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2 years ago

One number is 7 more than another. The difference between their squares is 161. What are the numbers?

1 answer:

3 0

Let's say the numbers are "a" and "b"

now, we know one of them is 7 more than the other, so hmmm let's say "b" is 7 more than "a", so b = a + 7

the difference of their squares is 161, alrite

$\bf a^2-b^2=161\qquad b=a+7\implies a^2-(a+7)^2=161
\\\\\\
a^2-(a^2+14a+49)=161\implies \underline{a^2-a^2}-14a-49=161
\\\\\\
-14a-49=161\implies -49-161=14a\implies -210=14a
\\\\\\
\cfrac{-210}{14}=a$

and surey you know how much that is

now, what's "b"? well, b = a + 7

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Step-by-step explanation:

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

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

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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2 years ago