Gints

Mathematics

1 answer:

artcher [175]3 years ago

8 0

Answer:

The correct statements are: B and D.

Step-by-step explanation:

Standard deviation is one of the measure of dispersion or spread of a dataset from the average value of the dataset. The standard deviation measures the variability of the distribution of values.

The higher the spread or variability in the data set, the greater is the standard deviation value and thus greater will be the magnitude of the deviance of the value from their average.

Then the standard deviation (S.D.) is given by,

$S.D.=\sqrt{\frac{1}{n-1}\sum\limits^{n}_{i=1} {(x_{i}-\bar x)^{2} }}$

The correct statements are:

B. It is a measure of variability.

D. It is calculated using the mean.

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Write 2% as a reduced fraction

ANTONII [103]

2% as a reduced fraction is 2/100 = 1/50

3 0

3 years ago

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In a village the number of houses and the number of flats are in the ratio 9:5 the number of flats and the number of bungalows T

Makovka662 [10]

Answer:

Total no. of houses = 140

Step-by-step explanation:

Let there are 9x flats and 5x bungalows.

Total bungalows = 50

It means,

5x = 50

x = 10

No of flats = 9x

= 9(10)

= 90

Total houses = no. of flats + no. of bungalows

= 50+90

= 140

So, there are 140 houses in the village.

5 0

3 years ago

(5x^3-8x^2+9x+12)/(x-3)<br><br> please answer with sentences on how to do it step by step. thanks!

Anastaziya [24]

Answer:

$5x^2+7x+30+\frac{102}{x-3}$

Step-by-step explanation:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}$