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3 years ago

Two decimals that have the sum of 16.825

Mathematics

2 answers:

Alexus [3.1K]3 years ago

6 0

Answer:

10.805 and 6.02

Step-by-step explanation:

10.805 + 6.02 = 16.825

svlad2 [7]3 years ago

3 0

16.826 - 0.001

17.825 - 1.

16.82 + 0.005

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4 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

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4 years ago

USA Today reports that about 25% of all prison parolees become repeat offenders. Alice is a social worker whose job is to counse

pychu [463]

Answer:

a) $P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039$

$P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469$

$P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211$

$P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422$

$P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316$

b) $E(X) = np = 4*0.75=3$

c) $Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866$

d) $P(X \geq 3) \geq 0.98$

And the dsitribution that satisfy this is $X\sim Binom(n=9,p=0.75$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=4, p=1-0.25=0.75)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Part a

$P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039$

$P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469$

$P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211$

$P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422$

$P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316$

Part b

The expected value is givn by:

$E(X) = np = 4*0.75=3$

Part c

For the standard deviation we have this:

$Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866$

Part d

For this case the sample size needs to be higher or equal to 9. Since we need a value such that:

$P(X \geq 3) \geq 0.98$

And the dsitribution that satisfy this is $X\sim Binom(n=9,p=0.75$

We can verify this using the following code:

"=1-BINOM.DIST(3,9,0.75,TRUE)" and we got 0.99 and the condition is satisfied.

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3 years ago

The average of three numbers is 22. the average of two numbers is also 22. what is the value of the third number in the original

IRINA_888 [86]

Answer:

Step-by-step explanation:

If the average wasn't shifted at all, that means the third value must've been the same as the average to not shift it.

8 0

4 years ago

Read 2 more answers