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aleksandr82 [10.1K]
2 years ago
15

PLS HELPPPPPPPPPPPPPPPP

Mathematics
1 answer:
seraphim [82]2 years ago
5 0

Answer:

Help is on the way we can rescue you

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You are having a pizza-making party. You will need 6 ounces of dough and 4 ounces of sauce for each person at the party (includi
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Dans la classe de Nina, il y a 12 filles et 13 gar-
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Tyrone makes a scale drawing of his backyard. The scale from the backyard to the drawing is 2 ft to 1 in. The width of the patio
Sphinxa [80]

Answer:

The width of Tyrone's actual patio is 16 feet

Step-by-step explanation:

\frac{2ft}{1in} =  \frac{x}{8}   If we look at what we know, we see that the scale tells us that 1 inch is proportional to 2 ft.  What do we need to multiple 1 inch by to get 8 inches?  8.  So, we multiply 2 feet by 8 to get the actual size of the patio which is 16 feet.

6 0
1 year ago
A card is drawn at random from a deck of 52 playing cards. Find the probability that the card drawn is:
melomori [17]

probability that the card drawn is:

  1. an ace or a king  = 0.1538
  2. a king or a diamond = 0.3269

<u>Step-by-step explanation:</u>

Here we have , A card is drawn at random from a deck of 52 playing cards. We need to find that Find the probability that the card drawn is:

  • an ace or a king

A deck of standard 52 cards contain four aces. There are four kings in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.

probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

⇒ probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

⇒ probability = \frac{8}{52}

⇒ probability = 0.1538

  • a king or a diamond

A deck of standard 52 cards contain four kings. There are 13 Diamonds in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.

probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

⇒ probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

⇒ probability = \frac{(13+4=17)}{52}

⇒ probability = \frac{(17)}{52}

⇒ probability =0.3269

Therefore, probability that the card drawn is:

  1. an ace or a king  = 0.1538
  2. a king or a diamond = 0.3269
8 0
3 years ago
Read 2 more answers
A manufacturing company produces 3 different products A, B, and C. Three types of components, i.e., X, Y, and Z, are used in the
Murljashka [212]

Answer:

Step-by-step explanation:

Using the Excel Formula:

Decision    Variable        Constraint              Constraint

A                     65                          65                         100

B                     80                          80                         80

C                     90                         90                          90

                      14100                    300                        300

= (150 *B3)+(80*B4) +(65*B5)-(100-B3+80-B4+90-B5)*90

Now, we have:

Suppose A, B, C represent the number of units for production A, B, C which is being manufactured

                             A              B                  C                Unit price

Need of X          2                 1                   1                     $20

Need of Y           2                3                  2                    $30

Need of Z           2                2                  3                    $25

Price of  

manufac -      $200          $240            $220      

turing

Now,  for manufacturing one unit of A, we require 2 units of X, 2 units of Y, 2 units of Z are required.  

Thus, the cost or unit of manufacturing of A is:

$20 (2) + $30(2) + $25(2)

$(40 + 60 + 50)

= $150

Also, the market price of A = $200

So, profit = $200 - $150 = $50/ unit of A

Again;

For manufacturing one unit of B, we require 1 unit of X, 3 units of Y, and 2 units of Z are needed and they are purchased at $20, $30, and 425 each.

So, total cost of manufacturing a unit of B is:

= $20(1) + $30(3) + $25(2)

= $(20 + 90+50)

= $160

And the market price of B = $240

Thus, profit = $240- $160  

profit = $80

For manufacturing one unit of C, we have to use 1 unit of X, 2 unit of Y, 3 units of Z are required:

SO, the total cost of manufacturing a unit of C is:

= $20 (1) + $30(2) + $25(3)

= $20 + $60 + $25

= $155

This, the profit = $220 - $155 = $65

However; In manufacturing A units of product A, B unit of product B & C units of product C.

Profit  --> 50A + 80B + 65C

This should be provided there is no penalty for under supply of there is under supply penalty for A, B, C is $40

The current demand is:

100 - A

80 - B

90 - C respectively

So, the total penalty

{(100 - A) + (80 - B) +(90 - C) } + \$40

This should be subtracted from profit.

So, we have to maximize the profit  

Z = 50A + 80B + 65C = {(100 -A) + (80 - B) + (90 - C)};

Subject to constraints;

we have the total units of X purchased can only be less than or equal to 300 due to supplies capacity

Then;

2A + B +C \le 300 due to 2A, B, C units of X are used in manufacturing A, B, C units of products A, B, C respectively.

Next; demand for A, B, C will not exceed 100, 80, 90 units.

Hence;

A \le 100

B \le 80

C \le 90

 

and A, B, C \ge 0 because they are positive quantities

The objective is:

\mathbf{Z = 50A + 80B + 65 C - (100 - A + 80 - B + 90 - C) * 40}

A, B, C \to Decision Varaibles;

Constraint are:

A \le 100 \\ \\  B \le 100 \\ \\ C \le 90 \\ \\2A + B + C \le 300 \\ \\ A,B,C \ge 0

6 0
3 years ago
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