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3 years ago

if you have 24 square piece of wood describe all the different ways you can make good rectangle by placing them side-by-side

1 answer:

3 0

This is an exercise you should actually do with real physical objects. The objects don't have to be wood and they don't have to be square for you to be able to arrange them into a rectangular shape. You can cut squares from paper, or use Scrabble™ tiles, poker chips, pennies, M&M™ candies, bottle caps, Legos™, Cheez-It™ crackers, pebbles, or any other objects of similar size and shape.

What you will find is that there are 8 possibilities, corresponding to the number of divisors of 24:

1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1

_____

If you make rectangles from numbers of pieces other than 24, you will find that only two rectangles are possible for prime numbers, like 23: 1×23 or 23×1. Even numbers of tiles will all have rectangles of 1×_ and 2×_ (as well as _×1 and _×2).

Doing this sort of playing with physical objects helps you develop number sense that can serve you well later. Don't discount it.

You might be interested in

In a math competition, every question that is answered correctly gets a score of 4. Wrong gets a score of -2. and not answered g

11111nata11111 [884]

Answer:

Step-by-step explanation:

First we need to find the score of the incorrect answers and the questions that were not answered.

Score for the incorrect answers = 4*(-2) = -8

Score for the questions not answered = 6*(-1) = -6

Score for the questions answered correctly = 106 - [ (-8) + (-6) ]

= 106 -[-14]

= 106 + 14

= 120

Number of questions answered correctly = total score for questions that were answered correctly ÷ score of 1 correct answer

= 120 ÷ 4

= 30

30 questions were answered correctly.

7 0

2 years ago

Please help and I need EXPLANATION too

lys-0071 [83]

a because

$x + 1 \geqslant 0 \\ \: \: \: - 1 \: \: \: \: - 1 \\ x \geqslant - 1$

4 0

3 years ago

question 31 pts a fair six-sided die is tossed 300 times. find the expected value of the sample mean of the values obtained in t

mojhsa [17]

The expected value of the sample mean of the values obtained in these 300 tosses is 3.5

Given :

A fair six sided die is tossed 300 times

We have to find the expected value of the sample mean of the values obtained in these 300 tosses.

A dice have six sides:

In a fair dice every side has equal chance of occurrence,

Now the probability of presence of each side = 1/6

Now the expected value of the sample mean of the values obtained in these 300 tosses is given by:

E(x) = Σx P(x =x)

= 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6

= 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6

= 1/6 (1+2+3+4+5+6)

= 1/6 (21)

= 21/6

E(x) = 3.5

Hence, the expected value of the sample mean of the values obtained in these 300 tosses is 3.5

To know more about sample mean check the below link:

brainly.com/question/29739571

#SPJ4

3 0

2 years ago

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )

P ( x ≥ 6) = 1 - 0.9707

P ( x ≥ 6) = 0.0293

4 0

3 years ago

How many pounds of raisins were in her supply when she started?

sp2606 [1]

Answer:

i think it's 15 because if you use 5 and you have to do it by 3 it should be 15

8 0

3 years ago