HELP PLEASE I HAVE 10 MINS

A dresser contains six pairs of shorts one each in the colors, red, black, blue, green, khaki, and gray. The dresser also contai

marissa [1.9K]

Answer: 1/24

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Work Shown:

A = selects green pair of shorts
B = selects gray t-shirt

P(A) = probability of selecting green shorts
P(A) = (number of green shorts)/(number of shorts total)
P(A) = 1/6
P(B) = probability of selecting gray t-shirt
P(B) = (number of gray t-shirts)/(number of t-shirts total)
P(B) = 1/4

P(A and B) = probability of selecting green shorts AND gray t-shirt
P(A and B) = P(A)*P(B) ... since A and B are independent events
P(A and B) = (1/6)*(1/4)
P(A and B) = (1*1)/(6*4)
P(A and B) = 1/24

Note: The fraction 1/24 is approximately equal to 0.041667

5 0

3 years ago

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

Which of the following diagrams represents 80%?<br><br>Help please. it'd be very appreciated. :)

Alenkasestr [34]

Answer:

H

Step-by-step explanation:

there is 10 parts to the shape so your denominator would be 10 then if you count the shaded area there's 8, so 8 would be your numerator

8/10 = 0.8 or 80%

5 0

3 years ago

The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is

bulgar [2K]

The integral is approximated by the sum,

$\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))$

where $f(x)=x^3$ and $x_n=\dfrac12n$ , giving you

$\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)$
$\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)$
$\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)$

Faulhaber's formulas make short work of computing the sum. You have

$\displaystyle\sum_{n=0}^k1=k+1$
$\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2$
$\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6$
$\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4$

which gives

$\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31$
$\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}$
$\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25$

4 0

3 years ago

Solve for x using common denominators. 3/4(x + 2)=x/(x+2)

IceJOKER [234]

Answer:

x = 3/4

Step-by-step explanation:

The lowest common denominator is 4(x + 2) so we multiply each side by this:

3 = 4x

x = 3/4.

4 0

3 years ago

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