.
To get the solution (2,-2) (x,y)she went positive 2 to the right and -2 down.