I need help with these two questions

Write out the base-x expansion of \(101_x\) :

\(101_x\) = 1 • x² + 0 • x¹ + 1 • x⁰

This number is equal to 17 (in base 10), so that

x² + 1 = 17

Solve this for x :

x² = 16

x = ±4

(Yes, negative bases are totally valid.)

Answer:

Step-by-step explanation:

b) the solution is B = {x ∈ R : x < -2 or x > 1}.

c)    The solution is C = {x ∈ R : -2 < x < -1 or 1 ≤ x < 2}.

(a) A = {x ∈ R : x ≤ 1.5}

We solve the inequality as follows:

2x + 3 ≤ 6

2x ≤ 3

x ≤ 1.5

(b) B = {x ∈ R : x < -2 or x > 1}

To solve the inequality x^2 + x > 2, we first find the roots of the equation x^2 + x - 2 = 0:

(x+2)(x-1) = 0

Thus, x = -2 or x = 1.

Then, we test the inequality for intervals around these roots:

For x < -2: (-2)^2 + (-2) > 2, so this interval is included in the solution.

For -2 < x < 1: The inequality x^2 + x > 2 is satisfied if and only if x > 1, which is not true for this interval.

For x > 1: (1)^2 + (1) > 2, so this interval is also included in the solution.

Therefore, the solution is B = {x ∈ R : x < -2 or x > 1}.

(c) C = {x ∈ R : -2 < x < -1 or 1 ≤ x < 2}

We solve x^2 ≥ 1 as follows:

x^2 ≥ 1

x ≤ -1 or x ≥ 1

Then we combine this with the inequality 1 ≤ x^2 < 4 to get:

-2 < x < -1 or 1 ≤ x < 2

Therefore, the solution is C = {x ∈ R : -2 < x < -1 or 1 ≤ x < 2}.

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Answer:

10.5

Explanation:

500×2, 1000×2 etc

The weight that the wrestler gain every 2 months will be 90.8 kg.

It should be noted that a function is used to illustrate the relationship that exists between the variables.

This is illustrated as the function is given as:

W = 80 + 5.4t

where w = weight

t = number of months

Therefore, the weight that the wrestler gain every 2 months will be:

W = 80 + 5.4t

W = 80 + 5.4(2)

W = 80 + 10.8

W = 90.8 kg.

Therefore, the weight that the wrestler gain every 2 months will be 90.8 kg.

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The pH value of water when the H₃O⁺ is 1.0 x 10⁻⁷ will be 7.

Algebra is the study of abstract symbols, while logic is the manipulation of all those ideas.

The acronym PEMDAS stands for Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction. This approach is used to answer the problem correctly and completely.

The pH value of any ion is given as,

pH = - log [H₃O⁺]

The ions of H₃O⁺ is 1.0 x 10⁻⁷. Then the pH value is given as,

pH = - log [H₃O⁺]

pH = - log [1.0 x 10⁻⁷]

pH = - log [10⁻⁷]

pH = - (-7) log 10

pH = 7

The pH value of water when the H₃O⁺ is 1.0 x 10⁻⁷ will be 7.

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Answer:

5x2x6

Step-by-step explanation:

Because you cut it in half so you have to divide the length and width by 2.

Answer:

Step-by-step explanation:

30/7

74/9

47/7

Answer:

\(\frac{3}{2}\)

Step-by-step explanation:

let n be the number , then

\(\frac{1}{3}\) n = \(\frac{1}{2}\) ( multiply both sides by 3 )

n = 3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)

Answer:

178

Step-by-step explanation:

200*11%

200*.11

22 employees will leave

200-22 = 178 left

The distance between the student's starting point and ending point is approximately 1.4 mi. The correct option is B.

We can use the Pythagorean theorem to find the distance between the student's starting point and ending point, which is the length of the straight line between the two points (i.e., the distance between the two points).

If we draw a diagram, we can see that the student's displacement forms the hypotenuse of a right triangle with legs of length 1.0 mi (the distance traveled due west) and 1.0 mi (the distance traveled due north).

So, we can use the Pythagorean theorem to find the length of the hypotenuse:

distance = √[(1.0 mi)² + (1.0 mi)²] ≈ 1.4 mi

Thus, the distance between the student's starting point and ending point is approximately 1.4 mi.

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Given the function f(x)=sin(2πx), with x = 0.3, f(x) = 0.951057. The objective is to approximate the value of f(0.2) using the first two terms in the Taylor series and Vx=0.1.

We know that the Taylor series for a function f(x) can be written as:f(x)=f(a)+f′(a)(x−a)+f′′(a)2(x−a)2+…+f(n)(a)n!(x−a)n+…The first two terms of the Taylor series are given by:f(x)=f(a)+f′(a)(x−a)The first derivative of f(x) is given by:f′(x)=2πcos(2πx)On substituting x = a = 0.1, we get:f′(0.1) = 2πcos(2π * 0.1) = 5.03118603447The value of f(x) at a=0.1 is given by:f(0.1) = sin(2π * 0.1) = 0.587785252292With a=0.1, the first two terms of the Taylor series become:f(x)=0.587785252292+5.03118603447(x−0.1) = 0.587785252292+0.503118603447x−0.503118603447×0.1Using x=0.2 and substituting the values of a and f(a) in the equation above, we get:f(0.2)=0.587785252292+0.503118603447*0.2−0.503118603447×0.1=0.712261After approximating the value of f(0.2) using the first two terms in the Taylor series,

we can conclude that the value of f(0.2) = 0.712261 with a = 0.1, with an error of approximately 0.012796.

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The number 57 belongs to the sets Z, Q, and R, but not N or W.

What are natural numbers ?

Natural numbers are the counting numbers that we use to represent quantities and perform arithmetic operations. These are the numbers that we use to count objects, such as apples, books, or people.

The number 57 belongs to the following sets of numbers:

N (the natural numbers): No, because 57 is not a counting number.

W (the whole numbers): No, because 57 is not a non-negative integer.

Z (the integers): Yes, because 57 is an integer (a positive integer, in fact).

Q (the rational numbers): Yes, because 57 can be written as a ratio of two integers (57/1).

R (the real numbers): Yes, because every rational number is also a real number.

Therefore, the number 57 belongs to the sets Z, Q, and R, but not N or W.

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Complete question -

Let a manager records the assemble output of three employees who each work on three different machines for three different days. The sample results are given below, and the Minitab results follow:

ABC

                                     Total

1           46     11     13       70

2         37       10    11        58

3          20      15    16        51

Total    103     36    40     179

Let that the number of items generate is not affected by an interaction between employee and machine.

Formula = Expected frequencies = (Row total" Column Total) /Grand Total.

Therefore,

                          58*103/179

                         = -33.374

The Expected Count for Shift = 2 and Machine - A is 33.374

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The distance traveled in 25 minutes is 32.34 miles.

Remember the relation:

distance = speed*time

Here we know that speed = 77mi/h

And the time is 25 min, but we should write this in hours, remember that:

1 hour = 60 min

then:

25 min = (25/60) h = 0.42 hours.

Then the distance is:

D = (77 mi/h)*0.42h = 32.34 miles.

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\(( ~~ \csc(\theta )-\cot(\theta ) ~~ )^2=\cfrac{1-\cos(\theta )}{1+\cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ \csc(\theta )-\cot(\theta ) ~~ )^2\implies \csc^2(\theta )-2\csc(\theta )\cot(\theta )+\cot^2(\theta ) \\\\\\ \cfrac{1^2}{\sin^2(\theta )}-2\cdot \cfrac{1}{\sin(\theta )}\cdot \cfrac{\cos(\theta )}{\sin(\theta )}+\cfrac{\cos^2(\theta )}{\sin^2(\theta )}\implies \cfrac{1}{\sin^2(\theta )}-\cfrac{2\cos(\theta )}{\sin^2(\theta )}+\cfrac{\cos^2(\theta )}{\sin^2(\theta )}\)

\(\cfrac{\cos^2(\theta )-2\cos(\theta )+1}{\sin^2(\theta )}\implies \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{\sin^2(\theta )} \\\\\\ \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{1-\cos^2(\theta )}\implies \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{-[\cos^2(\theta )-1]}\)

\(\cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{-[\cos^2(\theta )-1^2]}\implies \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{-[\cos(\theta )-1][\cos(\theta )+1]} \\\\\\ \cfrac{\cos(\theta )-1}{-[\cos(\theta )+1]}\implies \cfrac{-[\cos(\theta )-1]}{\cos(\theta )+1}\implies \cfrac{1-\cos(\theta )}{1+\cos(\theta )}\)

The relationship between the two confidence intervals will be that the width of i81 will be 3 times the width of i9. So, the correct answer is option D.

In statistics, a confidence interval is a range of values that is used to estimate the unknown population parameter. They are used to express the degree of uncertainty associated with a sample estimate of a population parameter.

Two 99 percent confidence intervals are going to be constructed to estimate the difference in means of two populations, r and w. One confidence interval, i9, is going to be constructed using samples of size 9 from each of r and w.

The other confidence interval, i81, is going to be constructed using samples of size 81 from each of r and w. When all other things remain the same, the width of 181 will be 3 times the width of Ig.

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Answer: Find the answer with the largest absolute value.

Step-by-step explanation:

Out of this list:

-25 -> |-25| = 25

-42 -> |-42| = 42

-16 -> |-16| = 16

-3 -> |-3| = 3

Your answer would be -42, as that is farthest from 0.

Out of this list:

40 -> |40| = 40

32 -> |32| = 32

96 -> |96| = 96

101 -> |101| = 101

Your answer would be 101, as that is the farthest from 0.

The Laplace transform of -6 for the given conditions is -6/s + 1/s^2

The Laplace transform is a mathematical operation that transforms a function of time into a function of a complex variable s, commonly used in solving linear ordinary differential equations. The Laplace transform of a function f(t), denoted by L{f(t)}, is defined as:

L{f(t)} = F(s) = ∫[f(t)e^(-st)]dt, where s is a complex variable.

In this case, the given function is -6 for 0<=x and x. Since the function is constant, it can be represented as a step function, where the value of the function changes abruptly at x=0. The step function is denoted as u(x), where u(x) = 0 for x<0, and u(x) = 1 for x>=0.

So, the given function can be written as -6u(x), where u(x) is the step function.

Now, applying the definition of the Laplace transform, we get:

L{-6u(x)} = ∫[-6u(x)e^(-sx)]dx

Since u(x) = 0 for x<0, the integral becomes:

L{-6u(x)} = ∫[0*e^(-sx)]dx = 0

Since u(x) = 1 for x>=0, the integral becomes:

L{-6u(x)} = ∫[-6*e^(-sx)]dx = -6∫[e^(-sx)]dx

Integrating e^(-sx) with respect to x, we get:

L{-6u(x)} = -6 * (-1/s) * e^(-sx) + C, where C is the constant of integration.

Finally, substituting back u(x) = 1, we get:

L{-6u(x)} = -6 * (-1/s) * e^(-sx) + C = 6/s * e^(-sx) + C

So, the Laplace transform of -6 for the given conditions is -6/s * e^(-sx) + C, which can also be written as -6/s + C/s, where C is a constant.

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Answer:

I didnt know that and now that I do idc caus math is well y'know math -_-

Answer: 7,8,9

Step-by-step explanation: all the numbers between 6 and 10

Answer: Total = $200

Step-by-step explanation:

Given information

Money spent on shoes = $20.00

Percent of Money spent on shoes = 10%

Given formula

The total amount of money spent × Percent = Money spent on shoes

                                                      ⇵    

The total amount of money spent = Money spent on shoes / Percent

Substitute values into the given formula

The total amount of money spent = Money spent on shoes / Percent

Total = (20.00) / (10%)

Convert percentage into decimal

Total = 20.00 / 0.1

Simplify the fraction

\(\Large\boxed{Total~=~200~Dollars}\)

Hope this helps!!  :)

Please let me know if you have any questions

Answer: You determine the coefficient by looking to see if there is a number in front of a variable. If there is a  number in front of the variable, then that is the coefficient. The coefficient is the number, and the variable is the letter.

Step-by-step explanation: Hope this helps

Answer:

If f=5 pounds of pressure, l=5.4 feet long

Step-by-step explanation:

Force needed to break a board varies inversely with the length of the board

Let

Force to break the board=f

Length of the board=l

f=k/l

Where k is a constant

If f=9 pounds of pressure and l=3 feet long

f=k/l

9=k/3

Cross product

9*3=k

27=k

If k=27, f=5 pounds of pressure. Find l?

f=k/l

5=27/l

Cross product

5l=27

l=27/5

=5.4

l=5.4 feet long

Using proportions, it is found that the board that requires 5 pounds of pressure to break is 1.67 inches long.

The rule of three is:

9 pounds - 3 feet

5 pounds - x feet

Applying cross multiplication:

\(9x = 3(5)\)

\(x = \frac{15}{9}\)

\(x = 1.67\)

The board that requires 5 pounds of pressure to break is 1.67 inches long.

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It is false that mobile banner ads perform significantly better than desktop banners.

There is no clear consensus on whether mobile banner ads or desktop banner ads perform better. The effectiveness of banner ads depends on various factors such as the placement of the ad, its design, and the target audience.

However, it is true that mobile usage has been increasing rapidly in recent years, and more people are accessing the internet through their mobile devices than through desktop computers. Therefore, it is important for advertisers to optimize their ads for mobile devices and ensure that they are mobile-responsive.

Nevertheless, it cannot be generalized that mobile banner ads are more effective than desktop banner ads. The effectiveness of an ad should be evaluated on a case-by-case basis, taking into account the specific objectives, target audience, and design of the ad.

Therefore, it is important for advertisers to test their banner ads on both desktop and mobile devices to determine which platform works best for their specific campaign. And the given statement is false.

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To find the margin of error for a 95% confidence interval using the t-distribution, we first need to calculate the t-score with degrees of freedom (df) equal to the sample size minus 1 (n-1).

In this case, df = 9. Using a t-table or calculator, the t-score for a 95% confidence level and df = 9 is approximately 2.262. The margin of error can be calculated by multiplying the t-score by the standard error of the mean, which is equal to the standard deviation divided by the square root of the sample size.

In this case, the standard error of the mean is 206 / sqrt(10) = 65.20. Multiplying by the t-score, we get a margin of error of 2.262 * 65.20 = 147.54. To construct the 95% confidence interval, we can use the formula:
sample mean +/- (t-score * standard error of the mean)

Substituting in the values, we get:
$450.50 +/- (2.262 * $65.20) = $450.50 +/- $147.54

So, the 95% confidence interval for the population mean mu is $302.96 to $598.04.

Interpreting the results, we can say with 95% confidence that the true population mean full retail price of cell phones is between $302.96 and $598.04. The margin of error is the amount by which the sample mean may differ from the true population mean, and in this case it is +/- $147.54.


The 95% confidence interval for the population mean (µ) is ($304.1, $596.9), and the margin of error is $146.4. This means we can be 95% confident that the true population mean price of cell phones is between $304.1 and $596.9.

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To solve the given linear programming (LP) problem using the graphical LP approach, we will plot the feasible region, identify the corner points, and determine the optimal solution.

Here are the steps:

Step 1: Plot the constraints:

- Plot the line 5x + y = 32, which represents the constraint 5x + y ≤ 32. To do this, find two points that lie on this line by assigning values to x and solving for y. For example, when x = 0, y = 32, and when x = 6, y = 2. Connect these points to draw the line.

- Plot the line x + 3y = 12, representing the constraint x + 3y ≥ 12. Again, find two points on this line and connect them. For instance, when x = 0, y = 4, and when x = 12, y = 0.

- Draw the lines representing x = 20 and y = 20 as vertical and horizontal lines passing through x = 20 and y = 20, respectively.

Step 2: Identify the feasible region:

- Shade the area that satisfies all the constraints. The feasible region is the intersection of the shaded regions.

Step 3: Identify the corner points:

- Find the coordinates of the corner points of the feasible region. These points are the intersections of the lines representing the constraints.

Step 4: Evaluate the objective function:

- Calculate the objective function Z = 3x + 12y for each corner point.

Step 5: Determine the optimal solution:

- Identify the corner point that gives the minimum value of the objective function Z. This point corresponds to the optimal solution of the LP problem.

Once you have completed these steps, you will have the complete solution to the LP problem using the graphical LP approach.

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