Round 70.59 to the nearest whole number

Mathematics

1 answer:

il63 [147K]3 years ago

5 0

the answer would be 71

You might be interested in

Complete the equation of the line through (-8,-2) (-4,6)<br> Use exact numbers.

kykrilka [37]

Answer:

y = 2x + 14

Step-by-step explanation:

First, you have to find the slope by using

y2-y1

x2-x1

In other words,

-2-6

-8+4

The slope is 2.

Then you plug the rest into point-slope form (you can use either of the points, I used the first one)

y-y1=m(x-x1)

Remember that m is the slope.

y+2=2(x+8)

Distribute the slope to the parenthesis

y+2=2x+16

Isolate the y variable

y=2x2x+14

please mark me as a brain list

4 0

4 years ago

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:

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

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