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Zepler [3.9K]
3 years ago
11

Find the surface area of each figure.

Mathematics
1 answer:
timurjin [86]3 years ago
6 0

Answer:

For a cylinder of radius R and height H, the surface area is given by:

A = 4*pi*R^2 + H*(2*pi*R)

Where pi = 3.14

We can just use that formula for each one of the given cylinders.

1) We can see that the diameter is 10 yd, and the radius is half of the diameter, then the radius is:

R = 10yd/2 = 5yd

And the height is 3yd, then H = 3yd

Replacing these in the area equation, we get:

A = 4*3.14*(5 yd)^2 + 3yd*(2*3.14*5yd) = 408.2 yd^2

2) Here we can see that the diameter is 24 cm, then the radius is:

R = 24cm/2 = 12cm

And the height is H = 10cm

Then the area is:

A = 4*3.14*(12 cm)^2 + 10cm*(2*3.14*12cm) = 2,562.24 cm^2

3) In this case we have a radius equal to 12 cm, and a height equal to 7 cm, then the area is:

A = 4*3.14*(12 cm)^2 + 7cm*(2*3.14*12cm) = 2,336.16cm^2

4) Here is hard to see the measures, I think that here we have:

diameter = 8m

Then R = 8m/2 = 4m

And the height is also 8m, H = 8m

Then the area is:

A = 4*3.14*(4 m)^2 + 8 m*(2*3.14*4m) =401.92 m^2

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Scores on a certain IQ test are knownto have a mean of 100. A random sample of 43 students attend a series of coaching classes b
NeX [460]

Answer:

Part 1

Type II error

Part 2

No ; is not ; true

Step-by-step explanation:

Data provided in the question

Mean = 100

The Random sample is taken = 43 students

Based on the given information, the conclusion is as follows

Part 1

Since it is mentioned that the classes are successful which is same treated as a null rejection and at the same time it also accepts the alternate hypothesis

Based on this, it is a failure to deny or reject the false null that represents type II error

Part 2

And if the classes are not successful so we can make successful by making type I error and at the same time type II error is not possible

Therefore no type II error is not possible and when the null hypothesis is true the classes are not successful

6 0
3 years ago
PLZ HELP!!!!!!! <br><br> Evaluate.<br><br> 43−4÷2⋅5<br><br> 20<br> 40<br> 54<br> 150
vitfil [10]
Isnt it 33 ////??? thats what i got.
6 0
3 years ago
Read 2 more answers
Find the common factor and factor out <br> 9ab^2-3abc
dsp73
Lets go step by step and factor out each as we go

first lets factor out a 3 since 3 is common in 9 and 3 

3(3ab^2 - abc) 

now lets factor out a since its common in both 

3a(3b^2 - bc) 

now lets factor out a b since its common in both 

3ab(b - c) 

we can't factor anything else out so we are done. 

4 0
3 years ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
Maru [420]

Parameterize S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k

with 0\le u\le1 and 0\le v\le1.

Take the normal vector to S to be

\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}

6 0
3 years ago
Mandy has $2.25 credit on her mobile phone. It costs $0.10 to send a text message. What is the largest amount of text messages s
Brilliant_brown [7]

Answer:

22.5

Step-by-step explanation:

since 2.25 divided by 0.10 is 22.5 so she can message 22.5 friends

8 0
2 years ago
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