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3 years ago

Writing equation of the line that passes through the given points (16, -4) ; m= -3/4

Mathematics

2 answers:

Schach [20]3 years ago

8 0

I think it’s m= -1/16

makvit [3.9K]3 years ago

5 0

Answer:

y = -3/4 - 12

Step-by-step explanation:

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What’s the mean I’m really dumb and can’t do it x

Julli [10]

Multiply the values across the rows. So you'll multiply each mark with its corresponding frequency

6*5 = 30

7*4 = 28

8*7 = 56

9*10 = 90

10*4 = 40

Then add up those results

30+28+56+90+40 = 244

Finally, you divide by 30 since this is the total number of scores (add up everything in the second column of the original table to get 30)

So we end up with the mean being 244/30 = 8.133 approximately

Round that value however you need to.

6 0

4 years ago

Find the average rate of change for f(x) = x2 + 9x + 18 from x = 10 to x = 20.

Montano1993 [528]

Answer:

The average rate of change is 39

Step-by-step explanation:

First, plug x=10 and x=20 into the formula:

10^2+90+18

190+18

208

20^2+180+18

400+198

598

598-208=390

390/10 = 39

The average rate of change is 39

please mark as Brainliest

7 0

3 years ago

Which data set COULD NOT be represented by the box plot shown? A) {13, 5, 8, 8, 6, 10, 15, 12, 4, 13, 11} B) {2, 10, 13, 7, 9, 5

siniylev [52]

Answer:

Step-by-step explanation:

From the box plots shown above:

The lowest number is 4

The highest number is 15

Therefore, the values in each boxes must lie between {4,15}.

In option A the values in the box include: {13,5,8,8,6,10,15,12,4,13,11}. The values of this data lies between 4 and 15 which makes it correct.

In the second option(B) the values in the boxes include: {2,10,13,7,9,5,15,12,7,13,11}.

The value 2 is not present in the interval, therefore the set of data is wrong.

In option C the values in the box include: {15,13,6,7,10,8,11,13,13,6,4}.

The values of this data lie between the interval of 4 and 15 which makes it correct.

In option D, the values in the box include: {4,13,6,7,15,12,11,10,5,13,9}·

The values of this data lie between the interval of 4 and 15 which makes it correct.

5 0

3 years ago

Plssss help meeeeeeee plssss

djyliett [7]

The correct answer is for the first image is B and the second image the answer will be D

6 0

3 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:

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

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

7 0

3 years ago