I neeeeed help this is my last question

Mathematics

1 answer:

uranmaximum [27]3 years ago

5 0

I'm pretty sure its the first answer because you put the 3x on the other side making it neg. and then you bring the neg. to the other side making the 3x positive and the 2 a neg.

You might be interested in

the class marks of a distribution are 37 42 and 47 the class limits corresponding to class mark 42 are

I am Lyosha [343]

Answer:

$39.5,44.5$

Step-by-step explanation:

$Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ first\ interval\ be\ u_1,v_1.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ second\ interval\ be\ u_2,v_2.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ third\ interval\ be\ u_3,v_3.\\Hence,\\As\ the\ class\ marks\ are\ uniform\ with\ an\ a\ difference\ of\ 5, the\\ observation\ is\ continuous\ and\ the\ class\ size\ is\ 5\ too.\\Hence,\\v_1=u_2, v_2=u_3\\$ $Now,\\Lets\ consider\ the\ first\ two\ classes.\\37=\frac{u_1+v_1}{2} \\42=\frac{u_2+v_2}{2}\\By\ adding\ the\ equations:\\79= \frac{u_1+v_1}{2}+\frac{u_2+v_2}{2}\\79=\frac{u_1+v_1+u_2+v_2}{2}\\79=\frac{(u_1+v_2)+(v_1+u_2)}{2}\\79=\frac{(u_1+v_2)}{2}+\frac{(v_1+u_2)}{2}\\Now,\\As\ the\ mid-point\ between\ two\ points\ can\ be\ calculated\ by\\ its\ average:\frac{u+v}{2} \\Hence,\\u_2\ lies\ in\ the\ mid-point\ of\ u_1\ and\ v_2.\\u_2=\frac{(u_1+v_2)}{2}\\$

$Hence,\\By\ substituting\ u_2=\frac{(u_1+v_2)}{2},\\79=u_2+\frac{v_1+u_2}{2}\\As\ v_1=u_2[Proven],\\79=u_2+ \frac{u_2+u_2}{2}\\79=u_2+\frac{2u_2}{2}\\79=2u_2\\Hence,\\u_2=\frac{79}{2}\\u_2=39.5\\Now,\ as\ we\ already\ know\ that\ the\ class\ size=5,\\v_2-u_2=5\\Hence,\\v_2=5+u_2\\Here,\\v_2=5+39.5\\v_2=44.5\\Hence,\\The\ upper\ limit\ of\ the\ second\ interval=44.5\\The\ lower\ limit\ of\ the\ second\ interval=39.5\\$

7 0

3 years ago

Find the median of the following data set. 1 1/4,5/8 ,3/5 ,1/2 ,1 1/2, 1 3/4

Anna35 [415]

Answer:

0.55

Step-by-step explanation:

(3/5+1/2)/2

3 0

3 years ago

6.2 poiThe width of a rectangle is 4 less than half thelength. If l represents the length, which equationcould be used to find t

Vilka [71]

We are told that l is the length of the rectangle. We are told that the widht is 4 less than half the lenght. So, first, we calculate the half of l, which would be