you are expected to design a circuit that can add two 2-digit bcd numbers, a1a0 and b1b0 to produce the three-digit bcd sum s2s1s0. following instructions given below:
1. Use switches SW15?8 and SW7?0 to represent 2-digit BCD numbers A1A0 and B1B0, respectively. The
value of A1A0 should be displayed on the 7-segment displays HEX7 and HEX6, while B1B0 should be on
HEX5 and HEX4. Display the BCD sum, S2S1S0, on the 7-segment displays HEX2, HEX1 and HEX0.

Answer:

1 1/7

Step-by-step explanation:

If you make  1 3/4 into a mixed number it is 7/4.

Then if you divide 2/1 (when you turn a whole number into a fraction you put the whole number over a 1)  by 7/4 you multiply by reciprocal which is 4/7 and 2/1 times /7 equals 1 1/7.

Could I please have BRAINLIEST.

Answer:

So all the possible solutions are:

Solution Steps:

First you need to solve the real inequality to understand how to find the rest of the possible equations.

Add 78 to both sides:

So now we know the real answer, but it ask for all possible answers.

(Means anything larger than 300 when you plug it into x - 78 > 300.)

Numbers that are greater than 378:

  1.)  \(377-78=299 > 300\)  (False)

  2.)  \(523-78=445>300\)  (True)

  3.)  \(0-78=-78>300\)  (False)

  4.)  \(-62-78=-140>300\)  (False)

  5.)  \(400-78=322>300\)  (True)

  6.)  \(379-78=301>300\)  (True)

  7.)  \(222-78=144>300\)  (False)

  8.)  \(378.6-78=300.6>300\) (True)

______________________________

 \(\bold{Hope~this~helps!}\\\bold{If~you~need~help~with~anything~else,~feel~free~to~ask!}\\\\\bold{~~~~~-TotallyNotTrillex}\)

The approximate height of the cylindrical pepper shakers on the right filled with pepper is equal to option b. 80mm.

Let us consider V be the volume of the cylindrical paper,

r be the radius, and h be the height.

To estimate the height of the pepper in the shaker on the right,

Use the fact that the same amount of pepper was poured into each shaker.

This implies volume of pepper in each shaker is the same.

Since the shaker on the left was filled to a height of 45 mm,

The formula for the volume of a cylinder is,

V = πr^2h

Radius of left hand shakers is equal to 32mm

And radius of right hand shakers is equal to 24mm

Substitute the values we have,

⇒ π (32)^2 x 45 = π (24)^2 x h

Simplifying, we get,

⇒ h = 46080 / 576

⇒ h = 80mm

Therefore, the approximate height of the pepper in the shaker on the right is option b. 80 mm.

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The above question is incomplete, the complete question is:

Mr. Getting poured the same amount of pepper into each of the cylindrical pepper shakers shown below. The pepper in the shaker on the left is filled to a height of 45 millimeters. To what approximate height is the shaker on the right filled with pepper?

A 60 millimeters

B. 80 millimeters

C. 32 mm

D. 24 mm-

E. 100 mm

Answer:

-3

Step-by-step explanation:

- 9/4 X 4/3

-9 X 1/3 =

-3

I hope this helps! If it does could you please mark me brainliest?

Answer:

$7.20

Step-by-step explanation:

20% of 36 is 7.2

The range of the function g(x) is given as follows:

[0, ∞).

From the graph of the function given in this problem, y assumes all real non-negative values, hence the interval notation representing the range of the function is given as follows:

[0, ∞).

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Answer:

k

Step-by-step explanation:

The Linnaean class system provides a framework for understanding the diversity of life on Earth by grouping similar organisms together based on shared characteristics.

By comparing and contrasting the definitions of each class, we can see how different groups of animals are related to each other and how they differ in terms of their biological traits.

The Linnaean class system is a way of organizing living things based on shared characteristics. Let's compare and contrast the definitions of the Linnaean classes using a Venn diagram.

First, we have the class Mammalia, which includes all animals that have hair or fur, produce milk to feed their young, and have specialized teeth. This class overlaps with the class Aves, which includes all birds, because some birds have specialized beaks and feathers that are similar to mammalian hair and teeth. However, birds do not produce milk.

Next, we have the class Reptilia, which includes animals that are cold-blooded, lay eggs, and have scales or plates on their skin. This class overlaps with both Mammalia and Aves in terms of species that lay eggs, such as monotremes (platypus and echidnas) and some birds (ostriches and emus). However, reptiles lack specialized teeth and do not produce milk.

Finally, we have the class Amphibia, which includes animals that are cold-blooded, breathe through their skin, and undergo metamorphosis from a water-dwelling larval stage to a land-dwelling adult stage. This class overlaps with Reptilia in terms of some shared characteristics, but Amphibia also lacks specialized teeth and does not lay eggs with hard shells like reptiles.

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The sum in summation notation of the expression (a) 1 + 2 + 3 + 4 + X5, where X takes the values 1-4, is ΣX from 1 to 4. In other words, it represents the summation of X over the range 1 to 4.

In summation notation, the expression (a) 1 + 2 + 3 + 4 + X5 can be written as ΣX, where X takes the values from 1 to 4. The capital sigma (Σ) represents the summation operator, and the variable X is summed over the range 1 to 4. This means that we need to substitute X with each value from 1 to 4 and add them together.

When we evaluate the expression, we have:

ΣX = 1 + 2 + 3 + 4 = 10

However, the expression also includes X5, indicating that X takes the value 5 as well. Therefore, we add 5 to the sum:

ΣX = 10 + 5 = 15

So, the sum of 1 + 2 + 3 + 4 + X5, where X takes the values 1-4, is equal to 15.

Moving on to expression (b) y2 + y3 + Y, where Y takes the values y₁ = -1, y₂ = -0.5, y₃ = 0, y₄ = 1.35, we can express it in summation notation as ΣY from 1 to 3. This represents the summation of Y over the range 1 to 3.

Since Y has a different value for each term, we simply substitute Y with each value from 1 to 3 and add them together:

ΣY = y₁ + y₂ + y₃ = -1 + (-0.5) + 0 = -1.5

Hence, the sum of y₂ + y₃ + Y, where Y takes the values y₁ = -1, y₂ = -0.5, y₃ = 0, y₄ = 1.35, is equal to -1.5.

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The test statistic for this hypothesis is 8.57, rounded to two decimal places.

A hypothesis is a proposed explanation for a phenomenon or set of observations that can be tested through experimentation or further observation. It is essential to scientific inquiry, as the hypothesis provides a starting point for further investigation. Hypotheses can be generated through observation, existing research, or logical deduction. Once a hypothesis is identified, it can be tested through experimentation or observation.

The test statistic for this hypothesis is calculated using the formula t = (M - μ) / (s/√n),
where M is the sample mean,
μ is the population mean (in this case, 0 days late),
s is the sample standard deviation and n is the sample size.
Therefore, the test statistic is calculated as:
t = (1.2 - 0) / (1.4 / √100)
t = 1.2 / 0.14
t = 8.57
Therefore, the test statistic for this hypothesis is 8.57, rounded to two decimal places.

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The height of the intersection of a line from the bottom of the first to the top of the second is a line from the top of the first to the bottom of the second, and it is 6 m.

In order to find the height of the point where a line from the top of the first to the bottom of the second intersects a line from the bottom of the first to the top of the second if two flagpoles are 10 m and 70 m tall and are 100 m apart, we can use similar triangles.

        The height can be found by calculating the length of the third side of a triangle whose other two sides are the height of the first flagpole and the length between the two flagpoles respectively.

        Let us now assume that a line is drawn from the top of the first flagpole to the bottom of the second flagpole. The point where it intersects with a line drawn from the bottom of the first flagpole to the top of the second flagpole is the point where we are to find the height. The flagpoles are 100 m apart, so the horizontal distance between these two lines is 100 m.

       The height of the first flagpole is 10 m, and the height of the second flagpole is 70 m, respectively.

       The height of the point where these two lines intersect can be found using similar triangles. Since these two lines intersect at right angles, we can use Pythagoras' theorem to find the distance between the two flagpoles.

        This distance is √(100² + 60²) = √10000 + 3600

                                                           = √13600

                                                           ≈ 116.62 m.

         Let x be the height that we are to find. Thus, we have a right triangle where one side is x, and another side is     116.62 - 100 = 16.62. The third side is the difference between the heights of the two flagpoles, which is 60.

       The similar triangles will be:

                          x/10 = (60 + x)/70x

                                  = 7x/6 + 1.7x

                                   = 10.2x

                                x = 6m

                  Therefore, the height of the point where a line from the top of the first to the bottom of the second intersects a line from the bottom of the first to the top of the second is 6 m.

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We have found a line segment joining two lattice points whose midpoint is also a lattice point. So, among any five lattice points, there must be a pair, the midpoint of which is also a lattice point.

Let’s assume that there are five lattice points on a plane and they are represented as follows:

(x1, y1), (x2, y2), (x3, y3), (x4, y4), (x5, y5)

To prove that among any five lattice points, there must be a pair, the midpoint of which is also a lattice point, we can follow the following steps.

Step 1: Let's consider any two points from the five lattice points, and let's call them P and Q.

Their coordinates are represented as (x1, y1) and (x2, y2), respectively.

Step 2: Let's apply the midpoint formula to find the midpoint of the line segment PQ. The midpoint formula is given by,

Midpoint of PQ = ( (x1+x2)/2, (y1+y2)/2 )

We know that the sum of two integers is always an integer, and the product of two integers is always an integer. Therefore, (x1+x2) and (y1+y2) are integers, and thus the midpoint of PQ is also a lattice point.

Step 3: Let's repeat step 2 with other pairs of points. There are a total of 10 pairs of points in five lattice points, and we can apply the midpoint formula to each pair. Therefore, we have 10 midpoints.

Step 4: Let’s observe that if one of these midpoints coincides with any of the five lattice points, then we are done. If not, then each midpoint must be a new point that is not among the five lattice points. And because the coordinates of each midpoint are the average of two integer coordinates, we know that each midpoint must be a point with integer coordinates (as mentioned in step 2).

Step 5: Let’s consider two midpoints, M1 and M2, that we calculated in step 3. Since M1 and M2 are each midpoints of a line segment joining two lattice points, we know that M1M2 is also a line segment. And because the coordinates of M1 and M2 are both integers, we know that the coordinates of the endpoints of M1M2 are integers too.

Hence Proved.

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The probability that exactly two of the four persons selected live in their own household and are income-qualified is approximately 0.1075 or 10.75%.

Probability of success (p): The probability that a person selected from the population lives in their own household and is income-qualified. From the given information, p = 0.84.

Probability of failure (q): The probability that a person selected from the population does not live in their own household and is not income-qualified. Since q is the complement of p, we have q = 1 - p = 1 - 0.84 = 0.16.

Number of trials (n): We are selecting four persons from the population, so n = 4.

Number of successes (k): We want exactly two persons to live in their own household and be income-qualified, so k = 2.

Now we can calculate the probability using the binomial probability formula:

P(X = k) = C(n, k) * p^k * q^(n-k)

where C(n, k) is the binomial coefficient, given by C(n, k) = n! / (k! * (n-k)!).

Plugging in the values, we get:

P(X = 2) = C(4, 2) * (0.84)²* (0.16)⁽⁴⁻²⁾

Calculating the binomial coefficient:

C(4, 2) = 4! / (2! * (4-2)!) = 4! / (2! * 2!) = 6.

Substituting the values:

P(X = 2) = 6 * (0.84)² * (0.16)⁽⁴⁻²⁾

Calculating the result:

P(X = 2) = 6 * 0.84² * 0.16²

= 6 * 0.7056 * 0.0256

= 0.1075 (approximately)

Therefore, the probability that exactly two of the four persons selected live in their own household and are income-qualified is approximately 0.1075 or 10.75%.

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Answer:

No

Step-by-step explanation:

It could only represent large colleges, not all colleges.  Sampling requires similar characteristics to represent the entire population of testing.

Answer:

It is different because perimeter is the outside and area is the inside.

Step-by-step explanation:

Answer:

y

2

​

=−

2

z

​

+7

Steps for Solving Linear Equation

z=18−2×2−2y2

Multiply 2 and 2 to get 4.

z=18−4−2y

2

​

Subtract 4 from 18 to get 14.

z=14−2y

2

​

Swap sides so that all variable terms are on the left hand side.

14−2y

2

​

=z

Subtract 14 from both sides.

−2y

2

​

=z−14

Divide both sides by −2.

−2

−2y

2

​

​

=

−2

z−14

​

Dividing by −2 undoes the multiplication by −2.

y

2

​

=

−2

z−14

​

Divide z−14 by −2.

y

2

​

=−

2

z

​

+7

Step-by-step explanation:

the given equation defines a paraboloid that lies below the plane z=0. Specifically, it is situated above the xy-plane, which means that the z-values of all points on the surface are greater than or equal to zero.

we can break down the equation z=18-2x^2-2y^2. This equation represents a paraboloid with its vertex at (0,0,18) and axis of symmetry along the z-axis. The first term 18 is the z-coordinate of the vertex and the last two terms -2x^2 and -2y^2 determine the shape of the paraboloid.

Since the coefficient of x^2 and y^2 terms are negative, the paraboloid is downward facing and opens along the negative z-axis. Therefore, all points on the paraboloid have z-values less than 18. Additionally, since the paraboloid is situated above the xy-plane, its z-values are greater than or equal to zero.

the paraboloid defined by the equation z=18-2x^2-2y^2 is situated below the plane z=0 and above the xy-plane. Its vertex is at (0,0,18) and it opens along the negative z-axis.

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The proofing of the triangle is illustrated below.

AB = c

BC = a

AC = b

h² = b² - x²

h² = a² - y²

a² - y² = b² - x²

a² = b² - x² + y²

Note that x = c - (c - x) = c

a² = b² - (-x)² + y²

a² = b² + c² + y²

Cos A = x/b

cos = adjacent/hypothenuse

x = bcosA

a² = b² + c² - 2bx

a² = b2 + c² - 2bccosA

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As per the probability of downloads  the capacity of Goo-gle's servers is designed to handle Ama-zon Ale-xa app downloads per day is equal to 4800.

Probability that there are 2000 or fewer downloads of Ama-zon Ale-xa in a day,

Standardize the value using the z-score

Probability using a standard normal distribution table,

z = (2000 - 2800) / 860

  = -0.9302

Using a standard normal distribution table,

Probability of a standard normal variable being less than or equal to -0.9302 is 0.1762.

Probability of 2000 or fewer downloads of Ama-zon Al-exa in a day is 0.1762.

Probability that there are between 1500 and 2500 downloads of Ama-zon Ale-xa in a day,

Standardize the values using the z-score

Probability using a standard normal distribution table,

z₁= (1500 - 2800) / 860

   = -1.5279

z₂ = (2500 - 2800) / 860

    = -0.3488

Using a standard normal distribution table,

Probability of a standard normal variable being less than or equal to -0.3488 is 0.3632.

Probability of a standard normal variable being less than or equal to -1.5279 is 0.0630.

Probability of between 1500 and 2500 downloads of Ama-zon Ale-xa in a day is,

0.3632 - 0.0630

= 0.3002

Probability that there are more than 3000 downloads of Ama-zon Ale-xa in a day.

Standardize the value using the z-score

Probability using a standard normal distribution table,

z = (3000 - 2800) / 860

  = 0.2326

Using a standard normal distribution table,

Probability of a standard normal variable being greater than 0.2326 is 0.4090.

Probability of more than 3000 downloads of Ama-zon Ale-xa in a day is 0.4090.

Capacity of Goo-gle's servers,

Probability of the number of Ama-zon Ale-xa app downloads in a day exceeding the capacity is 0.01.

Let's assume that the number of downloads per day follows a normal distribution ,

Mean =2800

Standard deviation = 860.

Let C be the capacity of Goo-gle's servers.

Let X is the number of Ama-zon Ale-xa app downloads in a day.

P(X > C) = 0.01

Using the standard normal distribution,

P(X > C)

= P((X - μ) / σ > (C - μ) / σ)

= P(Z > (C - 2800) / 860)

= 0.01

where Z is a standard normal variable.

Using a standard normal distribution table,

z-score corresponding to a probability of 0.01 is 2.3263.

This implies

(C - 2800) / 860 = 2.3263

Solving for C, we get,

⇒C = 4800.618

⇒C = 4800(Rounding to the nearest whole number)

Therefore, , the capacity of Goo-gle's servers is designed to handle Ama-zon Ale-xa app downloads per day as per the probability is equal to 4800.

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The above question is incomplete, the complete question is:

Ale-xa is the popular virtual assistant developed by Ama-zon. Ale-xa interacts with users using artificial intelligence and voice recognition. It can be used to perform daily tasks such as making to-do lists, reporting the news and weather, and interacting with other smart devices in the home. In , the Ama-zon Ale-xa app was downloaded some 2800 times per day from the Goo-gle Play store (AppBrain website). Assume that the number of downloads per day of the Ama-zon Ale-xa app is normally distributed with a mean of 2800 and standard deviation of 860. What is the probability there are 2000 or fewer downloads of Ama-zon Ale-xa in a day (to 4 decimals)? What is the probability there are between 1500 and 2500 downloads of Ama-zon Ale-xa in a day (to 4 decimals)? What is the probability there are more than 3000 downloads of Ama-zon Ale-xa in a day (to 4 decimals)? Assume that Goo-gle has designed its servers so there is probability 0.01 that the number of Ama-zon Ale-xa app downloads in a day exceeds the servers' capacity and more servers have to be brought online. How many Ama-zon Ale-xa app downloads per day are Goo-gle's servers designed to handle (to the nearest whole number)?

The factor of the expression 32z - 48 will be 16 and (2z - 3). And the greatest common factor is 16.

Simply said, it is the biggest common element. The highest consistent element in the above example is 15, making 15 the greatest common factor amongst 15, 30, and 105. The biggest chunk of the common factors is called the "Greatest Common Factor" (of two or more numbers).

The expression is given below.

⇒ 32z - 48

The expression can be rewritten as,

⇒ 32z - 48

⇒ 16 · 2z - 16 · 3

⇒ 16 (2z - 3)

The factor of the expression 32z - 48 will be 16 and (2z - 3). And the greatest common factor is 16.

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Rhonda has approximately 0.77 seconds to react before the volleyball hits the ground.

To calculate the time Rhonda has to react, we need to determine how long it takes for the volleyball to reach the ground after Marsha serves it. We can use the equation of motion, which relates the initial velocity (u), time (t), and height (h) of an object in free fall.

The initial velocity of the volleyball is 15 ft/s in the upward direction. Since the ball is 3.5 feet above the ground when struck, we consider the total distance traveled by the ball as the sum of the initial height and the height of the fall.

Using the equation h = ut + (1/2)gt^2, where g is the acceleration due to gravity (approximately 32.2 ft/s^2), we can rearrange the equation to solve for time (t):

3.5 = 15t - (1/2)(32.2)t^2

Simplifying this quadratic equation, we get:

16.1t^2 - 15t + 3.5 = 0

Solving this equation, we find two roots: t ≈ 0.23 s and t ≈ 1.08 s. Since we are interested in the time it takes for the ball to hit the ground, we discard the smaller root (0.23 s) because it corresponds to the time it takes for the ball to reach its maximum height.

Therefore, Rhonda has approximately 1.08 seconds - 0.31 seconds (the time it takes for the ball to reach its initial height) = 0.77 seconds to react before the volleyball hits the ground.

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Answer:

99 and 33

Step-by-step explanation:

let x and y be the 2 numbers, with x being the larger of the 2, then

x + y = 132 → (1)

x - y = 66 → (2)

add (1) and (2) term by term to eliminate y

(x + x) + (y - y) = 132 + 66

2x + 0 = 198

2x = 198 ( divide both sides by 2 )

x = 99

substitute x = 99 into (1) and solve for y

99 + y = 132 ( subtract 99 from both sides )

y = 33

the 2 numbers are 99 and 33

The control limits for 3-sigma x chart are

Upper Control Limit - 3.807 inches

Lower Control Limit - 2.291 inches

The control limits for the 3-sigma R-chart are Upper Control Limit- 2.373 inches

X-double bar = Sum of observations (x-bar) / Total number of observations

X-double bar = 73.17 / 24

X-double bar = 3.049

R-bar = Sum of observations (R) / Total number of observations

R-bar = 24.97 / 24

R-bar = 1.040

(a) The Upper Control Limit (UCLx) and Lower Control limit (LCLx) for the 3-sigma x-bar chart is calculated as,

UCLx = X-double bar + (A2 x R-bar)

LCLx = X-double bar - (A2 x R-bar)

From the table of factors for computing the control chart limits (3-sigma),

For n = 4 (4-pieces of precision cut wire)

A2 = 0.729

Putting these values in the above formulas, we get,

UCLx = 3.049 + (0.729 x 1.040)

UCLx = 3.807 inches

LCLx = 3.049 - (0.729 x 1.040)

LCLx = 2.291 inches

Based on the x-bar chart, the wire cutting process has been out of control, because the observation at 19th hour (4.51) lie outside the computed upper control limit and lower control limit.

(b) The Upper Control Limit (UCLr) and Lower Control limit (LCLr) for the 3-sigma  R-chart is calculated as,

UCLr = D4 x R-bar

LCLr = D3 x R-bar

From the table of factors for computing the control chart limits (3-sigma),

For n = 4 (4-pieces of precision cut wire)

D3 = 0, D4 = 2.282

Putting these values in the above formulas, we get,

UCLr = 2.282 x 1.040

UCLr = 2.373 inches

LCLr = D3 x R-bar

LCLr = 0 x 1.040

LCLr = 0 inches

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Answer:

3.4 or 3 and 2/5

Step-by-step explanation:

Answer:

Step-by-step explanation:

\(A=P(1+rt)\\ \\ t=\left(\frac{A}{P}-1\right)/r\\ \\ t=\left(\frac{5800}{5700}-1\right)/0.01=1.75y\)

The relative price is a useful measure for comparing the prices of different products or services, especially in the context of consumer preferences and demand.

Relative price refers to the price of a particular product or service in relation to other goods or services in the market.

It is calculated as the ratio of the price of a given product or service to the price of a reference product or service, commonly referred to as a base good or service.

Let the price of a pair of shoes be $5 and the price of a box of tea be $3.

Then the relative price of a pair of shoes is given by:

Relative price of shoes = Price of shoes / Price of tea

= $5 / $3

= 1.67

Thus, the relative price of a pair of shoes is 1.67.

Similarly,

The relative price of a box of tea can be calculated as follows:

Relative price of tea = Price of tea / Price of shoes

= $3 / $5

= 0.6

Therefore, the relative price of a box of tea is 0.6.

This means that the price of tea is relatively cheaper than that of shoes, as its relative price is less than one.

The relative price of shoes is greater than one, which indicates that shoes are relatively more expensive than tea.

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a.  The sample size must be at least 8 to have a 95% confidence interval estimate for the gain in a circuit on a semiconductor device with a length of not greater than 60. b. The sample size must be at least 22 to have a 99% confidence interval estimate for the gain in a circuit on a semiconductor device with a length of not greater than 60.

a. The formula for finding the width of a confidence interval is:
Width of confidence interval = 2 × Zα/2 × σ/√n

where Zα/2 is the Z-score for the desired level of confidence.

For 95% confidence level, Zα/2 = 1.96.

Width of 95% confidence interval = 60σ = 30Zα/2 = 1.96

Substituting the given values in the formula:60 = 2 × 1.96 × 30/√nn = (2 × 1.96 × 30/60)²n = 7.84≈ 8

Therefore, the sample size must be at least 8 to have a 95% confidence interval estimate for the gain in a circuit on a semiconductor device with a length of not greater than 60.

b. For a 99% confidence level, Zα/2 = 2.576

Width of 99% confidence interval = 60σ = 30Zα/2 = 2.576

Substituting the given values in the formula:

60 = 2 × 2.576 × 30/√nn = (2 × 2.576 × 30/60)²n = 21.16≈ 22

Therefore, the sample size must be at least 22 to have a 99% confidence interval estimate for the gain in a circuit on a semiconductor device with a length of not greater than 60.

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Answer:

25

Step-by-step explanation:

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Answer:

25 (third option)

Step-by-step explanation:

(4 + 6 × 3) + 3

(4 + 18) + 3

22 + 3

= 25

Answer: She will give 8 square meter area to school snap pea plant.

Step-by-step explanation:

Given: Area required by each planter box = \(\dfrac45\) square meter

Total number of snap pea seeds  = 10

The total area required =  (Area required by each planter box ) x (number of snap pea seeds)

\(=\dfrac45\times10\\\\=4\times2=8\)

Hence, she will give 8 square meter area to school snap pea plant.

For the inference in part b to be valid, the conditions required are reasonable satisfaction.

To determine if the conditions for the inference in part b are reasonably satisfied, we need to consider the specific details of the inference.

However, in general, the conditions for a valid inference typically include logical reasoning, accurate data, reliable sources, and relevant contextual information.

These conditions ensure that the conclusion drawn from the inference is supported by evidence and does not contain logical fallacies or biases. It is important to evaluate the quality and reliability of the information used in the inference to determine if the conditions are reasonably satisfied.

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