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alukav5142 [94]
3 years ago
9

What is the surface area of the cylinder?

Mathematics
1 answer:
loris [4]3 years ago
7 0
Each sides are equal so the answer 400 cubic in
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Solve for s.<br> S + 3s - 4 = 16<br> S =<br> Submit
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The answer is 5 because 16+4=20 and 3s + s = 4s so you’re left with 4s = 20 and I know that 4 times 5 is 20.
ANSWER: S=5
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Need help ASAP! Please show work
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The answer is f because if you find the average between 78 and 62 you get 70
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3 years ago
I still don't get how to do this: Solve for t then simplify your answer. -6/5t=9
stepladder [879]

Answer:

t= \frac{-15}{2}

Step-by-step explanation:

Multiply both sides by 5/(-6).

\frac{5}{-6} *\frac{-6}{5}=\frac{5}{-6}*9

ik the doesnt rlly make any sense but hope this helps

5 0
3 years ago
Read 2 more answers
Find the area of the triangle whose vertices are (2, 3); (-1, 0); (2, -4)...
Neko [114]

Answer:

Area = 14.5

Step-by-step explanation:

Let's label the given coordinates A, B, C;

Thus;

A = (2, 3)

B = (-1, 0)

C = (2, -4)

From the coordinate geometry formula, the formula for area of a triangle with 3 vertices is;

Area = [Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By)]/2

Area = [2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - (-4))]/2

Area = 29/2

Area = 14.5

8 0
3 years ago
Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are relat
anzhelika [568]

Answer:

(c) Probability that a failure is due to loose keys = 0.2376

(d) Probability that a failure is due to improperly connected or poorly welded wires = 0.078

Step-by-step explanation:

The Whole probability scenario is given for Computer Keyboard failures.

(a) Let F be the event of failure due to faulty electrical connects, P(F) = 0.12

 M be the event of failure due to mechanical defects, P(M) = 0.88

 LK be the event of mechanical defect due to loose keys, P(LK/M) = 0.27

 IA be the event of mechanical defect due to improper assembly, P(IA/M)   =0.73

 DW be the event of electrical connects due to defective wires,P(DW/F) = 0.35

 IC be the event of electrical connects due to improper connections,

  P(IC/F) = 0.13 .

PWW be the event of electrical connects due to poorly welded wires,

  P(PWW/F) = 0.52

(b)                                     <u> </u><u>Keyboard failures</u>

<h2>                              /               \</h2>

           <u> </u><u> Faulty electrical connects   </u>            <u>Mechanical Defects </u>          

                      P(F) = 0.12                                             P(M) = 0.88

<h2>        /            |             \                  /            \</h2>

<u><em>Defective wires</em></u>  <u><em>Improper</em></u>        <u><em>Poorly</em></u>                  <u><em>Loose Keys</em></u>      <u><em>Improper</em></u><em> </em>

P(DW/F)=0.35   <u><em>Connections</em></u>   <u><em>Welded wires</em></u>      P(LK/M)=0.27   <em> </em><u><em>Assembly</em></u>

                           P(IC/F)=0.13     P(PWW/F)=0.52                            P(IA/M)=0.73              

This is the required tree diagram.

(c) Probability that a failure is due to loose keys is given by:

  P(LK) =P(LK/M) * P(M) {This means mechanical failure is due to loose  

                                               keys}

    P(LK) = 0.27 * 0.88 = 0.2376 .

(d) Probability that a failure is due to improperly connected or poorly welded

     wires is given by P(IC \bigcup PWW) ;

 P(IC \bigcup PWW) = P(IC) + P(PWW) - P(IC \bigcap PWW) { Here P(IC \bigcap PWW) = 0 }

 P(IC) = P(IC/F) * P(F)  = 0.13 * 0.12 = 0.0156

 P(PWW) = P(PWW/F) * P(F) = 0.52 * 0.13 = 0.0676

Therefore, P(IC \bigcup PWW) = 0.0156 + 0.0676 - 0 = 0.078 .

8 0
3 years ago
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