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Dmitry [639]
3 years ago
5

What is 254 divided by 40

Mathematics
2 answers:
Zigmanuir [339]3 years ago
8 0
The answer is 6.35. Because you put in 254 / 40 and it says 6.35.

tangare [24]3 years ago
5 0
Solutions 

254 divided by 40 also know as 254 ÷ 40 

254 ÷ 40= <span>6 with a remainder of 35 (6.35)</span>
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Find the solution to the differential equation dx/dy+y/3=0 subject to the initial conditions y(0)=12.
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\dfrac{\mathrm dx}{\mathrm dy}+\dfrac y3=0\iff\mathrm dx=-\dfrac y3\,\mathrm dy
\implies\displaystyle\int\mathrm dx=-\frac13\int\mathrm y\,dy
\implies x=-\dfrac16y^2+C

Given that y(0)=12, we have

0=-\dfrac16(12)^2+C\implies C=24

So the particular solution to the initial value problem is

x=-\dfrac16y^2+24\iff y^2=144-6x
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3 years ago
g An urn contains 150 white balls and 50 black balls. Four balls are drawn at random one at a time. Determine the probability th
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Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Sampling with replacement:

I consider a success choosing a black ball, so p = \frac{50}{150+50} = \frac{50}{200} = 0.25

We want 2 black balls and 2 white, 2 + 2 = 4, so n = 4, and we want P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so N = 200

Sample of 4, so n = 4

50 are black, so k = 50

We want P(X = 2).

P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

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Answer:

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