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yarga [219]
3 years ago
14

Is it true that a smaller standard deviation means the data set tended to cluster around the mean? In the form of a paragraph, e

xplain in complete sentences the reasons that you believe this statement is true or false. Complete your work in the space provided or upload a file that can display math symbols if your work requires i
Mathematics
1 answer:
pav-90 [236]3 years ago
4 0
Yes, the smaller the standard deviation the closer the results are clustered around the mean.

Standard deviations measure the spread of the values in a data set. The more spread out the values, the higher the standard deviation.
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Find the area of the composite figure.
larisa86 [58]
Find the area of all the separate figures and add them together
area of rectangle = length x width
11 x 3 = 33 ft
area of square = a^2
3^2 = 9ft
area of triangle = base x height over 2
11 x 3 = 33
33 over 2 = 16.5 ft
33 + 9 + 16.5 = 58.5 ft ^ 2
8 0
3 years ago
Order from least to greatest: The square root of 64, 8.8, 26/3, 8 2/7
Eddi Din [679]

Square root 64 = 8

26/3 = 8 2/3 = 8.66666

8 2/7 = 8. 2857

So the order is sqrt(64) 8 2/7 26/3 8.8 <<<<< Answer

7 0
3 years ago
3.
coldgirl [10]
The answer is a I think that is correct
3 0
2 years ago
How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the
miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

<h3>8d.</h3>

\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}

___

<h3>9b.</h3>

\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}

___

<h3>10b.</h3>

\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}

Neither solution makes any denominator be zero, so both are good solutions.

8 0
3 years ago
Lisa spends part of her year as a member of a gym. She then finds a better deal at another gym
Tema [17]
I’m having trouble understanding this question.?
5 0
3 years ago
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