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Usimov [2.4K]
3 years ago
14

A blueprint for a building includes a rectangular room that measures 3 inches long and 5.5 inches wide. The scale for the bluepr

int says that 1 inch on the blueprint is equivalent to 10 feet in the actual building. What are the dimensions of this rectangular room in the actual building? Group of answer choices
Mathematics
2 answers:
Nataly_w [17]3 years ago
8 0

Answer:

hahajahahhahahahahahahahahahahahahha

Step-by-step explanation:

losiento mucho ❣❣☹☹✌✌☣☣☣⚠️⚠️⚠️⚠️⚠️⚠️

Elodia [21]3 years ago
4 0
<h3>Answer: 30 feet by 55 feet</h3>

======================================================

The blueprints say that 1 inch represents 10 feet. This means we can form the equation below

1 inch = 10 feet

Of course 10 feet is much larger than 1 inch, so they cannot be actually equal to one another, but it helps tie together the two quantities. And it also allows us to multiply both sides by 3 to get

3 inches = 30 feet

So 3 inches on paper means we have 30 feet in real life.

Similarly,

1 inch = 10 feet

5.5 inches = 55 feet (multiply both sides by 5.5)

The 3 inch by 5.5 inch rectangle on paper corresponds to a 30 feet by 55 feet rectangle in real life.

----------------

In short, you multiply each given dimension by 10

3 inches goes to 3*30 = 30 feet

5.5 inches goes to 5.5*10 = 55 feet

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Answer:

The estimate is  P__{hat}} \pm E  = 0.37 \pm 0.0348

Step-by-step explanation:

From the question we are told that  

    The sample size is  n =  522

    The sample proportion of students  would like to talk about school is  \r p__{hat}} =  0.37

  Given that the confidence level is  90 % then the level of significance can be mathematically evaluated as

                  \alpha  =  100 - 90

                  \alpha  =  10\%

                  \alpha  =  0.10

Next we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table, the value is  

               Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } =  1.645

Generally the margin of error can be mathematically represented as

               E =  Z_{\frac{\alpha }{2} } *  \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }

=>            E = 1.645 *  \sqrt{\frac{0.37 (1- 0.37  )}{522 } }

=>             E = 0.0348

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is  

                       P__{hat}} \pm  E

substituting values

                     0.37 \pm 0.0348

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