Question

The ages of the eight-member school board are listed below.

28, 40, 34, 46, 60, 34, 51, 57

If A is equal to the median age of the group and B is equal to the mode of the group/s age, what is the value of A-B?

Answer 1

Median is 43
Mode is 34

Answer 2

The ages of the eight-member school board are listed below.

28, 40, 34, 46, 60, 34, 51, 57

If A is equal to the median age of the group and B is equal to the mode of the group/s age, what is the value of A-B?

6
7
8
9
10

Answer 3

A - B (subtract B from A) at least that much u can do!!!!!!!!!!

Answer 4

idk bruh.... lol

Answer 5

im flutterin around wit yu

Answer 6

flutterin*

Answer 7

43 - 34 = 9..................who u r

Answer 8

1, 8,15, 22, . . .

What number is next in the above sequence?

37
30
29
27
none of the above

Answer 9

29

Answer 10

i doubt it may be 30

Answer 11

hmmmm may be 27

Answer 12

f 7a = (91)(13), what is the value of \[\sqrt{a}\] ?

Grid-in type answer.

Answer 13

13

Answer 14

\[2x ^{3}y ^{4}z ^{3}a ^{2}\over4x ^{2}y ^{5}za\]

Answer 15

may i know asl plz

Answer 16

18 m fl

Answer 17

kin you please answer the last question

Answer 18

\[\LARGE \begin{array}{l}
\left( {{{\left( {\frac{x}{7}} \right)}^2}\sqrt {\frac{{\left| {|x| - 3} \right|}}{{|x| - 3}}} + {{\left( {\frac{y}{3}} \right)}^2}\sqrt {\frac{{\left| {y + \frac{{3\sqrt {33} }}{7}} \right|}}{{y + \frac{{3\sqrt {33} }}{7}}}} - 1} \right) - \left( {\left| {\frac{x}{2}} \right| - \left( {\frac{{3\sqrt {33} - 7}}{{112}}} \right){x^2} - 3 + \sqrt {1 - {{(\left| {\left| x \right| - 2} \right| - 1)}^2}} - y} \right) \cdot \\
\left( {9\sqrt {\frac{{\left| {(\left| x \right| - 1)(\left| x \right| - 0.75)} \right|}}{{(1 - \left| x \right|)(\left| x \right| - 0.75)}}} - 8\left| x \right| - y} \right) \cdot \left( {3\left| x \right| + 0.75\sqrt {\frac{{\left| {(\left| x \right| - 0.75)(\left| x \right| - 0.5)} \right|}}{{(0.75 - \left| x \right|)(\left| x \right| - 0.5)}}} - 8\left| x \right| - y} \right) \cdot \\
\left( {2.25\sqrt {\frac{{\left| {(x - 0.5)(x + 0.5)} \right|}}{{(0.5 - x)(x + 0.5)}}} - y} \right) \cdot \left( {\frac{{6\sqrt {10} }}{7} + (1.5 - 0.5\left| x \right|)\sqrt {\frac{{\left| {\left| x \right| - 1} \right|}}{{\left| x \right| - 1}}} - \frac{{6\sqrt {10} }}{{14}}\sqrt {4 - {{(\left| x \right| - 1)}^2}} - y} \right) = 0
\end{array}\]