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SOVA2 [1]
3 years ago
7

The head of the math department compared the scores of students in two classes on the Chapter 2 test. He found that the mean sco

re of both classes was the same. He also found that there was greater variability in the scores of Ms. Biren’s students than in the scores of Mr. Hayes’s students. Which quality of the data sets would lead the department head to that conclusion? The median score in Ms. Biren’s class is greater than the median score in Mr. Hayes’s class.The median score in Ms. Biren’s class is less than the median score in Mr. Hayes’s class.The range of scores in Ms. Biren’s class is greater than the range of scores in Mr. Hayes’s class.The range of scores in Ms. Biren’s class is less than the range of scores in Mr. Hayes’s class.
Mathematics
2 answers:
Liono4ka [1.6K]3 years ago
5 0
Given:
Mean of Ms. Biren's students score are the same with the mean of Mr. Haye's students scores.

Ms. Biren's students scores have greater variability than Mr. Haye's students scores.

The quality of the data set that would lead the department head to that conclusion is:

<span>The range of scores in Ms. Biren’s class is greater than the range of scores in Mr. Hayes’s class.</span>
arlik [135]3 years ago
3 0

Answer:

he range of scores in Ms. Biren’s class is greater than the range of scores in Mr. Hayes’s class.

Step-by-step explanation:

The range is one measure of variability.  The larger the range is, the greater the distance is between the lowest and the highest values; this means the data with a greater range is more variable, and the data with a lower range is less variable.

Since Ms. Biren's class has a higher range, this means her class has greater variability.


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Can someone please help me with this question asap!?
katovenus [111]

\qquad \qquad  \bf \huge\star \:  \:  \large{  \underline{Answer} }  \huge \:  \: \star

  • HJ = 23.5 in

\textsf{  \underline{\underline{Steps to solve the problem} }:}

Take HJ = a, GH = b and GJ = c

  • a = b + 2

  • c = a + b - 17

  • a + b + c = 73

put the value of a from equation 1 in equation 2

  • c = (b + 2) + b − 17

  • c = 2b − 15

now, put the value of a and c in equation 3

  • b + 2 + b + 2b − 15 = 73

  • 4b − 13 = 73

  • 4b = 86

  • b = 21.5 in

Now, we need to find HJ (a)

  • a = b + 2

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{ \qquad \large \sf {Conclusion} :}

  • HJ = 23.5 in
3 0
1 year ago
Solve for X, (with steps please) <br> 0.46=0.527-[0.0562/2 * log(0.1+x/1.5-x)]
Ivanshal [37]
Rearrange the equation: 

\frac{0.0562}{2} * log( \frac{0.1+x}{1.5 - x}) = 0.527 - 0.46 = 0.067

Isolate the log on its own

log (\frac{0.1 + x}{1.5 - x}) = \frac{0.067*2}{0.0562} = 2.42 (2dp)

To get rid of the log you must do 10^2.42 

\frac{0.1 +x}{1.5-x} = 10^{2.42} = 242.29 (2dp)

Then multiply both sides by (1.5 - x)

0.1 +x = 242.29(1.5-x)

0.1 + x = 363.44 - 242.29x

Solve normally

243.29x = 363.34    x = 363.34/243.29    x= 1.49344    x= 1.49 (2dp)

Hope that helps. Please send me a message if there's patchy bits. Also I'm sure you'll figure this out, but just in case, anywhere I've put down a 2dp shows that I've rounded the number to 2 decimal places. It becomes a pain to deal with otherwise. 

3 0
3 years ago
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