Complete the comparison : 27> ?

The price of a toy usually costing £50 to £65 work out the percentage increase FAST!!

VLD [36.1K]

Answer:

50 * 1.3 = 65

130% increase

6 0

2 years ago

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A study indicates that 37% of students have laptops. You randomly sample 30 students. Find the mean and the standard deviation o

Brilliant_brown [7]

Answer:

The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.

Step-by-step explanation:

Let X = number of students who have laptops.

The probability of a student having a laptop is, P (X) = p = 0.37.

A random sample of n = 30 students is selected.

The event of a student having a laptop is independent of the other students.

The random variable X follows a Binomial distribution with parameters n and p.

The mean and standard deviation of a binomial random variable X are:

$\mu=np\\\sigma=\sqrt{np(1-p)}$

Compute the mean of the random variable X as follows:

$\mu=np=30\times0.37=1.11$

The mean of the random variable X is 1.11.

Compute the standard deviation of the random variable X as follows:

$\sigma=\sqrt{np(1-p)}=\sqrt{30\times0.37\times(1-0.37)}=\sqrt{0.6993}=0.836$

The standard deviation of the random variable X is 0.836.

5 0

3 years ago

Suppose that a company needs 1, 200,000 items during a year and that preparation for each production run costs $500. Suppose als

MAVERICK [17]

Answer:

The number of items in each production run so that the total costs of production and storage are minimized is 8165 items/run

Step-by-step explanation:

We will use the following variables:

Q = Quantity being ordered

Q* = the optimal order Quantity: the result being sought

D = annual Demand for the item, over the year

P = unit Production cost

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run)

H = annual cost to Hold one unit

It is important to note which variables are annualized, which are per-order and which are per-unit.

Using the variables, here are the components of the first equation

Total Cost, TC = PC + SC + HC

PC = P x D : Production Cost = unit Production cost times the annual Demand

SC = (D x S)/Q : Setting up Cost = annual Demand times cost per production setup, divided by the order Quantity (number of units)

HC = (H x Q)/2: Holding Cost = annual unit Holding cost times order Quantity (number of units), divided by 2 (because throughout the year, on average the warehouse is half full).

So TC = PC + SC + HC = (P x D) + ((D x S)/Q) + ((H x Q)/2) = PD + (DS/Q) + HQ/2

To obtain the optimal order quantity, Q* that minimizes TC, at the minimum TC, dTC/dQ = 0

dTC/dQ = (H/2) – (D x S)/(Q²) = 0

(H/2) – (D x S)/(Q²) = 0

Solving for Q, which is Q* at this point.

(Q*)² = 2DS/H

Q* = √(2DS/H)

D = annual demand for the item = 200000

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run) = $500

H = annual cost to Hold one unit = $3

Q* = √(2×200000×500/3) = 8164.97 = 8165 items.

3 0

3 years ago

A boy has 3 red , 2 yellow and 3 green marbles. In how many ways can the boy arrange the marbles in a line if: a) Marbles of the

Ilya [14]

Answer:

The total number of different arrangements is 560.

Step-by-step explanation:

A multiset is a collection of objects, just like a set, but can contain an object more than once.

The multiplicity of a particular type of object is the number of times objects of that type appear in a multiset.

Permutations of Multisets Theorem.

The number of ordered n-tuples (or permutations with repetition) on a collection or multiset of $n$ objects, where there are $k$ kinds of objects and object kind 1 occurs with multiplicity $n_1$ , object kind 2 occurs with multiplicity $n_2$ , ... , and object kind $k$ occurs with multiplicity $n_k$ is:

$\begin{equation*}\frac{n!}{n_1!*n_2!*\dots * n_k!}\end{equation*}$

We know that a boy has 3 red, 2 yellow and 3 green marbles. In this case we have n = 8.

If marbles of the same color are indistinguishable, then the total number of different arrangements is

${8 \choose 3, 2, 3} = \frac{8 !}{3 ! 2 ! 3 !} = \frac{8\cdot \:7\cdot \:6\cdot \:5\cdot \:4}{2!\cdot \:3!}=\frac{6720}{2!\cdot \:3!}=\frac{6720}{12}=560$

8 0

2 years ago

How many houses have an area less than 100 m² ?

dsp73

Answer:

This question overall depends on a continuously changing factor.

Step-by-step explanation:

There are many houses in this world that have an area of less than 100 $m^2$ , while many others have up to 700, 800, or even 1000 $m^2$ of land. Your question is impossible to calculate because of the massive amount of houses in this world. If you want a good estimate, say around 0.9 to 1.8 billion.

6 0

3 years ago

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