What is the class boundary between the first and second classes?

$Orange\ juice\ concentrate=4\ cups\\\\Water=6\ cups\\\\Tota\ content=4+6=10\ cups\\\\Percentage\ of\ orange\ juice\ concentrate=\frac{Orange\ juice\ concentrate}{Total\ amount}\times 100\\\\Percentage\ of\ orange\ juice\ concentrate=\frac{4}{10}\times 100=40\%$

Second:

$Orange\ juice\ concentrate=1\ cups\\\\Water=3\ cups\\\\Tota\ content=1+3=4\ cups\\\\Percentage\ of\ orange\ juice\ concentrate=\frac{Orange\ juice\ concentrate}{Total\ amount}\times 100\\\\Percentage\ of\ orange\ juice\ concentrate=\frac{1}{4}\times 100=25\%$

Third:

$Orange\ juice\ concentrate=2\ cups\\\\Water=4\ cups\\\\Tota\ content=2+4=6\ cups\\\\Percentage\ of\ orange\ juice\ concentrate=\frac{Orange\ juice\ concentrate}{Total\ amount}\times 100\\\\Percentage\ of\ orange\ juice\ concentrate=\frac{2}{6}\times 100=33.33\%$

First mixed has highest percentage of orange juice. Hence first mixed is the most orangey.

4 0

3 years ago

In how many ways can you spell the word FOOT in the grid below? You can start on any letter F, then on each step you can step on

muminat

There are six ways in which the word FOOT may be spelt starting at any letter.

<h3>What is combination?</h3>

The term combianntion offers a plausible way to carry out a selection as long as we do not give preference to the order in which the selection is made. In this case, we are told that we can start from any letter therefore order is not important.

There are four letters in FOOT and one letter is repeated twice therefore;

4C2 = 4!/(4-2)! 2!

4C2 = 4 * 3 * 2!/2! * 2 * 1

4C2 = 12/2

4C2 = 6

Learn more about combination: brainly.com/question/25351212

7 0

2 years ago

A builder of houses needs to order some supplies that have a waiting time Y for delivery, with a continuous uniform distribution

xenn [34]

Answer:

The Expected cosy of the builder is $433.3

Step-by-step explanation:

$400 is the fixed cost due to delay.

Given Y ~ U(1,4).

Calculating the Variable Cost, V

V = $0 if Y≤ 2

V = 50(Y-2) if Y > 2

This can be summarised to

V = 50 max(0,Y)

Cost = 400 + 50 max(0, Y-2)

Expected Value is then calculated by;

Waiting day =2

Additional day = at least 1

Total = 3

E(max,{0, Y - 2}) = integral of Max {0, y - 2} * ⅓ Lower bound = 1; Upper bound = 4, (4,1)

Reducing the integration to lowest term

E(max,{0, Y - 2}) = integral of (y - 2) * ⅓ dy Lower bound = 2; Upper bound = 4 (4,2)

E(max,{0, Y - 2}) = integral of (y) * ⅓ dy Lower bound = 0; Upper bound = 2 (2,0)

Integrating, we have

y²/6 (2,0)

= (2²-0²)/6

= 4/6 = ⅔

Cost = 400 + 50 max(0, Y-2)

Cost = 400 + 50 * ⅔

Cost = 400 + 33.3

Cost = 433.3

5 0

3 years ago

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