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2 years ago

Which of the factors of 120 are composite numbers?

Mathematics

1 answer:

Pani-rosa [81]2 years ago

6 0

Answer:

Find a common prime factor in each of these pairs of numbers

a.21 and 411

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A manager at a local company asked his employees how many times they had given blood in the last year. The results of the survey

Lubov Fominskaja [6]

Answer:

$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

Step-by-step explanation:

For this case we have the following distribution given:

X 0 1 2 3 4 5 6

P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02

For this case we need to find first the expected value given by:

$E(X) = \sum_{i=1}^n X_i P(X_I)$

And replacing we got:

$E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61$

Now we can find the second moment given by:

$E(X^2) =\sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97$

And the variance would be given by:

$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

8 0

3 years ago

You deposit $2200 Into three separate bank accounts for each pay 3% annual interest. How much interest does each account earn af

Hatshy [7]

The formula to find the amount is

$A=P(1+r)^n$

Here A = amount

P is the principal

r is the rate

n is the number of years

Then to find the interest we subtract principal from amount.

Interest = A - P

Here

P= 2200, r = 3% = 0.03 , n = 6 years

$A = 2200 (1+0.03)^6=2626.92$

Hence the interest earned = 2626.92-2200 = $426.92

Now if the total of $2200 was deposited in three banks then each account earns $\frac{426.92}{3}= 142.31$

Each account earns $142.31

5 0

3 years ago

The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is

bulgar [2K]

The integral is approximated by the sum,

$\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))$

where $f(x)=x^3$ and $x_n=\dfrac12n$ , giving you

$\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)$
$\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)$
$\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)$

Faulhaber's formulas make short work of computing the sum. You have

$\displaystyle\sum_{n=0}^k1=k+1$
$\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2$
$\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6$
$\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4$

which gives

$\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31$
$\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}$
$\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25$