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3 years ago

For data at the interval level you cannot calculate meaningful differences between data entries.

Mathematics

1 answer:

Alex Ar [27]3 years ago

6 0

Answer: A. FALSE

Step-by-step explanation: The interval scale presents a measurement level for representing numerical variables in an ordered and fixed manner. Measurement are represented on the scale at equal intervals with each difference between consecutive intervals being the same. With this attribute, data represented on an interval scale gives meaningful difference between the data entries also allowing the US to know the size of the interval. Interval scales allows us to make subtraction and addition on data it's data. Examples of data represented on interval scale include, Temperature, ph and other numeric data with arbitrary zero levels

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What is 1/3 divided by 5 I’m so confused lol

Vera_Pavlovna [14]

The answer should be 0.06

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3 years ago

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Please answer this correctly without making mistakes I want ace expert and genius people to answer this correctly without making

ivanzaharov [21]

Answer:

(9³×m)3

Step-by-step explanation:

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3 years ago

Find all pairs of real numbers (a,b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

Mrrafil [7]

Answer:

$(3,0)$

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: $(u+v)^2=u^2+2uv+v^2$ or $(u-v)^2=u^2-2uv+v^2$ .

Left-hand side:

$(x-a)^2+(2x-b)^2$

$(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)$

$x^2-2ax+a^2+4x^2-4bx+b^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)$

$5x^2+(-2a-4b)x+(a^2+b^2)$

Right-hand side:

$(x-3)^2+(2x)^2$

$x^2-2(3)x+9+4x^2$

$x^2-6x+9+4x^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9$

$5x^2-6x+9$

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values $a$ and $b$ such the following are true equations:

$-2a-4b=-6$

$a^2+b^2=9$

So if we solve the system we can find the values $a$ and $b$ such that the left=right.

Let's solve the first equation for $a$ in terms of $b$ .

Add $2a$ on both sides:

$-4b=-6+2a$

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

$a^2+b^2=9$

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity $(u-v)^2=u^2-2uv+v^2$ )

Multiply both sides by 4 to clear the fractions from the problem:

$4a^2+(9-6a+a^2)=36$

Combine like terms on left hand side:

$4a^2+a^2-6a+9=36$

$5a^2-6a+9=36$

Subtract 36 on both sides:

$5a^2-6a-27=0$

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace $-6a$ with $-15a+9a$ and factor by grouping.

$5a^2-15a+9a-27=0$

$5a(a-3)+9(a-3)=0$

$(a-3)(5a+9)=0$

This implies $a-3=0$ or $5a+9=0$ .

Solving the first is easy. Just ad 3 on both sides to get: $a=3$ .

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

$5a=-9$

$a=\frac{-9}{5}$ .

So let's go back to finding $b$ now that we know the $a$ values.

If $a=3$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-3}{2}=0$ .

So one ordered pair $(a,b)$ that satisfies the equation is:

$(3,0)$ .

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-\frac{-9}{5}}{2}$ .

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair $(a,b)$ that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$ .

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4 years ago

an item is regularly priced at $90. John bought it on sale for 20% off regular price. How much did John pay

Svetlanka [38]

Answer:

John paid $72

Step-by-step explanation:

To find the discount, we take the regular price and multiply it by the discount percentage

Discount = 90 * .2

= $18

The new price is the regular price minus the discount

New price = 90 -18

= $72

5 0

3 years ago

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Subtract 2xy from 6xy<br>

dmitriy555 [2]

Answer:

4xy

Step-by-step explanation:

We need to subtract 2xy from 6xy, So that is,

=> 6xy - 2xy

=> 4xy

6 0

3 years ago

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