An electronics company finds that 5 out of 75 capacitors produced are damaged. How
many damaged items can they expect if the company produces 1,500 capacitors?

According to the given data set:

values10,11,12,13,14,17,17,18,18,19,19,19,21,23,23,25,26,27,27,27,31,32,33,33,36,37,37,38,38more than 20 and less than or equal to 29 values21,23,23,25,26,27,27,27more than 20 and less than or equal to 29 values.

A data set is a group of figures or values related to a specific topic. A data set might include, for instance, each student's test results in a certain class. A data set is the number of fish that each dolphin consumes in an aquarium.

There are eight values that are both bigger than 20 and less than 29.The data collection has 12 values that are both higher than or equal to 30 and lower than or equal to 39.Values greater than nor equal to 30 and less than or equal to 3931,32,33,33,35,36,37,37,37,38,38The data collection has 12 values that are both higher than or equal to 30 and lower than or equal to 39.

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Answer:

7 no =option B

8 no =option

9 no=option D

10 no=option c

Step-by-step explanation:

The correct answer for part c is option d: (XA1 XOR XA2). This Boolean expression uses XOR (exclusive OR) which means that either XA1 = 1 and XA2 = 0, or XA1 = 0 and XA2 = 1,

Indicating class A is scheduled during exactly one of the two periods.
For part d, the correct answer is option a: (XA1 XOR XB1)(XA2 XOR XB2). This expression ensures that classes A and B are not scheduled during the same period.

If A is in period 1 and B is not, or vice versa (XA1 XOR XB1 = 1), and if A is in period 2 and B is not, or vice versa (XA2 XOR XB2 = 1), then the overall expression will be true.

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Answer:

Step-by-step explanation:

circumference is 2πr

r=d/2

r=4.46m

C=2π(4.46)=28.02

Answer:

a

Step-by-step explanation:

i got this question and got a 100 on my quiz

Jennifer has 48 nickels, which means she has 144 quarters.

Let's call the number of nickels Jennifer has "n".

Since Jennifer has three more dimes than nickels, she has n + 3 dimes.

And since she has three times as many quarters as nickels, she has 3n quarters.

We can start by figuring out the total value of the nickels and dimes in terms of cents:

n * 5 cents + (n + 3) * 10 cents = 750 cents

Expanding the right side of the equation and solving for n:

5n + 10n + 30 cents = 750 cents

15n + 30 cents = 750 cents

15n = 720 cents

n = 48

So Jennifer has 48 nickels, which means she has 3 * 48 = 144 quarters.

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The equation represents the total cost of turkey and amount is y= 3.25x.

We have,

The turkey costs $3. 25 per pound.

let x be the amount in pounds and y be the total cost in dollar.

Then, the relation between x and y

Total cost = 3.25 (amount in  poind)

y = 3.25 x

Thus, the required equation is y= 3.25x.

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Answer:

1. 7+62+5+4

2. 18+62+5

3.15+4+1

4.5 families

6. three

Step-by-step explanation:

Answer:

The coordinates of the vertices are ′(,−), e prime . open 1 comma negative 3 close comma ′(,−), f prime . open 5 comma negative 2 close comma and ′(,−). g prime . open 4 comma negative 8 close .

Step-by-step explanation:

The breadth of the rectangle is 6 cm.

To find the breadth, we can use the formula for the perimeter of a rectangle:

Perimeter = 2 × (length + breadth)

Substituting the given values, we get:

18 cm = 2 × (3 cm + breadth)

Simplifying the right side of the equation, we get:

18 cm = 6 cm + 2 × breadth

Subtracting 6 cm from both sides of the equation, we get:

12 cm = 2 × breadth

Dividing both sides by 2, we get:

breadth = 6 cm

Therefore, the breadth of the rectangle is 6 cm.

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Answer:

80.

Step-by-step explanation:

When multiplying if the tenth place is " 7" then we look " next door" which is the ones and 8 is great then 5 so we count to from 8 to see when the next number ending in 0 would be in this case its 80. ( if the number is 4 or less count backwards to a number ending with 0  which in that case would have been 70 ) but the answer is 80.

Answer:

  C.  142°

Step-by-step explanation:

You want the angle between vectors u=3i+√3j and v=-2i-5j.

There are a number of ways the angle between the vectors can be found. For example, the dot-product relation can give you the cosine of the angle:

  u•v = |u|·|v|·cos(θ) . . . . . . where θ is the angle of interest

You can find the angles of the vectors individually, and subtract those:

  u = |u|∠α

  v = |v|∠β

  θ = α - β

When the vectors are expressed as complex numbers, the angle between them is the angle of their quotient:

  \(\dfrac{\vec{u}}{\vec{v}}=\dfrac{|\vec{u}|\angle\alpha}{|\vec{v}|\angle\beta}=\dfrac{|\vec{u}|}{|\vec{v}|}\angle(\alpha-\beta)=\dfrac{|\vec{u}|}{|\vec{v}|}\angle\theta\)

This method is used in the calculation shown in the first attachment. The angle between u and v is about 142°.

A graphing program can draw the vectors and measure the angle between them. This is shown in the second attachment.

__

Additional comment

The approach using the quotient of the vectors written as complex numbers is simply computed using a calculator with appropriate complex number functions. There doesn't seem to be any 3D equivalent.

The dot-product relation will work with 3D vectors as well as 2D vectors.

<95141404393>

Step-by-step explanation:

I can use ur calculator to get the co ordinates

The two possible answer is  3(5.5a - 2.4b + 12) that Laine could have written for expression 16.5a - 77.2b + 36 .

Laine could just just used another common number to write it down as,

LCM(165, 36, 72) = 3960

Prime factorization of the numbers:

165 = 3 × 5 × 11

36 = 2 × 2 × 3 × 3

72 = 2 × 2 × 2 × 3 × 3

LCM(165, 36, 72)

= 2 × 2 × 2 × 3 × 3 × 5 × 11

= 3960

GCD(165, 36, 72) = 3

So we get three as a common multiple of them so after removing 3 common from them we can write 16.5a - 7.2b + 36 as

3(5.5a - 2.4b + 12)  

Hence, The two possible answer is  3(5.5a - 2.4b + 12) that Laine could have written for expression 16.5a - 77.2b + 36.

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Answer:Mate fank yew mate

Step-by-step explanation:

A gardener is filling 5 planter boxes (cubical pots) with potting soil. Each box is 1.5 feet long on each side. The

gardener has 3 bags of potting soil, each with a volume of 5 cubic feet.

Is there enough potting soil for the 5 planter boxes?

O There is not enough potting soil. The gardener needs 1.875 cubic feet more potting soil.

O There is enough potting soil. The gardener needs only 3.375 cubic feet of potting soil.

O There is not enough potting soil. The gardener needs 3.75 cubic feet more potting soil.

O There is enough potting soil. The gardener needs only 4.5 cubic feet of potting soil.

O There is not enough potting soil. The gardener needs 7.5 cubic feet more potting soil.

Answer:



Step-by-step explanation:



Answer:

k= negative 4

Step-by-step explanation:

k=

Answer:

C

Step-by-step explanation:

8^2-3^2=x^2

root x= root 55

Answer:

44 degrees.

Step-by-step explanation:

The 3 small triangles are isosceles  so m < a = m < b , therefore

m < a = (180 - 92) / 2 = 44 degrees.

m < f = m < a ( they are both subtended on the circle by the same chord (cd)),

So m < f = 44.

Answer:

y = 12/5 = x + 7

hope this helped

lol

Answer:

Step-by-step explanation:

a.)

\(4880.76=4000(1+.04/4)^{4x}\\\\1.22019=1.01^{4x}\\\frac{\ln{1.22019}}{\ln{1.01}}=4x\\x= 4.999999= 5\)

b.)

\(5395.4=4000(1+x/12)^{12*5}\\1.34885=(1+x/12)^{60}\\\sqrt[60]{1.34885} =1+x/12\\x= 0.0599999772677= .06\)

c.)

\(\i)\\100*(1+.1255)= 112.55\\\\2)\\100*(1+.1218/2)^2= 112.550881= 112.55\)

Answer:

B

Step-by-step explanation:

using the identity

sin²x + cos²x = 1 ( subtract 1 from both sides )

sin²x + cos²x - 1 = 0 ( subtract sin²x from both sides )

cos²x - 1 = - sin²x

then

\(\frac{cos^20-1}{4sin^20}\)

= \(\frac{-sin^20}{4sin^20}\) ← cancel sin²Θ on numerator and denominator

= \(\frac{-1}{4}\)

= - \(\frac{1}{4}\)

Answer:3

Step-by-step explanation:

Answer:

3

Step-by-step explanation:

CLT

5+7= 4a

12=4a

Divide through by 4

12÷4= 3

Answer: 5c represents the cost of 'c' large cheese pizzas and 6p represents the cost of 'p' large pepperoni pizzas.

Step-by-step explanation:

Given: Large cheese pizzas cost $5 each and large pepperoni pizzas cost $6 each.

The expression 5c+6p models the total cost of large cheese pizzas.

here c represents Number of large cheese pizzas and p represents number of  large pepperoni pizzas such that

5c represents the cost of 'c' large cheese pizzas and 6p represents the cost of 'p' large pepperoni pizzas.

[Total cost = Cost of each item x Number of items ]

Answer:

2.3=v+0.47v=

We move all terms to the left:

2.3-(v+0.47v)=0

We add all the numbers together, and all the variables

-(+1.47v)+2.3=0

We get rid of parentheses

-1.47v+2.3=0

We move all terms containing v to the left, all other terms to the right

-1.47v=-2.3

v=-2.3/-1.47

v=1+0.83/1.47

Answer:

\(v = 1.564\)

Step-by-step explanation:

\(1v + 0.47v = 2.3\)

\( = > 1.47v = 2.3\)

\( = > v = \frac{2.3}{1.47} = 1.564\)

Answer:

b

Step-by-step explanation:

Answer:

d Because I just took the test and got it right

Answer: 6000 millimeters

Step-by-step explanation:

Basically multiply 6 by 100 because there are 1000 millimeters in 1 liter.

       For C1(Q) = 3 - 2Q.

        For C2(Q) =  4.

     2. The AVC function

           For C1(Q) = 5/Q + 3 + 20/Q - Q.

           For C2(Q) = 3/Q + 4 + 2/Q.

     3. The AFC function

              For C1(Q)= 5/Q - 20/(5 + 3Q + 20/Q - Q)

              For C2(Q) = 0.

     4.  To find the AC function

        For C1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

         For C2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

    5.To find the ATC function

    For C1(Q)= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²)

     For C2(Q)= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

1. To find the MC function, we take the derivative of the cost functions with respect to Q.

For C1(Q) = 5 + 3Q + 202 - Q^2, MC1(Q) = 3 - 2Q.

For C2(Q) = 3 + 4Q + 2, MC2(Q) = 4.

2. To find the AVC function, we divide the cost functions by Q.

For C1(Q), AVC1(Q) = (5 + 3Q + 202 - Q^2)/Q = 5/Q + 3 + 20/Q - Q.

For C2(Q), AVC2(Q) = (3 + 4Q + 2)/Q = 3/Q + 4 + 2/Q.

3. To find the AFC function, we subtract the AVC function from the ATC function.

For C1(Q), AFC1(Q) = (5 + 3Q + 202 - Q^2)/Q - (5 + 3Q + 202 - Q^2)/(5 + 3Q + 20/Q - Q)

= 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AFC2(Q) = (3 + 4Q + 2)/Q - (3 + 4Q + 2)/(3/Q + 4 + 2/Q) = 0.

4. To find the AC function, we add the AVC function to the AFC function.

For

C1(Q), AC1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AC2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

5. To find the ATC function, we divide the AC function by Q.

For

C1(Q), ATC1(Q) = [(5 + 3Q + 202 - Q²)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q)]/Q

= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²).

For

C2(Q), ATC2(Q) = [(3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q]/Q

= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

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(i) Euler's method: At x = 0.05, y ≈ 1.05, At x = 0.1, y ≈ 1.107, At x = 0.15, y ≈ 1.174

(ii) Fourth-order Runge-Kutta method: At x = 0.05, y ≈ 1.05 , At x = 0.1, y ≈ 1.107, At x = 0.15, y ≈ 1.174

(i) Euler's Method:

Euler's method is a simple numerical method for approximating solutions to differential equations. It uses the formula:

yᵢ₊₁ = yᵢ + h * f(xᵢ, yᵢ),

where yᵢ is the approximate value of y at xᵢ and f(x, y) is the given differential equation.

Let's start by calculating the approximate solution using Euler's method.

Step size, h = 0.05

At x = 0.05:

x₀ = 0

y₀ = 1

Using the given differential equation: dy/dx = 2x + eˣʸ

We have: f(x, y) = 2x + eˣʸ

Using Euler's method:

x₁ = x₀ + h = 0 + 0.05 = 0.05

y₁ = y₀ + h * f(x₀, y₀)

   = 1 + 0.05 * (2(0) + e^(0*1))

   = 1 + 0.05 * (0 + e^0)

   = 1 + 0.05 * (0 + 1)

   = 1 + 0.05

   = 1.05

At x = 0.1:

x₂ = x₁ + h = 0.05 + 0.05 = 0.1

y₂ = y₁ + h * f(x₁, y₁)

 \(= 1.05 + 0.05 * (2(0.05) + e^(0.05*1.05)) = 1.05 + 0.05 * (0.1 + e^(0.0525))\)

   = 1.05 + 0.05 * (0.1 + 1.05383959925)

   = 1.05 + 0.05 * (1.15383959925)

   = 1.107691979963

At x = 0.15:

x₃ = x₂ + h = 0.1 + 0.05 = 0.15

y₃ = y₂ + h * f(x₂, y₂)

  \(= 1.107691979963 + 0.05 * (2(0.1) + e^(0.1*1.107691979963)) = 1.107691979963 + 0.05 * (0.2 + e^(0.1107691979963))\)

   = 1.107691979963 + 0.05 * (0.2 + 1.12281597877)

   = 1.107691979963 + 0.0661407989385

   = 1.173832778901

Approximate solution using Euler's method:

At x = 0.05, y ≈ 1.05

At x = 0.1, y ≈ 1.107

At x = 0.15, y ≈ 1.174

(ii) Fourth-order Runge-Kutta Method:

The fourth-order Runge-Kutta method is a more accurate numerical method

for approximating solutions to differential equations. It uses the formula:

k₁ = h * f(xᵢ, yᵢ)

k₂ = h * f(xᵢ + h/2, yᵢ + k₁/2)

k₃ = h * f(xᵢ + h/2, yᵢ + k₂/2)

k₄ = h * f(xᵢ + h, yᵢ + k₃)

yᵢ₊₁ = yᵢ + (k₁ + 2k₂ + 2k₃ + k₄)/6

Let's calculate the approximate solution using the fourth-order Runge-Kutta method.

Step size, h = 0.05

At x = 0.05:

x₀ = 0

y₀ = 1

k₁ = 0.05 * f(x₀, y₀)

   = 0.05 * (2(0) + e^(0*1))

   = 0.05 * (0 + e^0)

   = 0.05 * (0 + 1)

   = 0.05

k₂ = 0.05 * f(x₀ + 0.05/2, y₀ + k₁/2)

   = 0.05 * f(0 + 0.025, 1 + 0.025/2)

   = 0.054123529771

k₃ = 0.05 * f(x₀ + 0.05/2, y₀ + k₂/2)

   = 0.05 * f(0 + 0.025, 1 + 0.054123529771/2)

   = 0.05 * (1.08260731316)

   = 0.054130365658

k₄ = 0.05 * f(x₀ + 0.05, y₀ + k₃)

   = 0.05 * f(0 + 0.05, 1 + 0.054130365658)

   = 0.05 * (1.15381700056)

   = 0.057690850028

y₁ = y₀ + (k₁ + 2

k₂ + 2k₃ + k₄)/6

   = 1 + (0.05 + 2(0.054123529771) + 2(0.054130365658) + 0.057690850028)/6

   = 1 + (0.05 + 2(0.054123529771) + 2(0.054130365658) + 0.057690850028)/6

   = 1.04999993086

At x = 0.1:

x₁ = x₀ + h = 0 + 0.05 = 0.05

k₁ = 0.05 * f(x₀, y₁)

   = 0.05 * (2(0.05) + e^(0.05*1.04999993086))

   = 0.05 * (1.15382091332)

   = 0.057691045666

k₂ = 0.05 * f(x₀ + 0.05/2, y₁ + k₁/2)

   = 0.05 * f(0 + 0.025, 1.04999993086 + 0.057691045666/2)

   = 0.05 * f(0.025, 1.076845501196)

   = 0.05 * (1.08293606756)

   = 0.054146803378

k₃ = 0.05 * f(x₀ + 0.05/2, y₁ + k₂/2)

   = 0.05 * f(0 + 0.025, 1.04999993086 + 0.054146803378/2)

   = 0.05 * (0.05 + 1.03297648754)

   = 0.05 * (1.08297648754)

   = 0.054148824377

k₄ = 0.05 * f(x₀ + 0.05, y₁ + k₃)

   = 0.05 * f(0 + 0.05, 1.04999993086 + 0.054148824377)

   = 0.05 * f(0.05, 1.104148755237)

   = 0.

05 * (2(0.05) + e^(0.05*1.104148755237))

   = 0.05 * (1.15636272702)

   = 0.057818136351

y₂ = y₁ + (k₁ + 2k₂ + 2k₃ + k₄)/6

   = 1.04999993086 + (0.057691045666 + 2(0.054146803378) + 2(0.054148824377) + 0.057818136351)/6

   = 1.04999993086 + (0.057691045666 + 2(0.054146803378) + 2(0.054148824377) + 0.057818136351)/6

   = 1.10730401829

At x = 0.15:

x₂ = x₁ + h = 0.1 + 0.05 = 0.15

k₁ = 0.05 * f(x₁, y₂)

   = 0.05 * (2(0.1) + e^(0.1*1.10730401829))

   = 0.066134153561

k₂ = 0.05 * f(x₁ + 0.05/2, y₂ + k₁/2)

   = 0.05 * f(0.1 + 0.025, 1.10730401829 + 0.066134153561/2)

   = 0.05 * (1.40252551209)

   = 0.070126275604

k₃ = 0.05 * f(x₁ + 0.05/2, y₂ + k₂/2)

   = 0.05 * f(0.1 + 0.025, 1.10730401829 + 0.070126275604/2)

   = 0.05 * f(0.125, 1.13726868037)

   = 0.05 * (1.40260973286)

   = 0

.070130486643

k₄ = 0.05 * f(x₁ + 0.05, y₂ + k₃)

   = 0.05 * f(0.1 + 0.05, 1.10730401829 + 0.070130486643)

   = 0.05 * (1.49307437508)

   = 0.074653718754

y₃ = y₂ + (k₁ + 2k₂ + 2k₃ + k₄)/6

   = 1.10730401829 + (0.066134153561 + 2(0.070126275604) + 2(0.070130486643) + 0.074653718754)/6

   = 1.17379707103

Approximate solution using the fourth-order Runge-Kutta method:

At x = 0.05, y ≈ 1.05

At x = 0.1, y ≈ 1.107

At x = 0.15, y ≈ 1.174

Therefore, the approximate solutions using Euler's method and the fourth-order Runge-Kutta method at x = 0.05, 0.1, and 0.15 are as follows:

Euler's method:

At x = 0.05, y ≈ 1.05

At x = 0.1, y ≈ 1.107

At x = 0.15, y ≈ 1.174

Fourth-order Runge-Kutta method:

At x = 0.05, y ≈ 1.05

At x = 0.1, y ≈ 1.107

At x = 0.15, y ≈ 1.174

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