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Andrew [12]
3 years ago
5

Hi pls answer itd be really nice

Mathematics
1 answer:
emmainna [20.7K]3 years ago
4 0

Answer:

-12/5×4/3

(because if anything is in divide if we have to convert it in multiply So we have to reciprocal it

therefore option A is correct -16/5

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What is the mean absolute deviation of 275, 252, 248, 240, 240, 225, 215, 214, 189, 184
Sati [7]
Hi , if we want to find the M.A.D of the numbers shown , then we have to add them .
275+252+248+240+240+225+215+214+189+184=2,282
Then we have to divide the number by the total of numbers that we have.
2,282 ÷ 10 =228.2
Then we have to subtract 228.2 by each number that we have.
228.2-275=-46.8
228.2 - 252=-23.8
228.2 - 248=-19.8
228.2 - 240=-11.8
228.2-240=-11.8
228.2-225=3.2
228.2-215=13.2
228.2-214=14.2
228.2-189=39.2
228.2-184=44.2
Then we add those numbers , that gives us 0 , if you don't work with negatives numbers then your answer would be 228 , finally we divide that number by the total of numbers that we have , your answer is 22.8 , if you do work with negative numbers then your answer is 0.
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Viktor [21]
A. If it has a 60% mark up price of $120 that would be $60 so the total would be $180. B. If there selling the bike at what the store is selling it for it would be $126 because 30% of $180 is $54 so subtract $54 from $180 and you will get $126 C. The store would make a total profit of $6.
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Answer should be A, I hope this helps! Btw you shouldn't be asking about something thats on a test haha but thats fine! We all do it sometimes! 
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Need help with this trigonometry word problem
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Answer:

\huge \boxed{ \boxed{ \sf  6\sqrt{3}  \: or \: 10.4}}

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • trigonometry
  • PEMDAS
<h3>let's solve:</h3>

to find <u>how far the ladder is from the foot of the </u><u>wall </u>

we will use tan function because we are given opposite and angle

we need to find adjacent

\quad \tan( \theta)  =  \dfrac{opp} {adj}

let adjacent be BC

\boxed{ \red{ \boxed{accoding \: to \: the \: question : }}}

\quad \:  \tan(  {60}^{ \circ} )  =  \dfrac{6}{BC}

we will use a little bit algebra to figure out BC

  1. \sf substitute \: the \: value \: of \:  \tan(  {60}^{ \circ} )   \: i.e \:  \sqrt{3}  :  \\  \sqrt{3}  =  \frac{BC}{6}
  2. \sf cross \: multiplication :  \\  \therefore \: BC =  6\sqrt{3}  \: or \: 10.4

\text{And we are done!}

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