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2 years ago

Find the area of the composite figure.

Mathematics

1 answer:

Bad White [126]2 years ago

7 0

Answer:

Area = 139.27 m²

Step-by-step explanation:

Area of the composite figure = area of rectangle + area of semicircle

= (L*W) + ½(πr²)

L = 10 m

W = 10 m

r = ½ of 10 = 5 m

Plug in the values

Area = (10*10) + ½(π*5²)

Area = 100 + 39.27

Area = 139.27 m²

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Find the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)

Eva8 [605]

The coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4) is $(\frac{-17}{5} ,\frac{2}{5} )$

Explanation :

the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)

Apply section formula to find coordinates of point Q

$(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} )$

Ratio m:n is 2:3 and point R is (x1,y1) , point S is (x2,y2)

Substitute all the values inside the formula

$(\frac{2(2)+3(-7)}{2+3} , \frac{2(4)+3(-2)}{2+3} )\\\\(\frac{-17}{5} ,\frac{2}{5} )$

The coordinates of point Q is $(\frac{-17}{5} ,\frac{2}{5} )$

Learn more : brainly.com/question/13878373

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3 years ago

there are 3 teams of 18 employees working today company policy says that we need to have one supervisor for every 8 employees on

Vadim26 [7]

The <span> answer is 6.75. Since if you divide the number by 8 you get 6.75.</span>

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Two brothers, Kevin and Greg, each inherit S43000. Kevin invests his inheritance in a savings account with an annual return of 1

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Answer:

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Greg: After 1 year he has 43,000 * 0.053 = S2279

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A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

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Answer:

The 90% confidence interval that estimate the true proportion of students who receive financial aid is

$0.533 < p < 0.64$

$n = 1789$

Step-by-step explanation:

Considering question a

From the question we are told that

The sample size is n = 200

The number of student that receives financial aid is $k = 118$

Generally the sample proportion is

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{118}{200}$

=> $\^ p = 0.59$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

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

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }$

=> $E = 0.057$

Generally 90% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.533 < p < 0.64$

Considering question b

From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

$[\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

=> $n = 1789$

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3 years ago