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3 years ago

The LCM of 5 and 6 is ____.

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

5 0

15 is da answer so ya

djverab [1.8K]3 years ago

4 0

Answer:

The lcm of 5 and 6 is 30

Step-by-step explanation:

It is 30 because 5 * 6 equals 30

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3 years ago

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3 years ago

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal 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.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

$\mu = \lambda = 4(3) = 12$

$\sigma = \sqrt{\lambda} = \sqrt{12}$

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

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

$Z = \frac{20 - 12}{\sqrt{12}}$

$Z = 2.31$

$Z = 2.31$ has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

5 0

3 years ago

What can expressions be simplified by combining?!

Nina [5.8K]

Answer:

you can combine like term such as x or 2x, but not variables with different exponents like x and x squared.

Step-by-step explanation:

5 0

3 years ago

Can some one plz help me?! The LATERAL Surface Area for the shape below is:

love history [14]

Answer:

430 cm2

Step-by-step explanation:

The Lateral Surface Area of it is equal to:

Area of ABCD + Area of A'B'C'D' + Area of ABB'A' + Area of DCC'D' + Area of ADD'A' + Area of BCC'B'

As ABCDA'B'C'D' is a rectangle prism, so that:

+) Area of ABCD = Area of A'B'C'D'

+) Area of ABB'A' = Area of DCC'D'

+) Area of ADD'A' = Area of BCC'B'

=> Total lateral surface area = 2. Area ABCD + 2. Area ABB'A + 2. Area ADD'A

As ABCDA'B'C'D' is a rectangle prism, so that we have: ABCD, ABB'A', ADD'A' are rectangles.

The formula to calculate the area of rectangle is: Area = Length x Width

We have:

+) Area ABCD = AB x BC =11 x 10 = 110 cm2

+) Area ABB'A = AB x AA' = 11 x 5 = 55 cm2

+) Area ADD'A = AA' x A'D' = 5 x 10 = 50cm2

Total lateral surface area = 2. Area ABCD + 2. Area ABB'A + 2. Area ADD'A

= 2 x 110 + 2 x 55 + 2 x 50 = 430 cm2

4 0

3 years ago