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Komok [63]
3 years ago
15

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
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