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elena-s [515]
3 years ago
8

[find the area please explain steps]​

Mathematics
1 answer:
RSB [31]3 years ago
4 0

Length of base of the top triangle = 23 inches

Length of height of the top triangle = 47 inches

Area of the top triangle :

=  \frac{1}{2}   \: \times \:  base  \: \times \:  height

=  \frac{1}{2}  \times 23 \times 47

=  \frac{1}{2}  \times 1081

=  \frac{1081}{2}

= \color{hotpink} \: 540.5 \: \:sq.  inches \:

Thus, area of the top triangle = 540.5 sq.inches

Length of base of the bottom triangle = 47 inches

Length of height of the bottom triangle = 46 inches

Area of the bottom triangle :

=  \frac{1}{2}   \: \times  \: base \:  \times height

=  \frac{1}{2}  \times 47 \times 46

=  \frac{1}{2}  \times 2162

= \color{hotpink}1081 \:sq. inches

Area of the bottom triangle = 1081 sq.inches

Area of the total figure :

= Area of top triangle + Area of bottom triangle

= 540.5 + 1081

= \color{hotpink}\bold{1621.5 \:  \:  sq.inches}

Therefore, the area of the figure = 1621.5 sq.inches

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Okay so every part of sugar requires 4 parts of water 

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A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th
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Answer:

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

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Sample mean = \bar{x}=\frac{\sum x}{n}

Sample mean = \bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}

Sample mean = \bar{x}=23

Sample standard deviation = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}

Sample standard deviation = \sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table t_{\frac{\alpha}{2} = 3.883

Formula of confidence interval : \bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}

Substitute the values in the formula

Confidence interval : 23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 23 -3.883 \times \frac{1.72}{\sqrt{20}} to 23 + 3.883 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

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kodGreya [7K]

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this means 2y + 6 = 30

we can rearrange this as 30 - 6 = 2y

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We can rearrange this as 24 + 3 = 3x

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