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4 years ago

Oof

Mathematics

2 answers:

Marizza181 [45]4 years ago

8 0

Answer:

triangle C has the largest area.

Step-by-step explanation:

From the the three triangles A, B, C based on the area occupied by the triangles, it can be seen that triangle C occupies the largest area.

On the centimetre grid, let each box represent one unit. For triangle A, the base is 2 boxes (i.e 2 units) and the height is 3 units, For triangle B, the base is 2 units and the height is 3 units while for triangle C, the base is 3 units and the height is 3 unit. The area of a triangle is given by:

Area = $\frac{1}{2}base*height$

For triangle A:

Area = $\frac{1}{2}base*height$ = $\frac{1}{2} *2*3=3 \ unit^2$

For triangle B:

Area = $\frac{1}{2}base*height$ = $\frac{1}{2} *2*3=3 \ unit^2$

For triangle C:

Area = $\frac{1}{2}base*height$ = $\frac{1}{2} *3*3=4.5 \ unit^2$

Therefore, triangle C has the largest area.

Elina [12.6K]4 years ago

5 0

Answer:

triangle C

Step-by-step explanation:

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3 years ago

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4 years ago

The number of different species of plants in some gardens is recorded.

PolarNik [594]

(a) The mean of the number of different species of plants in some gardens is 4.5 .

(b) The standard deviation of the number of different species of plants in some gardens is 2.12 (Approximately)

It is given in the question that the number of different species recorded are:-

1 2 3 4 4 5 5 6 7 8

We have to find the mean and standard deviation of the given data of the number of plants of different species.

(a) We know that, the formula for mean is :-

Sum of all the observations/ Total number of observations

Hence,

Mean = $\frac{1+2+3+4+4+5+5+6+7+8}{10} =\frac{45}{10}=4.5$

(b) Standard deviation = $\sqrt{\frac{Sum(X-m)^{2} }{N} }$

Where,

X = each value

m = mean

N = number of values

Hence,

$Sum(X-m)^{2} =(1-4.5)^{2}+(2-4.5)^{2}+ (3-4.5)^{2}+ (4-4.5)^{2}+ (4-4.5)^{2}+ (5-5.5)^{2}+ (5-5.5)^{2}+ (6-5.5)^{2}+ (7-5.5)^{2}+ (8-5.5)^{2}\\ \\ Sum(X-m)^{2}=(-3.5)^{2} +(-2.5)^{2}+ (-1.5)^{2}+ (-0.5)^{2}+ (-0.5)^{2}+ (0.5)^{2}+ (0.5)^{2}+ (1.5)^{2}+ (2.5)^{2}+ (3.5)^{2} \\\\$

$Sum(X-m)^2=12.25+6.25+2.25+0.25+0.25+0.25+0.25+2.25+6.25+12.25\\\\=42.5$

Standard deviation = ${\sqrt{\frac{{42.25}}{10} }} =\frac{6.7}{\sqrt{10} }=\frac{6.7}{3.16}=2.12$ (approximately).

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1 year ago