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3 years ago

What is LCM of 6 and 30

Mathematics

2 answers:

Anettt [7]3 years ago

7 0

Answer:

Step-by-step explanation: Find the prime factorization of 6

6 = 2 × 3

Find the prime factorization of 30

30 = 2 × 3 × 5

Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:

LCM = 2 × 3 × 5

LCM = 30

Damm [24]3 years ago

5 0

Multiples of 6: 6, 12, 18, 24, 30

Multiples of 30: 30, 60, 90, 120

The smallest number that is common to both lists is the LCM.

So, what number is found on both lists and is the smallest?

I see 30. Do you see 30 on both?

Answer: 30

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A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)=72,000+60x

myrzilka [38]

Answer:

Step-by-step explanation:

Given

Cost Price $c(x)=72000+60x$

Price $p(x)=300-\frac{x}{20}$

Revenue generated $R(x)=P(x)\times x$

where x=no of units

$R(x)=300x-\frac{x^2}{20}$

To get maxima and minima differentiate R(x)

$\frac{\mathrm{d} R(x)}{\mathrm{d} x}=0$

$\frac{\mathrm{d} R(x)}{\mathrm{d} x}=300-2\times \frac{x}{20}=0$

$300=2\times \frac{x}{20}$

$x=3000$

maximum Revenue $R(x)=(300-\frac{300}{20})\times 300=4,50,000$

(b)Profit=Revenue - cost

$Profit=xp(x)-c(x)$

$Profit=300x-\frac{x^2}{20}-72000-60x$

$Profit(z)=240x-\frac{x^2}{20}-72000$

differentiate Profit to get maximum value

$\frac{\mathrm{d} z}{\mathrm{d} x}=240-2\times \frac{x}{20}$

$x=2400$

maximum Profit $z=2,16,000$

(c)Now company decided to tax the company $ 55 for each set

Profit $(z_1)=xp(x)-c(x)-55x$

$z_1=300x-\frac{x^2}{20}-72000x-60x^2-55x$

$z_1=185x-\frac{x^2}{20}-72,000$

differentiate Profit to get maximum value

$\frac{\mathrm{d} z_1}{\mathrm{d} x}=0$

$\frac{\mathrm{d} z_1}{\mathrm{d} x}=185-\frac{2x}{20}=0$

$x=1850$

$P(z_1\ at\ x=1850)=99125$

company should charge 207.5 $ for each set

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What is the X values such that y equals 15 - 3 X and Y equals 0 question

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2 years ago

describe the dimensions of a triangular prism that has a surface area between 550 square inches and 700 inches.

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Read 2 more answers

How would i solve a problem like this?

SVETLANKA909090 [29]

The number of combinations of size $k$ that you can make with $n$ items is given by the so-called binomial coefficient,

$\dbinom nk = \dfrac{n!}{k!(n-k)!}$

• $n!$ is the number of ways of permuting $n$ items.

• $(n-k)!$ is the number of ways of permuting all but $k$ of the $n$ items.

Dividing $n!$ by $(n-k)!$ then gives the number of ways of permuting only $k$ of the total $n$ items.

• $k!$ is the number of ways of permuting $k$ items.

Dividing $\frac{n!}{(n-k)!}$ by $k!$ then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

$\dbinom 63 = \dfrac{6!}{3!(6-3)!} = 20$

ways of selecting any 3 girls from the total 6 girls in the entire group of people.

There are

$\dbinom 72 = \dfrac{7!}{2!(7-2)!} = 21$

was of selecting any 2 boys from the total 7 boys.

Then there are

$\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}$

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

$\dbinom{13}5 = \dfrac{13!}{5!(13-5)!} = 1287$

Then the probability of selecting such a committee at random is

$\dfrac{\binom 63 \binom72}{\binom{13}5} = \dfrac{420}{1287} = \dfrac{140}{429} \approx 0.3263$

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1 year ago