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Paladinen [302]

3 years ago

9

What is the answer if 202^2 base3 - 112^2 base3?

1 answer:

yuradex [85]3 years ago

6 0

Answer:

Hello,

answer is 204 in base 10 or 21120 in base 3

Step-by-step explanation:

$((202)_3)^2=(202*202)_3=(112211)_3\\\\((112)_3)^2=(112*112)_3=(21021)_3\\\\(112211-21021)_3=(211120)_3=(204)_{10}\\\\\\\\\\Other\ way\\\\((202)_3)^2=((20)_{10})^2=(400)_{10}\\\\((112)_3)^2=((14)_{10})^2=(196)_{10}\\\\(400-196)_{10}=(204)_{10}=(21120)_3$

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3 years ago

Name an equivalent ratio for with a denominator of 8.

const2013 [10]

Answer:

(A)

Step-by-step explanation:

4+4=8,4/8 is an equalivent ratio. So the best answer would be A!

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6 0

2 years ago

6r-r + 8(15-r)+23-6 is it equivalent to 3r+137

natita [175]

Answer:

No, it's not equivalent

Step-by-step explanation:

Try and solve the one with the bracket first. Expand 8(15-r) : 8x15 and 8x-r = 120 - 8r

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Rearrange the like terms together and solve the equation:

6r-r-8r + 120+23-6 = -3r + 137

7 0

3 years ago

Read 2 more answers

An aquarium has display tanks that each contain 175 fish.how many fish are on display in 6 tanks

wel

Answer:

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Step-by-step explanation:

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3 0

3 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

So

$P(X \geq 2) = 0.93803$

5 0

2 years ago