How much metal is needed to cast a cubical metal box with outer side 10 inches long if the thickness of its walls should be exa
ctly 2 inches?
1 answer:
Notice the picture below
so... a cube has all sides equal, so is just a rectangular prism with equal sides
and a rectangular prism is just 6 rectangles stacked up to each other at the edges
now, notice in the graph at the bottom, one of the rectangles with a thickness of 2inches
so... just get the volume of each, and add them up, that's how much metal is needed
so... a cube has all sides equal, so is just a rectangular prism with equal sides
and a rectangular prism is just 6 rectangles stacked up to each other at the edges
now, notice in the graph at the bottom, one of the rectangles with a thickness of 2inches
so... just get the volume of each, and add them up, that's how much metal is needed
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Answer:
P = 0.42
Step-by-step explanation:
This probability problem can be solved by building a Venn like diagram for each probability.
I say that we have two sets:
-Set A, that is the probability of receiving this virus through the email.
-Set B, that is the probability of receiving it through the webpage.
The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that = 0.17.
By email only
The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:
where A is the probability that the system receives the virus just through the email.
The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.
By webpage only
There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:
where B is the probability that the system receives the virus just through the webpage.
The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.
What is the probability that the virus does not enter the system at all?
So, we have the following probabilities.
- The virus does not enter the system: P
- The virus enters the system just by email: 23% = 0.23
- The virus enters the system just by webpage: 18% = 0.18
- The virus enters the system both by email and by the webpage: 17% = 0.17.
The sum of the probabilities is 100% = 1. So:
P + 0.23 + 0.18 + 0.17 = 1
P = 1 - 0.58
P = 0.42
There is a probability of 42% that the virus does not enter the system at all.