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2 years ago

What is the value of x, when x2 + 5 = 21?

Mathematics

1 answer:

djverab [1.8K]2 years ago

4 0

Hey there!

$\bold{x^2+5=21}$
Firstly, subtract by the number $\bold{5}$ on each of your sides
$\bold{x^2+5-5=21-5}$
$\bold{Cancel:5-5}$ because it gives us the result of $\bold{1}$
$\bold{Keep:21-5}$ because it equals: $\bold{16}$
$\bold{x^2=16}$
Nextly, we have to take our square root
$\bold{x=}$ ± $\bold{\sqrt{16}}$
Solve the square root and you have found your answer
$\boxed{\boxed{\bold{Answer:x=4[or]x=-4}}}$ $\bold{\checkmark}$

Good luck on your assignment and enjoy your day!

~ $\bold{LoveYourselfFirst:)}$

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Answer:

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Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

For this case we have the following distribution given:

X 3 4 5 6

P(X) 0.07 0.4 0.25 0.28

We can calculate the mean with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$

In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$

And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

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