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Trava [24]
3 years ago
10

The surface area of three dimonsensed face is the sum of all the the surface areas of each of the face given below are the nets

of some solid derive the formula for calculating its curved surface area and total surface area
Please answer the question it's very urgent​

Mathematics
1 answer:
Ann [662]3 years ago
6 0

Answer:

cylinder:

curved sa = 2πrh

=> rectangle, breath = h, width = perimeter of circle =  2πr

total sa = 2πrh+2πr^2

=> curved sa + area of 2 circles

cone:

curved sa = πrl

=> treated as a triangle, base = circumference of circle, height = slant height

total sa = πr^2+πrl

=> curved sa + area of base (circle)

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melamori03 [73]
The answer to this question is z=-5 
6 0
3 years ago
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(SAT Prep) How many distinct triangles are for which the lengths of the sides are 5,9 and n, where n is an integer and 10
Nana76 [90]
The answer is C because you can have a triangle with 5,9,n   5,10,n   9,10,n   and 5,10,9

The reason why is because the sum of two sides is always larger than the value of the third side.

I hope that this helps.

7 0
3 years ago
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Please hurry I need the answer
timama [110]

Answer:

4.12

Step-by-step explanation:

A^2+b^2=c^2 so 8^2+b^2=9^2 so 64+b^2=81 so b^2=17 so b=4.12

7 0
2 years ago
Which of these points are on a line that has a slope of -3?
Ira Lisetskai [31]
Simple....

using the formula \frac{ y_{1}- y_{2}  }{ x_{1}- x_{2}  } to find slope....

1.)  (9,2)(4,3)

\frac{2-3}{9-4}= -\frac{1}{5} (No)

2.)  (9,2)(3,4)

\frac{2-4}{9-3} = -\frac{2}{6} =- \frac{1}{3} (No)

3.)  (2,9)(3,4)

\frac{9-4}{2-3} = \frac{5}{-1} =-5 (No)

4.)  (2,9)(4,3)

\frac{9-3}{2-4} = \frac{6}{-2} =-3 (YES!)

Thus, your answer.
3 0
3 years ago
Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
Zarrin [17]

Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population​ mean is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

             \mu = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.878 < t_1_8 < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 22.4 -2.878 \times {\frac{3.8}{\sqrt{19} } , 22.4 +2.878 \times {\frac{3.8}{\sqrt{19} } ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

6 0
3 years ago
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