All categories

2 years ago

A board is 3.5 m long. Four pieces, each.4 m long, are cut off. How much is left?

Mathematics

1 answer:

lys-0071 [83]2 years ago

8 0

Answer:

1.9 m

Step-by-step explanation:

3.5 - (4*0.4)

3.5 - 1.6 = 1.9

You might be interested in

Generalize the pattern by finding the nth term.

Veseljchak [2.6K]

Answer:

4n+2

Step-by-step explanation:

The common difference between the terms is 4 which means that the sequence is an arithmetic sequence.

The general formula for arithmetic sequence is:

$a_{n}=a_{0}+(n-1)d$

where a_n is the nth term, a_0 is the first term and d is the common difference between them.

We know that

$a_{0}=6\\d=4$

Putting the value in general formula:

$a_{n}=6+4(n-1)\\a_{n}=6+4n-4\\a_{n}=4n+6-4\\a_{n}=4n+2\\$

So the generalized pattern is 4n+2 ..

6 0

3 years ago

The product of x and y is less than or equal to 4

Trava [24]

Product means multiplication. So the equation would be xy ≤ 4.

3 0

3 years ago

If you are given the following stem and leaf display and asked to construct a frequency distribution chart, what would be the wi

Nuetrik [128]

Answer:

Width of intervals: 8

Step-by-step explanation:

We first look at how data is represented in a stem-leaf diagram.

Any number of the left (before -) is the stem and all numbers on right (after -) are the leaves. Each combination of stem and leaf represents one number. For example: 1 - 332 represents: 13, 13, 12.

Our data is as follows:

13, 13, 12, 24, 25, 31, 31, 35, 37, 42, 43, 41, 52, 51, 51, 52

To calculate the width of the frequency distribution chart, we have the following formula:

$Class\ width = \frac{Range}{Number\ of\ classes}$

The range of any data set = Maximum value in the data set - Minimum value in the data set

Maximum value in this case as seen from the data is 52 and minimum is 12.

Range = 52 - 12 = 40

Since we had only 5 stems in the data, we shall use that as the number of classes required in the frequency distribution chart.

$Class\ width = \frac{40}{5} = 8$

Hence, the class width in this data set will be 8.

To make the intervals, we begin from the minimum value and add 8 to it. The intervals will be:

12 - 20

20 - 28

28 - 36

36 - 44

44 - 52

Observe, that all the values of the stem lie within each interval.

For example, there are 3 values for stem 1: 12, 13, 13 and each lie in the first interval 12 - 20.

Next, the values of stem 2 are 24 and 25. Each of these value lie in the second interval 20 - 28; and henceforth.

8 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

For questions 3-5 solve the equation for the given variable:

lyudmila [28]

1 would be the right answer !!!!!!!!!!!!!!!!!!!!!

3 0

3 years ago

A board is 3.5 m long. Four pieces, each.4 m long, are cut off. How much is left?​

A board is 3.5 m long. Four pieces, each.4 m long, are cut off. How much is left?