Math, asked by revivestephen, 4 months ago

9. The following scores were obtained by a group of 40 students
on an achievement test:
32 78 27 65 88 83 63 52
86 70 42 66 56 44 63 59
73 52 43 69 59 46 71
71 65
42 55 39 70 57 49 78 70
34 61 62 77 81 72 79 69

Prepare a frequency distribution table and extend it to a
cumulative frequency distribution and cumulative percentage
frequency distribution tables for the above data by using a
class interval of 5.​

Answers

Answered by Anonymous
0

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answered by divyaprakash14060
0

Answer:

Answer:</p><p>\begin{gathered}\begin{gathered} \begin{array}{|c|c|c|} \bf{Observation} &amp; \bf \: Tally\:marks&amp; \bf{frequency} \\ 15 &amp;  \mid \:\mid \:\mid \:&amp; \: \bf \: 3    \\ \\16 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6  \\ \\17 &amp; \mid \:&amp; \: \bf \: 1 \\  \\18 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6\\  \\20 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6\\  \\24 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:} \:&amp; \: \bf \: 5\\  \\25 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6 \\  \\27 &amp;  \mid \:\mid \:&amp; \: \bf \: 2\\  \\28 &amp;  \mid \:\mid \:\mid \: \mid \:&amp; \: \bf \: 4 \\ \\30 &amp; \mid \:&amp; \: \bf \: 1 \end{array}\end{gathered} \\ \\ \end{gathered}Observation15 16 17  18  20  24  25  27  28 30 Tallymarks ∣∣∣ ∣∣∣∣∣ ∣ ∣∣∣∣∣ ∣∣∣∣∣ ∣∣∣∣ ∣∣∣∣∣ ∣∣ ∣∣∣∣ ∣frequency3  6 16656 241 </p><p>Step-by-step explanation:</p><p>Given that, scores obtained by 40 students in a test are</p><p>24,16,17,30,28,20,15,24,16,18,18,15,16,20,25,24,25,20,16,15,18,25,20,18,28,27,25,24,24,18,18,25,20,16,25,20,27,28,28,16.</p><p>Let's first arrange the data in ascending order.</p><p>So, above data in ascending order as</p><p>15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 18, 18, 18, 18, 18, 18, 20, 20, 20, 20, 20, 20, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 27, 27, 28, 28, 28, 28, 30</p><p>\begin{gathered}\begin{gathered} \begin{array}{|c|c|c|} \bf{Observation} &amp; \bf \: Tally\:marks&amp; \bf{frequency} \\ 15 &amp;  \mid \:\mid \:\mid \:&amp; \: \bf \: 3    \\ \\16 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6  \\ \\17 &amp; \mid \:&amp; \: \bf \: 1 \\  \\18 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6\\  \\20 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6\\  \\24 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:} \:&amp; \: \bf \: 5\\  \\25 &amp;  \cancel{\mid \:\mid \:\mid \:\mid \:}\mid \:&amp; \: \bf \: 6 \\  \\27 &amp;  \mid \:\mid \:&amp; \: \bf \: 2\\  \\28 &amp;  \mid \:\mid \:\mid \: \mid \:&amp; \: \bf \: 4 \\ \\30 &amp; \mid \:&amp; \: \bf \: 1 \end{array}\end{gathered} \\ \\ \end{gathered}Observation15 16 17  18  20  24  25  27  28 30 Tallymarks ∣∣∣ ∣∣∣∣∣ ∣ ∣∣∣∣∣ ∣∣∣∣∣ ∣∣∣∣ ∣∣∣∣∣ ∣∣ ∣∣∣∣ ∣frequency3  6 16656 241 </p><p>\rule{190pt}{2pt}</p><p>Additional Information :-</p><p>1. Mean using Direct Method</p><p>\begin{gathered}\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ \end{gathered}Mean=∑fi∑fixi</p><p>2. Mean using Short Cut Method</p><p>\begin{gathered}\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ \end{gathered}Mean=A+∑fi∑fidi</p><p>3. Mean using Step Deviation Method</p><p>\begin{gathered}\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ \end{gathered}Mean=A+∑fi∑fiui×h</p><p>

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