Answer:
D) 678.58
Step-by-step explanation:
just plug it into the formula which is V=πr2h
Answer:
First, we need to know when the function is increasing, and when is decreasing.
When a function is increasing, it's because to higher <em>x-values </em>belong higher <em>y-values. </em>On the other hand, a function is decreasing when to higher <em>x-values </em>belong lower <em>y-values. </em>The effect in the graph will be a upwards direction of the function when is increasing, and a downwards direction when is decreasing.
So, according to the given graph, we see that between 0 and 4 is increasing, the higher x-values are, higher y-values are. Between 4 and 6 is decreasing, is downwards. Between 6 and 8 is increasing. Between 8 and 10 is decreasing. In finally, between 10 and 14 is neither decreasing or increasing, it remains horizontal.
All these interpretations we can expressed using math language, specifically inequalities:
- 0 < x < 4: increasing.
- 4 < x < 6: decreasing.
- 6 < x < 8: increasing.
- 8 < x < 10: decreasing.
- 10 < x < 14: neither increasing or decreasing.
Answer:
6/250, 9/375, etc...
Step-by-step explanation:
multiply(GIANT ONE)
3/125 x 2/2 = 6/250
3/125 x 3/3 = 9/375
Answer:
The probability of founding exactly one defective item in the sample is P=0.275.
The mean and variance of defective components in the sample are:
![\mu=0.375\\\\\sigma^2=0.347](https://tex.z-dn.net/?f=%5Cmu%3D0.375%5C%5C%5C%5C%5Csigma%5E2%3D0.347)
Step-by-step explanation:
In the case we have a lot with 3 defectives components, the proportion of defectives is:
![p=3/40=0.075](https://tex.z-dn.net/?f=p%3D3%2F40%3D0.075)
a) The number of defectives components in the 5-components sample will follow a binomial distribution B(5,0.075).
The probability of having one defective in the sample is:
![P(k=5)=\binom{5}{1}p^1(1-p)^4=5*0.075*0.925^4=0.275](https://tex.z-dn.net/?f=P%28k%3D5%29%3D%5Cbinom%7B5%7D%7B1%7Dp%5E1%281-p%29%5E4%3D5%2A0.075%2A0.925%5E4%3D0.275)
b) The mean and variance of defective components in the sample is:
![\mu=np=5*0.075=0.375\\\\\sigma^2=npq=5*0.075*0.925=0.347](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D5%2A0.075%3D0.375%5C%5C%5C%5C%5Csigma%5E2%3Dnpq%3D5%2A0.075%2A0.925%3D0.347)
The Chebyschev's inequality established:
![P(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2}](https://tex.z-dn.net/?f=P%28%7CX-%5Cmu%7C%5Cgeq%20k%5Csigma%29%5Cleq%20%5Cfrac%7B1%7D%7Bk%5E2%7D)
Answer:
Step-by-step explanation:
3(0.3x + 1.3) = 2(0.4x - 0.85)
0.9x + 3.9 = 0.8x - 1.70
0.9x - 0.8x = - 1.70 - 3.9
0.1x = - 5.6
x = - 5.6 / 0.1
x = - 56