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Contact [7]
3 years ago
10

How do you estimate 586-321

Mathematics
2 answers:
aniked [119]3 years ago
4 0
I would estimate about 240
Lostsunrise [7]3 years ago
4 0
You would do 580 - 320 = 260
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HELP I WILL MARK BRAINLIEST
Ede4ka [16]

Answer:

the answer is A.

Step-by-step explanation:

7 0
2 years ago
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A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has
bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113  

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711  

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] =  1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0
3 years ago
Three moons are in the same circular orbit around a planet. The moons are each 120,000 kilometers from the surface of the planet
VladimirAG [237]

Answer:

Step-by-step explanation:

Alright, lets get started.

Please refer the diagram I have attached.

MN is the diameter of the planet which is MN = 60000

AM and NC are the distance of moons from the surface of the planet.

AM = NC = 120000

Since, angle ABC is given as 90 degree, it means, line AC will pass from the diameter of the planet.

So, distance between moon A and moon C is =

AC=AM + MN + NC

AC = 120000+ 60000+ 120000

AC = 300000

Hence the distance between point A and point C is 300000 Km.   :   Answer

Hope it will help :)

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3 years ago
There are as many 5p coins as 1p coins in the bag.
nekit [7.7K]

Answer:

160

Step-by-step explanation:

In math, the word of means multiply. :)

3/8 * 640 = 240, so there are 240 1p coins.

there are the same number of 5P coins, so there are 240 5p.

640 - 1p - 5p = 2p

640-240-240 = 160, so there are 160 2p coins.

6 0
2 years ago
Determine the domain of the following graph
sp2606 [1]

Answer:

[-9,3]

Step-by-step explanation:

The domain is all x-values or inputs of a function from left to right is -9 to + 3

6 0
2 years ago
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