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3 years ago

Convert 198 grams to kilograms

Mathematics

2 answers:

aleksandr82 [10.1K]3 years ago

7 0

Answer:

0.198

Step-by-step explanation:

divide 198 by 1,000

cluponka [151]3 years ago

5 0

Answer:

it's 0.198

Step-by-step explanation:

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Can you tell me how to factorise 6b+24bc

Kamila [148]

When we factorise an expression, we are looking for simple factors that multiply to get the original expression. Usually it is very natural to factorise something like a quadratic in x. For example:

x^2 + 3x + 2 = (x+1)(x+2)

But there are other situations where factorisation can be applied. Take this quadratic:

x^2 - 9x = x(x-9)

This second example is closer to the question in hand. Just like x was a common factor to both x^2 and -9x, we are looking for a common factor to both 6b and 24bc. The common factor is 6b.

Hence 6b + 24bc = 6b(1 + 4c).

I hope this helps you :)

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2 years ago

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What is the area of a triangle with a base of 2 5/8 mm and a height of 10 mm?

MrRissso [65]

Answer:

b looks like the more reziable awnser

Step-by-step explanation:

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3 years ago

7 (x + 3) - 3(4x + 9) = 14

Paul [167]

Answer:

-4

Step-by-step explanation:

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3 years ago

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A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

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2 years ago

Subtract. 23.87 - 1.12 A. 13.75 B. 22.75 C. 24.99 D. 12.67

Shtirlitz [24]

Answer:

B. 22.75

Step-by-step explanation:

23.87

- 1.12

22.75

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3 years ago

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