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Aliun [14]
3 years ago
8

Which number is closest to 51.4?

Mathematics
2 answers:
svetoff [14.1K]3 years ago
7 0
I think your answer would be D
shepuryov [24]3 years ago
5 0

Answer:

D

Step-by-step explanation:

Because its not exactly 40 and its .002 far away from beign 51.4

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PLEASEE HELPP!!! DUE TODAYY!!!
Alika [10]

Answer:

b 5/12

Step-by-step explanation:

4 0
3 years ago
A rectangular television screen has an area of w^2 + 2w square inches if the width of the television screen is w inches, what is
gregori [183]

 

The area of a rectangle is:

A =  length × width

A rectangular television screen has an area:

A = w² + 2w inches²

width = w

length = A / width  = (w² + 2w) / w = w(w + 2) / w = <u>w + 2</u>

Ansver:

<h2><u>length = w + 2</u></h2><h2><u></u></h2><h2><u></u></h2>
3 0
3 years ago
HELP ILL GIVE BRAINLISTTT!
olga nikolaevna [1]

Answer: √0.16

Step-by-step explanation:

√0.16 = 0.4

3 0
3 years ago
A triangle has sides with lengths of 60 feet, 63 feet, and 87 feet. Is it a right triangle?
Sedaia [141]

Answer:

Yes

Step-by-step explanation:

We can use the Pythagorean Theorem to check if this triangle is a right triangle:

a^2+b^2=c^2

Note that a and b are the legs of the triangle and c is the hypotenuse:

Substitute the lengths of the sides into the equation:

60^2+63^2=87^2

3600+3969=7569\\7569=7569

Therefore this triangle is a right triangle.

3 0
2 years ago
Read 2 more answers
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
3 years ago
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