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Dmitry_Shevchenko [17]

2 years ago

8

A rectangle has area 64 m2. Express the perimeter of the rectangle as a function of the length L of one of its sides. State the

domain of P

1 answer:

svp [43]2 years ago

3 0

A rectangle is a two-dimensional shape with two sets of equal, parallel sides. These dimensions are the length (L) and the width (W). The formula for the rectangle's area is the product of the two dimensions. The formula for perimeter is

P = 2L + 2W

Since
A = LW = 64
W = 64/L
Substituting to the formula for perimeter would be,

P = 2L + 2(64/L)
P = 2L + 128/L

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This background applies to the next several questions. Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has

nalin [4]

Answer:

1) $Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

2) $Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

3) $Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

And for the new yield we need to take in count the increase of 15% for the area and we got this:

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

4) $DR_{old}=\frac{1}{\sqrt{0.92}} -1$ =0.0426 defects/cm^2

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$ =0.0260defects/cm^2

Step-by-step explanation:

Part 1

For this part first we need to find the die areas with the following formula:

$Area= \frac{W area}{Number count}$

$Area_1 = \frac{\pi (7.5cm)^2}{84}=2.104 cm^2$

$Area_2 = \frac{\pi (10cm)^2}{100}=3.1415 cm^2$

Now we can use the yield equation given by:

$Yield=\frac{1}{(1+ DR\frac{Area}{2})^2}$

And replacing we got:

$Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

Part 2

For this part we can use the formula for cost per die like this:

$Cost/die = \frac{Cost per day_i}{Number count_i x Yield_i}$

And replacing we got:

$Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

Part 3

For this case we just need to calculate the new area and the new yield with the same formulas for part a, adn we got:

$Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

And for the new yield we need to take in count the increase of 15% for the area and we got this:

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

Part 4

First we can convert the area to cm^2 and we got 2 cm^2 the yield would be on this case given by:

$Yield= \frac{1}{(1+DR\frac{2cm^2}{2})^2}=\frac{1}{1+(DR)^2}$

And if we solve for the Defect rate we got:

$DR= \frac{1}{\sqrt{Yield}}-1$

Now we can find the previous and new defect rate like this:

$DR_{old}=\frac{1}{\sqrt{0.92}} -1$ =0.0426 defects/cm^2

And for the new defect rate we got:

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$ =0.0260defects/cm^2

5 0

2 years ago

A restaurant serves custom-made omelets, where guests select meat, cheese, and vegetables to be added to their omelet. There are

kondor19780726 [428]

Answer: The number of different combinations of 2 vegetables are possible = 15 .

Step-by-step explanation:

In Mathematics , the number of combinations of selecting r values out of n values = $^nC_r=\dfrac{n!}{r!(n-r)!}$

Given : Number of available vegetables = 6

Then, the number of different combinations of 2 vegetables are possible will be :

$^6C_2=\dfrac{6!}{2!(6-2)!}=\dfrac{6\times5\times4!}{2\times4!}=15$

Hence , the number of different combinations of 2 vegetables are possible = 15 .

5 0

2 years ago

Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e

Komok [63]

Answer:

0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost

Step-by-step explanation:

Independent events:

If two events, A and B, are independent, then:

$P(A \cap B) = P(A)*P(B)$

Conditional probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: At least one tag is lost

Event B: Exactly one tag is lost.

Each tag has a 40% = 0.4 probability of being lost.

Probability of at least one tag is lost:

Either no tags are lost, or at least one is. The sum of the probabilities of these events is 1. Then

$p + P(A) = 1$

p is the probability none are lost. Each one has a 60% = 0.6 probability of not being lost, and they are independent. So

p = 0.6*0.6 = 0.36

Then

$P(A) = 1 - p = 1 - 0.36 = 0.64$

Intersection:

The intersection between at least one lost(A) and exactly one lost(B) is exactly one lost.

Then

Probability at least one lost:

First lost(0.4 probability) and second not lost(0.6 probability)

Or

First not lost(0.6 probability) and second lost(0.4 probability)

So

$P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48$

Find the probability that exactly one tag is lost, given that at least one tag is lost (write it up to second decimal place).

$P(B|A) = \frac{0.48}{0.64} = 0.75$

0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost

8 0

2 years ago

If h(x) is the inverse of f(x), what is the value of h(f(x))? 0 1 x f(x) A.0 B.1 C.X D.f(x)

Semmy [17]

If f and h are inverses, then:

f (h(x) ) = x and also

h( f(x) ) = x

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Dorothy and Gray live 8 miles from each other.

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2 years ago