In a factory, a sample of 40 light bulbs was selected from a box of 1,000 light bulbs. In the sample, 2 light bulbs would not work. Based on this sample, how many non-working light bulbs would be expected in the group of 1,000 light bulbs?

The expected number of hits

= [tex]\sum xP(X=x) = 0.96[/tex]

Let X be a Binomial random variable that denotes the number of hits

where the parameters are n = 3, p = 0.320

P(X = x) = [tex]\binom{3}{x}*0.32^{x}*0.68^{3-x} , x = 0,1,2,3[/tex]

(a) The probability distribution of X is

P(X = 0) = 0.3144

P(X = 1) = 0.4439

P(X = 2) = 0.2089

P(X = 3) = 0.0328

X 0 1 2 3

P(X = x) 0.3144 0.4439

0.2089

0.0328

(c) Expected number of hits

= [tex]\sum xP(X=x) = 0.96[/tex]

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The three different values of n to demonstrate that 2n + 3n is equivalent to 5n are 1, 4 and 10.

Given data ,

Let the equation be represented as A

Now , the value of A is

Substituting the values in the equation , we get

A = 2n + 3n

On simplifying the equation , we get

A = 5n

when n = 1 , we get

2n + 3n = 2(1) + 3(1) = 2 + 3 = 5

5n = 5(1) = 5

Therefore, both expressions give the same result of 5.

when n = 4 , we get

2n + 3n = 2(4) + 3(4) = 8 + 12 = 20

5n = 5(4) = 20

Again, both expressions give the same result of 20.

when n = 10 , we get

2n + 3n = 2(10) + 3(10) = 20 + 30 = 50

5n = 5(10) = 50

Hence , the equation is solved

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Class 1 wiring is required to exceed standards for power and lighting wiring. Class 3 wiring is functionally similar to Class 2 wiring, but with higher voltage and power limitations.

It should be noted that a Class 2 circuit associated with the electrical equipment that is part of a residential oil furnace, for example, usually receives power from a 24-V transformer whose primary connects to a 120-V premises branch circuit.

In conclusion, a class 1 must sit in metal or non-metallic raceway or be metal-sheathed wiring as compared to jacketed cable such as type NM.

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Electricians know about Class 1, 2, and 3 wiring because these terms are spelled out in the National Electrical Code. Electronics engineers, however, tend to be in the dark about these terms. So here is a short primer on classes of wiring for the curious.

The NEC in Article 725 makes a sharp distinction between wiring for “ordinary” power and light circuits and specialized circuits where the power and/or voltage is limited. These circuits are designated Class 1, 2 and 3, not to be confused with the Class I, II and III hazardous locations – an entirely different matter. Class 1, 2 and 3 circuits are defined primarily in terms of the power supply to which they are connected. Power supplies are generally batteries, transformers or electronic power supplies. When working on an existing installation, it is a simple matter of identifying the power source and checking its marking. A Class 2 circuit associated with the electrical equipment that is part of a residential oil furnace, for example, usually receives power from a 24-V transformer whose primary connects to a 120-V premises branch circuit. The relatively high impedance of this small device ensures that its voltage and power output will be in the Class 2 range. Such a device will be marked as suitable as a Class 2 power source. Describe how the results from Class 1, Class 2, and Class 3 differ.

Answer:

To solve for k, we first need to isolate k on one side of the equation by subtracting 61.1 from both sides:

K/7.8 + 61.1 - 61.1 = 17.8 - 61.1

K/7.8 = -43.3

Next, we multiply both sides by 7.8 to isolate k.

K = -43.3 * 7.8

K = -338.34

Therefore, k is approximately equal to -338.34.

Answer:

The distribution for the number of weeks pregnancy lasts for a mother is typically a unimodal distribution, with a single peak. The majority of pregnancies last around 38-42 weeks, with fewer pregnancies lasting shorter or longer than this period. This creates a bell-shaped curve, with the peak at the most common duration and a gradual decrease in frequency as the duration becomes shorter or longer than the mode.

However, it's worth noting that there can be some variations in this distribution based on different factors, such as the age of the mother, health status, and other demographic characteristics. For example, teenage pregnancies and pregnancies in older women may have slightly different distributions with different peaks due to biological and social factors. Nonetheless, a unimodal distribution is the most commonly observed distribution for the number of weeks pregnancy lasts for a mother.

3

Sydney has enough ribbon to make 3 bows

Answer:

3

Step-by-step explanation:

I multiplied 1 1/5 x 2/5 because each bow is made from a piece of ribbon that is 2/5 yards long. 1 1/5 x 2/5 = 12/25 I simplified to get the product of 3.

The measure of angle θ is 45 degrees.

The measure of angle θ is calculated by applying trigonometry ratio as follows;

SOH CAH TOA

SOH = sin θ = opposite /hypothenuse side

TOA = tan θ = opposite side / adjacent side

CAH = cos θ = adjacent side / hypothenuse side

The height and the adjacent side is given, so the value of θ is calculated as follows;

tan θ = (206 - 7 ) / (199)

tan θ = 1

θ = arc tan (1)

θ = 45⁰

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The statement is false, a triangle can't have two angles of 90°.

Remember that for any triangle, the sum of the 3 interior angles should be equal to 180°.

So if the 3 interior angles are A, B, and C we have.

A + B + C = 180°

In this case we know that two of the angles are of  90°, so let's say that:

A = 90°

B = 90°

Replacing that we will get:

90° + 90° + C = 180°

C = 180° - 90° - 90°

C = 0°

And we can't have a 0° degree angle, so the statement is false, triangles with two 90° don't exist.

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The inverse of the function f(x) = 15x - 45 is g(x) = x + 45 / 15

An inverse function or an anti function is defined as a function, which can reverse into another function. In simple words, if any function “f” takes x to y then, the inverse of “f” will take y to x. If the function is denoted by ‘f’ or ‘F’, then the inverse function is denoted by f-1 or F-1.

In this problem, f(x) = 15x - 45, we can find the inverse as;

To determine the inverse function, switch x and y and solve the resulting equation for x

If the initial function is not one-to-one, then there will be more than one inverse.

y = 15x - 45

x = 15y - 45

Let's solve for y

15y = x + 45

y = x + 45 / 15

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The first derivative of the function f(x) = x^4 + 3x^2 + 2 at x=2 using Richardson's method with h=0.01 is approximately 33.8933.

To calculate the first derivative of the function f(x) = x^4 + 3x^2 + 2 using Richardson's method, we need to first calculate the approximate values of the function at x=2 and x=2+h, where h=0.01.

So, we have:

f(2) = 2^4 + 3(2^2) + 2 = 24

f(2+h) = (2+h)^4 + 3(2+h)^2 + 2

= 16 + 32h + 24h^2 + 8h^3 + h^4 + 12 + 12h + 3h^2 + 2

= h^4 + 8h^3 + 24h^2 + 35h + 30

Now, we can use Richardson's extrapolation formula to calculate the first derivative of the function at x=2:

f'(2) ≈ (4f(2+h) - f(2)) / (3h)

Substituting in the values we have:

f'(2) ≈ (4(h^4 + 8h^3 + 24h^2 + 35h + 30) - 24) / (3h)

= (4h^4 + 32h^3 + 96h^2 + 140h + 96) / (3h)

= (4h^3 + 32h^2 + 96h + 140) / 3

Now, we can substitute h=0.01 to get the value of the first derivative at x=2:

f'(2) ≈ (4(0.01^3) + 32(0.01^2) + 96(0.01) + 140) / 3

≈ (4(0.000001) + 32(0.0001) + 0.96 + 140) / 3

≈ (0.000004 + 0.0032 + 0.96 + 140) / 3

≈ 140.9632 / 3

≈ 33.8933

Therefore, the first derivative of the function f(x) = x^4 + 3x^2 + 2 at x=2 using Richardson's method with h=0.01 is approximately 33.8933.

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The time it will take for your account balance to reach $10,000 is given as follows:

t = 10.8 years.

The balance of an account after t years, using continuous compounding, is given as follows:

A(t) = A(0)e^(kt).

In which:

The parameters for this problem are given as follows:

A(0) = 6500, k = 0.04.

Hence the time needed for a balance of $10,000 is obtained as follows:

10000 = 6500e^(0.04t)

e^(0.04t) = 10000/6500

0.04t = ln(10000/6500) -> ln is the inverse operation of the exponential

t = ln(10000/6500)/0.04

t = 10.8 years.

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The calculated area of the figure is 59/2 square units

From the question, we have the following coordinates that can be used in our computation:

A (1, 2) B (1, 7) C (9, 7) D (9, 3) E (6, 5)

The area is then calculated as

Area = 1/2 * | (x1y2 + x2y3 + ... + xny1) - (y1x2 + y2x3 + ... + ynx1) |

So, we have

Area = 1/2 * | (1 * 7 + 1 * 7 + 9 * 3 + 9 * 5 + 6 * 1) - (2 * 1 + 7 * 9 + 7 * 9 + 3 * 6 + 5 * 1) |

Evaluate

Area = 1/2 * 59

So, we have

Area = 59/2

Hence, the area is 59/2 square units

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The equation (x - 3)² + (y - 7)² = 36 in general form will be x² + y² - 6x - 14y + 22 = 0

Given that:

(x - 3)²+(y - 7)² = 36

Expanding the square in the given equation, we get:

(x - 3)² + (y - 7)² = 36

(x - 3)(x - 3) + (y - 7)(y - 7) = 36

x² - 6x + 9 + y² - 14y + 49 = 36

x² + y² - 6x - 14y + 22 = 0

Therefore, the equation of the circle (x - 3)² + (y - 7)² = 36 in general form will be x² + y² - 6x - 14y + 22 = 0

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1. radius = 13cm

2. Diameter = 27.2cm

3. Diameter = 24m

4.  Radius = 24. 5m

The formula used for determining the area and also the circumference of a circle is given thus;

For the  area of a circle, we have;

A = πr²

Given that the parameters are;

For the circumference of a circle, we have;

C = 2πr

To calculate the  diameter of a circle, we have the formula;

Diameter = 2Radius

Now, substitute the values for each

1. radius = diameter/2 = 26/2

radius = 13cm

2. When the radius = 13.6m

diameter = 2(13.6) = 27.2m

3. When the radius = 12m

diameter = 2(12) =  24m

4. When the diameter  = 49m

Then, radius 49/2 = 24. 5m

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The factored expression would have two brackets as shown.

We can see that the Polynomial that we have to deal with in the questions that we have here are quadratic in nature and we have to deal with them as such. We are going to have that when we deal with the quadratic;

r^2 - 25r - 150

This can be factored into (r + 5) (r - 30)

Then;

64y^2 - 49z^2

This can be factorized to give;

(8y + 7z) (8y - 7z)

Also;

16y^2 - 88y + 121

Can be factorized to;

(4y - 11) (4y - 11)

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The lowest average he could obtain on his next two tests and still receive an A is 93 or higher.

To receive an average of 90 or better, W. Orrier needs to score a total of 540 points or more on all six tests

(90 x 6 = 540).

He has already scored a total of 88.5 x 4 = 354 points on his first four tests.

Therefore, he needs to score a total of 540 - 354 = 186 points on his next two tests.

To find the lowest average he could obtain on his next two tests, he needs to divide 186 by 2

186 ÷ 2 = 93.

Therefore, the lowest average he could obtain on his next two tests and still receive an A is 93 or higher.

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The solution is : d. 63% is the probability that the next marble Simran selects, rounded to the nearest percent, will be white.

We have,

The probability of white = P (w) = 41/65= 0.63

The  probability of yellow = P (y)= 24/65= 0.369=0.37

The probability of choosing white is 0.63 . When rounded to  nearest percent gives

0.63*100/100

=0.63*100 percent

= 63 percent

= 63%

the probability of getting   the next marble white is the same as the probability of getting a white.

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complete question:

Simran has a bag containing white and yellow marbles. Simran randomly selects one marble from the bag,

records the result, and returns the marble to the bag. The results of the first 65 selections are shown below.

A white marble was selected 41 times.

A yellow marble was selected 24 times.

Based on these results, what is the probability that the next marble Simran selects, rounded to the nearest

percent, will be white?

A41% b50% c59% d63%

answer:

15

step-by-step explanation:

-4x - 6y = 8

+ (6x + 5y = 7)

__________

2x - y = 15

The system of linear equations with infinite solutions is:

2y = x - 8 and y = (1/2)*x - 4

A system of linear equations  :

y = ax + b

y = cx + d

Has infinite solutions if and only if both equations represent the same line (then the lines intercept at infinite points)

The equations given are:

y = 2x + 7

2x + y = -7

y = (1/2)x - 4

y = -2x + 7

2y - x = -8

If we take the last one:

2y - x = -8

We can rewrite this as:

2y = x - 8

y = (1/2)*x - (1/2)*8

y = (1/2)*x - 4

That is other of the equations no the list, so the system with infinite solutions is:

2y = x - 8 and y = (1/2)*x - 4

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After 200 years, there will be 47.43 pounds left of the material.

We know that we start with 68 pounds, and after 40 years there are 63.3 pounds left, so we have an exponential decay:

63.3 = 68*(b)^40

Where b defines how it decays, we can solve that equation for b to get:

(63.3/68)^(1/40) = b

0.9982 = b

Then the decay is:

f(x) = 68*(0.9982)^x

The amount left after 200 years will be:

f(200) = 68*(0.9982)^200 = 47.43

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The value of the given summation is:

[tex]\sum (29a_i - 13b_i) ]= -30[/tex]

Here we know that the sums are:

[tex]\sum a_i = -10\\\\\sum b_i = -20[/tex]

And we want to find the value of the sum:

[tex]\sum (29a_i - 13b_i)[/tex]

First we can distribute the sum to get:

[tex]\sum 29a_i + \sum - 13b_i[/tex]

And take the coefficients out of the sum:

[tex]29\sum a_i -13\sum b_i[/tex]

Now replace the known values:

[tex]29\sum a_i -13\sum b_i = 29*-10 - 13*-20 = -290 + 260 = -30[/tex]

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Step-by-step explanation:

45 minutes is 3/4 of an hour.

how often does 3/4 fit into 3 ?

3 / 3/4 = 3/1 / 3/4 = (3×4)/(1×3) = 4

so, it will be rung 4 times.

Answer: 4 times.

Step-by-step explanation:

We can divide 3 hours by 45 minutes, to get the number of times the bell will ring in 3 hours:

3 hours ÷ 45 minutes = 4

Therefore, the bell will be rung 4 times in 3 hours.

We can also think of it like this: 3 hours is 180 minutes, and the bell rings every 45 minutes.

So, we can divide 180 by 45 to get the number of times the bell will ring:

180 minutes ÷ 45 minutes = 4

Therefore, the bell will be rung 4 times in 3 hours.

The statements about the attached daily temperature data that are true include the following:

A. the average temperature was below 90°F on less than 75% of the days.

C. the average temperature was below 75°F on less than 25% of the days.

In Mathematics and Statistics, a box plot is sometimes referred to as box-and-whisker plot and it can be defined as a type of chart that can be used to graphically or visually represent the five-number summary of a data set with respect to locality, skewness, and spread.

Based on the information provided about the given box-and-whisker plot for the daily temperature data, the five-number summary for the given data set include the following:

Minimum (Min) = 65.

First quartile (Q₁) = 70.

Median (Med) = 75.

Third quartile (Q₃) = 85.

Maximum (Max) = 90.

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The probability that a randomly selected ounce of chocolate chips will contain between 62 and 66 chocolate chips is 0.6644.

Using the z-score formula to standardize the values and find the probability using a standard normal distribution table:

z1 = (62 - 63) / 1.8 = -0.56

z2 = (66 - 63) / 1.8 = 1.67

Using the standard normal distribution table, we find that the area to the left of z1 is 0.2881 and the area to the left of z2 is 0.9525.

Therefore, the probability that a randomly selected ounce of chocolate chips will contain between 62 and 66 chocolate chips is:

P(62 < x < 66) = P(z1 < z < z2) = P(z < z2) - P(z < z1)

= 0.9525 - 0.2881

= 0.6644

Therefore, the probability is approximately 0.6644.

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Answer:

The final price of the stapler is 11.97 dollars and they would get 4.49 dollars in revenue.

Answer:

-x/5+6y-9

Step-by-step explanation:

This is simplified. I hope this helped you!

If the Federal Reserve sells $60,000 in Treasury bonds to a bank at 8% interest, the immediate effect on the money supply is B. It is decreased by $60,000.

One of the monetary factors that decreases the money supply in the economic is when the Federal Reserve raises the banks' reserve requirements.

Another factor that can decrease the money supply is the sale of treasury bonds to banks.

Raising the interest rates also decreases the money supply.

Thus, by selling $60,000 in Treasury bonds to a bank at 8% interest, it decreases the money supply.

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a) The mean number of paintings owned by these five people is 40. b) No, the mean is not a good average to use to describe the number of paintings that these people own

a) To find the mean number of paintings owned, we add up all the values and divide by the total number of people:

Mean = (1+0+2+196+1)/5 = 40

Therefore, the mean number of paintings owned by these five people is 40.

b) No, The mean is not necessarily a good average to use to describe the number of paintings that these people own.

This is because the data is skewed by the one outlier value of 196, which is much larger than the other values.

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