# Manuel ate 1 and 3 of a crackers on a plate. his brother ate 1 and 4 of the crackers.there where 5 crackers left on the plate.how many crackers were on the plate to begin with?

1/3 + 1/4 = 4/12 + 3/12 = 7/12

7/12 =  both brothers ate 7 crackers out of 12 crackers

there were 5 crackers left on the plate means 5/12 crackers left on the plate

answer: 12 crackers at the beginning

Step-by-step explanation:

## Related Questions

6x

5
y
=
14

Find
x
when
y
=
2

4

Step-by-step explanation:

Line segment of length k is divided into 3 equal parts. What is distance between midpoints of first and third segments?A) 2k/3 B) k C) k/6 D) 2k

The distance between the midpoints of the first segment and the third segment is 2k/3. Hence, option A is the right choice.

### What is the mid-point of a line segment?

The mid-point of a line segment is the point from which the distance to both ends of the line segment is equal.

### How to solve the question?

In the question, we are given a line segment of length k units, which is divided into 3 equal parts.

We are asked to find the distance between the midpoints of the first and third segments.

Firstly, we divide the line segment at points k/3 and 2k/3, to get three equal parts of lengths k/3 each.

Now, the mid-point of the first segment = (0 + k/3)/2 = k/6.

The mid-point of the third segment = (2k/3 + k)/2 = 5k/6

Therefore, the distance between the midpoints of the first segment and the third segment is (5k/6 - k/6) = 4k/6 = 2k/3. Hence, option A is the right choice.

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Line segment of length k is divided into 3 equal parts.

so first segment is 0-k/3 and third segment is 2/3k-k

so mid-pt of 1st = k/6 and 3rd = 5/6k

so the distance in between = 5/6k-k/6 = 4/6k = 2/3k

ans is A

0.78 of air is nitrogen and 0.93% is argon. Is there more nitrogen or argon in air?

There is more argon in the air.

Sanderson Manufacturing produces ornate, decorative wood frame doors and windows. Each item produced goes through three manufacturing processes: cutting, sanding, and finishing. Each door produced requires 1 hour in cutting, 30 minutes in sanding, and 30 minutes in finishing. Each window requires 30 minutes in cutting, 45 minutes in sanding, and 1 hour in finishing. In the coming week Sanderson has 40 hours of cutting capacity available, 40 hours of sanding capacity, and 60 hours of finishing capacity. Assume all doors produced can be sold for a profit of \$500 and all windows can be sold for a profit of \$400.Required:
a. Formulate an LP model for this problem.
b. Sketch the feasible region.
c. What is the optimal solution?

Let X1 be the number of decorative wood frame doors and X2 be the number of windows.

The profit earned from selling each door is \$500 and the profit earned from selling of each window is \$400.

The Sanderson Manufacturer wants to maximize their profit. So for this model, the objective function is

Max: 500X1 + 400X2

Now the total time available for cutting of door and window are 2400 minutes.

so the time taken in cutting should be less than or equal to 2400.

60X1 + 30X2 ≤ 2400

The total available time for sanding of door and window are 2400 minutes. Therefore, the time taken in sanding will be less than or equal to 2400.   30X1 + 45X2 ≤ 2400

The total time available for finishing of door and window is 3600 hours. Therefore, the time taken in finishing will be less than or equal to 3600. 30X1 + 60X2 ≤ 3600

As the number of decorative wood frame door and the number of windows cannot be negative.

Therefore, X1, X2 ≥ 0

so the questions

a)

The LP mode for this model is;

Max: 500X1 + 400X2

Subject to:

60X1 + 30X2 ≤ 2400

]30X1 +45X2 ≤ 2400

30X1 + 60X2 ≤ 3600

X1, X2 ≥ 0

b) Plot the graph of the LP

Max: 500X1+ 400X2

Subject to:

60X1 + 30X2 ≤ 2400

30X1 + 45X2 ≤ 2400

30X1 + 60X2 ≤ 3600

X1,X2

≥ 0

In the uploaded image of the graph, the shaded region in the graph is the feasible region.

c) Consider the following corner point's (0,0), (0, 53.33), (20, 40) and (40, 0) of the feasible region from the graph

At point (0, 0), the objective function,

500X1 + 400X2 = 500 × 0 + 400 × 0

= 0

At point (0, 53.33), the value of objective function,

500X1 + 400X2 = 500 × 0 + 400 × 53.33 = 21332

At point (40, 0), the value of objective function,

500X1 + 400X2 = 500 × 40 + 400 × 0 = 20000

At point (20, 40), the value of objective function

500X1 + 400X2 = 500 × 20 + 400 × 40 = 26000

The maximum value of the objective function is

26000 at corner point ( 20, 40 )

Hence, the optimal solution of this problem is

X1 = 20, X2 = 40 and the objective is 26000

Penny reads 13 pages in 1/2 hour. What is the unit rate for pages per hour? For hours per page? How many hours per page

13 pages : 1/2 hour Multiply both sides by 2

26 pages : 1 hour ------>> 26 pages/hr

Flip over 26 pages/hr ----->>> hr / 26 pages = 1 hr / 26 pages

Split up the fraction: (1/26) hr / page

26

Step-by-step explanation:

13 in 1/2 hour

13:1/2

26:1

There are 3 bags each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles. Bag 3 has 45 red and 55 blue marbles. Now a bag is chosen at random and a marble is also picked at random. 1) What is the probability that the marble is blue? 2) What is the probability that the marble is blue when the first bag is chosen with probability 0.5 and other bags with equal probability each? Make sure to clearly define your probabilistic events and mathematically show how different probability laws and rules that you learned in class could be applied to solve the problems.

0.4 ; 0.6125

Step-by-step explanation:

Given the following :

Bag 1 : 75 red ; 25 blue

Bag 2: 60 red ; 40 blue

Bag 3: 45 red ; 55 blue

Probability = (required outcome / Total possible outcomes)

A) since the probability of choosing each bag is equal :

BAG A:

P(choosing bag A) = 1 / total number of bags = 1/3 ; P(choosing blue marble) = number of blue marbles / total number of marbles = 25/100

HENCE, choosing a blue marble from bag A : = (1/3 × 75/100) = 25/300

BAG B:

P(choosing bag B) = 1/3 ;

P(choosing blue marble) = number of blue marbles / total number of marbles = 40/100

HENCE, choosing a blue marble from bag A : = (1/3 × 40/100) = 40/300

BAG C:

P(choosing bag C) = 1/3

P(choosing blue marble) = number of blue marbles / total number of marbles = 55/100

HENCE, choosing a blue marble from bag A : = (1/3 × 55/100) = 55/300

= (25/300) × (40/300) × (55/300) = (25 + 40 + 55)/300 = 120/300 = 0.4

2) What is the probability that the marble is blue when the first bag is chosen with probability 0.5 and other bags with equal probability each?

BAG A:

P(choosing bag A) = 0.5 ; P(choosing blue marble) = number of blue marbles / total number of marbles = 25/100

HENCE, choosing a blue marble from bag A : = (0.5 × 75/100) = (0.5 * 0.75) = 0.375

BAG B:

P(choosing bag B) = (1-0.5) / 2 = 0.25 ;

P(choosing blue marble) = number of blue marbles / total number of marbles = 40/100

HENCE, choosing a blue marble from bag A : = (0.25 × 40/100) = (0.25 × 0.4) = 0.1

BAG C:

P(choosing bag C) = (1 - (0.5+0.25)) = 0.25

P(choosing blue marble) = number of blue marbles / total number of marbles = 55/100

HENCE, choosing a blue marble from bag A : = (0.25 × 55/100) = 0.25 × 0.55 = 0.1375

= 0.1375 + 0.1 + 0.375 = 0.6125