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3 years ago

Select all names that apply to the Number √4

Mathematics

2 answers:

Alexxandr [17]3 years ago

8 0

<h2>ANSWER✔</h2>

$\sf\therefore \sqrt{4}=?$

$\sf\implies \sqrt{ 2 \times2}$

$\sf\implies \sqrt{2^2}$

$\sf\implies 2$

$\large{\boxed{\bf{ \implies \:2\:}}}$

never [62]3 years ago

7 0

Answer:

$\sqrt{4} = 2$

Step by step explaination:

$\sqrt{4} \\ = \sqrt{ {2}^{2} } \\ = 2$

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According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e

slamgirl [31]

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1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that $p = 0.09$

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225$

$P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827$

$P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082$

$P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982$

Finally

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180$

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

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2 years ago

Factor x^4 + x y^3 + x^3 + y^3 completely. Show your work.

andrey2020 [161]

$\bf x^4+xy^3+x^3+y^3\impliedby \textit{now let's do some grouping}
\\\\\\
(x^4+xy^3)~~+~~(x^3+y^3)\implies x(x^3+y^3)~~+~~(x^3+y^3)
\\\\\\
\stackrel{\stackrel{common}{factor}}{(x^3+y^3)}(x+1)\\\\
-------------------------------\\\\
\textit{now recall that }\qquad \qquad \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\
-------------------------------\\\\
(x+y)(x^2-xy+y^2)(x+1)$

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