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3 years ago

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:

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:

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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

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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

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

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

$P(X \geq 2) = 0.93803$

5 0

3 years ago