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3 years ago

The length and the breadth of a rectangular piece of land are 500m and 300m respectively. Find it’s area

Mathematics

2 answers:

garik1379 [7]3 years ago

8 0

It’s area would be 150000m because area is LxW so (500x300)

katrin [286]3 years ago

6 0

Area of a rectangle = Length x Breadth
The required area = 500 x 300
= 150000 m^2

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Mackenzie has a loyalty card good for a 20% discount at her local hardware store. What would her total in dollars and cents be,

8_murik_8 [283]

Answer: $14.0

Step-by-step explanation:

For us to calculate this question, we have to find 20% of $17.45 and then subtract the value gotten from $17.45. This will be:

= $17.45 - (20% × $17.45)

= $17.45 - (0.2 × $17.45)

= $17.45 - $3.49

= $13.96

= $14.0 to nearest cent

6 0

3 years ago

Read 2 more answers

Find the sum <br><br> 3/4 + 5/16

Dmitry_Shevchenko [17]

3/4 also equals 12/16 so 12/16+5/16= 17/16 or 1 1/16

5 0

3 years ago

Read 2 more answers

There is a pie eating contest with 14 participants. There are 29 pies that are all eaten. Which of the following best describes

kotegsom [21]

Hmm...I tried to divide 29 by 14, and I got 2.07142857143.... Does it let you round the decimal..?

3 0

3 years ago

The length of the hypotenuse of a right angled triangle exceeds the length of the base by 2cm and exceeds twice the length of th

insens350 [35]

Answer: a = 8; b = 15; c = 17

Step-by-step explanation:

c = b + 2 and c = 2a + 1

b = c − 2 and a = $\frac{c-1}{2}$

c² = b² + a²

$c^{2} = (c-2)^{2} + \frac{c-1}{2}\\$

$c^{2}= \frac{4 (c-2)^{2}+(c - 1)^{2} }{4}$

$4c^{2}={4 (c^{2} + 4 -4c)+c ^{2} + 1-2c$

$4c^{2}={4 c^{2} + 16 -16c)+c ^{2} + 1-2c$

$c^{2}-18c+17=0$

$c(c -17)-1(c-17)=0$

$(c -1)(c-17)=0$

c = 1 or c = 17

If c = 1 then b = 1 - 2 = -1, that's not possible

So x = 17

b = 17 - 2 = 15

a = $\frac{17-1}{2} = 8$

Length sides of the triangle are 17 cm, 15 cm and 8 cm.

I saw some of this on a site but I want to make sure its correct so we're gonna apply the pythagorean theory and see if its correct.

$a^{2} + b^{2} = c^{2} \\8^{2} + 15^{2} = 17^{2} \\64 + 225 = 289\\289 = 289\\$

This shows that this is correct.

Hope this helped! Again, some of this was from a site, not from me.

3 0

3 years ago

In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that

attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable $X\sim Poisson(\lambda = 0.62)$ .

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...$

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252$

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: $\lambda=7\times0.62=4.34$

Compute the probability of no deaths in a corps over 7 years as follows:

$P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130$

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0

3 years ago