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3 years ago

Any large random sample from a group of people will be representative of the entire group.true false

Mathematics

1 answer:

ddd [48]3 years ago

7 0

Answer:

true

Step-by-step explanation:

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A diagonal of a parellelogram measures 37 inches. The sides measure 35 inches and 1 foot. Is the parellelogram a rectangle? Expl

Monica [59]

Answer:

Yes, it is a rectangle

Step-by-step explanation:

12 ^2 +35^2 =37^2

3 0

4 years ago

Which statement correctly compares the function shown on this graph with the function y = -x+ 4?

s344n2d4d5 [400]

Answer:

I think the answer for this question is C

6 0

3 years ago

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Last year, Edward spent a total of $626.14 on his power bill. Rounding the total bill to the nearest tens place, approximate how

zlopas [31]

Answer:

He spent around $ 52.17 in each month.

Step-by-step explanation:

Last year the power bill price = $ 626.14 ≈ $ 626

We know that in a year 12 months. so power bill of each month in last year is calculated as follows

$Power \ bill \ of \ each \ month = \frac{Total \ bill \ of \ the \ year}{12}$

$Power \ bill \ of \ each \ month = \frac{626}{12}$

$Power \ bill \ of \ each \ month =$ $ 52.17

Therefore he spent around $ 52.17 in each month.

7 0

4 years ago

Find the slope of the line y=5x+4

Tatiana [17]

Answer:

Slope = 5

Step-by-step explanation:

y = mx + b

m = slope

Slope = 5

6 0

3 years ago

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Which expression is a sum of cubes?

erik [133]

we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

Let's verify each case to determine the solution

case A) $-64x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-64=-4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case A) is not a sum of cubes

case B) $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

case C) $32x^{6} y^{12} +125x^{9} y^{3}$

we know that

$32=2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

case A) $64x^{6} y^{12} +125x^{9} y^{3}$

we know that

$64=4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

$4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}$

$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

the answer is

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

6 0

3 years ago

Read 2 more answers