Factorise completely. 4 x y squared

A storage container at the arena is 3 metres wide, 6 metres long and 4 metres high. How many 50 cm by 50 cm by 50 cm boxes could

Ugo [173]

576 boxes can fit into a storage container

5 0

3 years ago

Read 2 more answers

What is the answerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

Alecsey [184]

Answer:

$1559.40

Step-by-step explanation:

1. Find the perimeter of the garden

$14\frac{2}{3} + 14\frac{2}{3} + 10\frac{1}{3} + 10\frac{1}{3} = 50 yd$

2. Convert to feet

1 yd = 3ft

50 x 3 = 150

50 yd = 150 ft

3. Divide by 2.5 to find out how many sections are needed

150/2.5 = 60 sections

4. Multiply by 25.99

60 x 25.99 = 1559.4

It will cost $1559.40

3 0

3 years ago

[GEOMETRY] true or false; All rectangles are parallelograms.

SVETLANKA909090 [29]

Yes, true because all rectangles have 4 sides with opposite sides that are parallel.

6 0

3 years ago

11:16 1

VikaD [51]

Answer:

268

Step-by-step explanation:

Divide the figure into 3 pieces, a triangle, a rectangle, and another rectangle.

Triangle area=18*(15+6+1) * 1/2, or 189

Rectangle 1= 4*11= 44

Rectangle 2= 7*5= 35

Add all up, get 268

3 0

3 years ago

The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below

nlexa [21]

Answer:

The true statement about Class B is that Class B has a smaller median and the same inter quartile range.

Step-by-step explanation:

We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;

Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40

Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42

1) Firstly, we will calculate Median for Class A;

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 27

Hence, the median of class A is 27.

2) Now, we will calculate Median for Class B;

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 26

Hence, the median of class B is 26.

3) Now, we will calculate the Inter quartile range for Class A;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $21+ 0.5[22- 21]$

= 21.5

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $34+ 0.5[35- 34]$

= 34.5

Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.

4) Now, we will calculate the Inter quartile range for Class B;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $18+ 0.5[20- 18]$

= 19

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $30+ 0.5[34- 30]$

= 32

Therefore, Inter quartile range for Class B = 32 - 19 = 13.

Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.

4 0

4 years ago

Factorise completely. 4 x y squared - 6 y cubed