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kodGreya [7K]
3 years ago
14

The question says “The image shows the front of Bob’s barn.Bob plans on painting the front of his barn,not including the doors o

f the barn.How many square feet of surface will Bob need to paint?

Mathematics
1 answer:
AleksandrR [38]3 years ago
6 0

Bob need to paint 205 square feet.

Solution:

Length of the door = 7 ft

Breadth of the door = 5 ft

Area of the door = length × breadth

                            = 7 × 5

                            = 35

Area of the door = 35 ft²

The given image is a trapezoid.

Top base = 12 ft

Bottom base = 20 ft

Area of the trapezoid = \frac{1}{2} \times\text{sum of the parallel sides}\times\text{height}

                                    $=\frac{1}{2}\times(12+20)}\times15

                                    $=\frac{1}{2}\times 32\times15

                                    = 240

Area of the trapezoid =  240 ft²

Surface area to paint = Area of the trapezoid – Area of the door

                                    = 240 ft² – 35 ft²

                                    = 205 ft²

Hence Bob need to paint 205 square feet.

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Suppose that we are testing a coin to see if it is fair, so our hypotheses are: H0: p = 0.5 vs Ha: p ≠ 0.5. In each of (a) and (
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Answer:

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a. We get 56 heads out of 100 tosses.

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x = 56

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\widehat{p}=\frac{56}{100}

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Formula of test statistic =\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

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refer the z table for p value

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a.  We get 560 heads out of 1000 tosses.

We will use one sample proportion test  

x = 560

n = 1000

\widehat{p}=\frac{x}{n}

\widehat{p}=\frac{560}{1000}

\widehat{p}=0.56

Formula of test statistic =\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

                                       =\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}

                                       =3.794

refer the z table for p value

p value = .000148

p value of part B is less than Part A because part B have 10 times the number the tosses.

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3 years ago
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