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Marysya12 [62]
3 years ago
13

What is the missing length

Mathematics
2 answers:
spayn [35]3 years ago
6 0

Is 10 because you multiple 5 by 2

Andrej [43]3 years ago
4 0

Answer:

10

Step-by-step explanation:

Utilize the proportions given to you. 3.3 is 1/2 of 6.6, and 3.7 is 1/2 of 7.4. That must mean that 5 is 1/2 of side h, making side h 10.

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For skewed data displays, the median is often a better esitmate of the center of distribution than the mean, but
bazaltina [42]

For skewed data displays, the median is often a better estimate of the center of distribution than the mean because the former is unaffected by large numbers.

<h3>What is mean?</h3>

Mean refers to the average of set of two or more numbers.

Mean of a set having 'n'  numbers = \frac{Sum Of 'n' Numbers}{n}

<h3>What is median?</h3>

Median refers to the middle-most value of a list of numbers, arranged either in ascending or descending order.

Median = \left \{ {{\frac{n}{2}^{th}  term, if n = even } \atop( (\frac{n-1}{2}+\frac{n+1}{2})/2)^{th} term, if n = odd   }} \right.

Now,

  • Since it takes the average of all the values in the data set, the mean is the most widely used measure of central tendency.
  • Because it is unaffected by exceptionally big numbers, the median performs better than the mean when analyzing data from skewed distributions.

Hence, For skewed data displays, the median is often a better estimate of the center of distribution than the mean.

To learn more about mean and median, refer to the link:brainly.com/question/6281520

#SPJ4

7 0
1 year ago
PLEASE HELP @) POINTS!!!<br><br> Which of these is true of a chemical reaction?
lakkis [162]
Energy is either released or absorbed
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Robert can complete 10 inspections in 8 hours. Which represents the unit rate for the number of inspections he completes per
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Me inspecting ok me is gonna inspect 54 hurs
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Can u help with 17 and 18 (IF U KNOW BOTH OF THEM) Please thanks
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6 0
3 years ago
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Estimate the limit. <br> Picture below
vova2212 [387]

Answer:

Hence, the limit of the expression:\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1} is:

\dfrac{-1}{9}

Step-by-step explanation:

We are asked to estimate the limit of the expression:

\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}

We will simplify the expression by first taking the l.c.m of the terms in the numerator to obtain the expression as:

\dfrac{3-(x+2)}{3(x+2)}\\\\=\dfrac{3-x-2}{3(x+2)}\\\\=\dfrac{1-x}{3(x+2)}

\lim_{x \to 1} \dfrac{1-x}{3(x+2)(x-1)}\\\\= \lim_{x \to 1} \dfrac{-(x-1)}{3(x+2)(x-1)}\\\\\\= \lim_{x \to 1} \dfrac{-1}{3(x+2)}

since the same term in the numerator and denominator are cancelled out.

Now the limit of the function exist as the denominator is not equal to zero when x→1.

Hence,

\lim_{x \to 1} \dfrac{-1}{3(x+2)}\\\\=\dfrac{-1}{3(1+2)}\\\\=\dfrac{-1}{3\times 3}\\\\=\dfrac{-1}{9}

Hence, the limit of the expression:\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1} is:

\dfrac{-1}{9}

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3 years ago
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