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Luba_88 [7]
3 years ago
6

Cubes and cubes roots 27000​

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

lcm for cuberoot

that is 2*2*2*5*5*5*3*3*3

make pairs of three common numbers and write them as one.

2*5*3=30

ans =30

check 30*30*30=27000

cube

multiply 27000 three times with itself

27000*27000*27000

=19683000000000.0

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1.33333333333333333333333333333333333333333333333
or 1.33
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It costs luis 5$ to park his car at a parking meter for 2 hours. What is the price to park for 1 hour? Draw a bar model or write
Aliun [14]

Answer:

$2.50

Step-by-step explanation:

If it costs $5 for two hours and one hour is half of two hours, you take half of $5 and use it as your answer, so your answer is $2.50

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3 years ago
Helppp me plsssssssss<br><br>​
Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

7 0
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krok68 [10]

Answer:

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Step-by-step explanation:

We can use either angle, but I'm going to use the one on the bottom. So, in order to find x, we need to use tangent. One side we know is the adjacent, and the side we don't know is the opposite, therefore we need tangent. Here's the equation:

tan(45)=\frac{x}{\sqrt{2}}

Obviously, we can't have a root in our denominator, so we need to get rid of it somehow. Here's how:

We multiply the denominator of the fraction by \sqrt{2}. \sqrt{2} multiplied by itself is simply 2. Try it! We also want to multiply the numerator by \sqrt{2}, but there isn't really a number we can use with that, so we'll just add it to the side. The equation you have now is:

tan(45)=\frac{x\sqrt{2} }{2}

Let's try to work this out now. Since the denominator is 2, we have to multiply both sides by it to find x.

\frac{x\sqrt{2} }{2*2} =x\sqrt{2}

tan(45)*2=2

We can plug 2 in for the x in the numerator now:

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2 and 2 cancel out, so you get 1 in both the numerator and denominator. That's how we get our answer of x=\sqrt{2}

Also, because this is a 45-45-90 triangle, you don't really have to do all that work. If it's a 45-45-90 triangle, both legs should be the same length. :)

7 0
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S_A_V [24]

Answer:

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Step-by-step explanation:

(7^2)^4= n^8

We know that a^b^c = a^(b*c)

7^(2*4) = n^8

7^8 = n^8

Since the exponents are the same, the bases must be the same

n=7

4 0
2 years ago
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