The <em>total surface</em> area of the three figures are 702 square feet.
<h3>How to estimate the amount of metal needed to produce a given amount of figures</h3>
According to this problem, we are required to determine the amount of metal needed to build three pieces with the dimensions presented in the figure.
In this case, we need to estimate the <em>total</em> surface area (
), in square feet, required for the production of the three pieces, that is, the product of the number of pieces (<em>n</em>), no unit, and the surface area of each piece (<em>A</em>).
Now we proceed to calculate the total surface area:
![+(5\,ft)\cdot (8\,ft)+2\cdot (3\,ft)\cdot (5\,ft) + 2\cdot (4\,ft)\cdot (2\,ft) + (3\,ft)\cdot (4\,ft) ]](https://tex.z-dn.net/?f=%2B%285%5C%2Cft%29%5Ccdot%20%288%5C%2Cft%29%2B2%5Ccdot%20%283%5C%2Cft%29%5Ccdot%20%285%5C%2Cft%29%20%2B%202%5Ccdot%20%284%5C%2Cft%29%5Ccdot%20%282%5C%2Cft%29%20%2B%20%283%5C%2Cft%29%5Ccdot%20%284%5C%2Cft%29%20%5D)
The <em>total surface</em> area of the three figures are 702 square feet. 
To learn more on surface areas, we kindly invite to check this verified question: brainly.com/question/2835293
22. $31.25 divided by 25 = $1.25 for each ticket.
23. $1.50 times 5 = $7.50 to climb the rock wall.
Answer:
my hands hurt, pls give brainilist and hope this helps
Step-by-step explanation:
Before leaving for work, Victor checks the weather report in order to decide whether to carry an umbrella. The forecast is “rain" with probability 20% and “no rain" with probability 80%. If the forecast is “rain", the probability of actually having rain on that day is 80%. On the other hand, if the forecast is “no rain", the probability of actually raining is 10%.
1. One day, Victor missed the forecast and it rained. What is the probability that the forecast was “rain"?
2. Victor misses the morning forecast with probability 0.2 on any day in the year. If he misses the forecast, Victor will flip a fair coin to decide whether to carry an umbrella. (We assume that the result of the coin flip is independent from the forecast and the weather.) On any day he sees the forecast, if it says “rain" he will always carry an umbrella, and if it says “no rain" he will not carry an umbrella. Let U be the event that “Victor is carrying an umbrella", and let N be the event that the forecast is “no rain". Are events U and N independent?
3. Victor is carrying an umbrella and it is not raining. What is the probability that he saw the forecast?
It means you multiply 6.2 x 0.48
which equals = 2.976
( ) mean multiply
