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4 years ago

Do two natural numbers always have a gcf

Mathematics

2 answers:

Vladimir79 [104]4 years ago

4 0

Yes, two natural numbers always have a GCF, and the smallest possible GCF is 2. 1 is not the prime factor.

ruslelena [56]4 years ago

3 0

Answer:

Yes natural numbers always have a gcf

Step-by-step explanation:

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Find all polar coordinates of point p if p=1,-x/6

Alecsey [184]

Answer:

All polar coordinates of point P are; (1 , -x/6 + 2nn) and (-1, -x/6 + (2n+1)n)

Step-by-step explanation:

The polar coordinates can be represented as;

(r,Ф)= (r,Ф+2nn) or (r,Ф) =(-r,Ф+(2n+1)n)

Where n is any integer value.

From the question, we are given that; p=(1,-x/6)

Comparing it with P(r,Ф), then r = 1 and Ф = -x/6

The polar coordinates will now be;

P(r,Ф) = (r,Ф+2nn) = (1 , -x/6 + 2nn)

Where; n is any positive integer and the value of r is positive.

While;

P(r,Ф) =(-r,Ф+(2n+1)n) = (-1, -x/6+(2n+1)n) and n is any positive integer and the value of r is negative.

Thus, all polar coordinates of point P are (1 , -x/6 + 2nn) and (-1, -x/6 + (2n+1)n)

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4 years ago

The scale on a map shows that 5 centimeters = 2 kilometers. What number of centimeters on the map represents an actual distance

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Given that 5 centimeters equals 2 kilometers, we are asked to find how many centimeters equals to 5 kilometers.

So, we know the rate is 5 centimeters is equal to 2 kilometers.

We should find out how many centimeters equals 1 kilometer.

To do this, we can divide 5 centimeters by 2 to find out how many centimeters is 1 kilometer.

5/2 = 2.5

So, the new rate is 2.5cm = 1km.

Then, all we need to do is multiply the new rate by 5 to find out how many centimeters is 5 kilometers.

5 x 2.5 = 12.5

Therefore, 12.5 centimeters equals a distance of 5 kilometers.

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A motel has a policy of booking as many as 150 guests in a building that holds 140. Past studies indicate that only 85% of booke

iren2701 [21]

Answer:

0.0015 = 0.15% probability that if the motel books 150 guests, not enough seats will be available.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ , if $np \geq 10$ and $n(1-p) \geq 10$ .

150 guests booked:

This means that $n = 150$

85% of booked guests show up for their room.

This means that $p = 0.85$

Is the normal approximation suitable:

$np = 150(0.85) = 127.5$

$n(1-p) = 150(0.15) = 22.5$

Both greater than 10, so yes.

Mean and standard deviation:

$\mu = E(X) = np = 150*0.85 = 127.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150*0.85*0.15} = 4.3732$

Find the probability that if the motel books 150 guests, not enough seats will be available.

More than 140 show up, which, using continuity correction, is $P(X > 140 + 0.5) = P(X > 140.5)$ , which is 1 subtracted by the p-value of Z when X = 140.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{140.5 - 127.5}{4.3732}$

$Z = 2.97$

$Z = 2.97$ has a p-value of 0.9985.

1 - 0.9985 = 0.0015.

0.0015 = 0.15% probability that if the motel books 150 guests, not enough seats will be available.

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