simplify the expression I-30I

To illustrate the definition of a limit, we need to find values of δ for different values of ε. We can solve inequalities involving the expression (5x − 3) - 2 to find suitable values of δ for ε = 0.1, ε = 0.05, and ε = 0.01.

Limits are an important concept in calculus. In this case, we are given that the limit as x approaches 1 of the expression (5x − 3) is 2. To illustrate the definition, we need to find values of δ that correspond to different values of ε. We can start by setting ε to 0.1 and then solve for δ.

When ε = 0.1, we want to find δ such that |(5x − 3) - 2| < 0.1 whenever 0 < |x - 1| < δ. We can solve this inequality by manipulating it and simplifying it to find a suitable value of δ. Similarly, we can find values of δ corresponding to ε = 0.05 and ε = 0.01 by solving the inequalities |(5x − 3) - 2| < 0.05 and |(5x − 3) - 2| < 0.01, respectively.

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Answer: 304.

Step-by-step explanation:

The formula to calculate the sample standard deviation is given by :-

[tex]s=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n-1}}[/tex]

, where x = sample element.

[tex]\overline{x}[/tex] = Sample mean

s=sample standard deviation.

n= Number of observations.

[tex]\sum(x-\overline{x})^2[/tex] = sum of the squared deviations from the sample mean

As per given , we have

s=4

n= 20

Substitute theses values in the above formula , we get

[tex]4=\sqrt{\dfrac{\sum(x-\overline{x})^2}{20-1}}[/tex]

[tex]4=\sqrt{\dfrac{\sum(x-\overline{x})^2}{19}}[/tex]

Square root on both sides , we get

[tex]\Righatrrow\ 16=\dfrac{\sum(x-\overline{x})^2}{19}\\\\\Righatrrow\ \sum(x-\overline{x})^2=16\times19=304[/tex]

Hence, the sum of the squared deviations from the sample mean is 304.

Sum of squared deviations: [tex]\( 16 \times (20 - 1) = 304 \),[/tex]  given a standard deviation of 4 and sample size of 20.

To find the sum of the squared deviations from the sample mean, we'll use the formula for variance. Since the standard deviation is the square root of the variance, we'll square the standard deviation to get the variance. Then, we'll multiply by the sample size minus 1 to get the sum of the squared deviations.

Given:

Standard deviation (σ) = 4

Sample size (n) = 20

First, let's find the variance:

[tex]\[ \text{Variance} = \sigma^2 = 4^2 = 16 \][/tex]

Now, we'll use the formula for the sum of squared deviations:

[tex]\[ \text{Sum of squared deviations} = \text{Variance} \times (n - 1) \]\[ \text{Sum of squared deviations} = 16 \times (20 - 1) \]\[ \text{Sum of squared deviations} = 16 \times 19 = 304 \][/tex]

So, the sum of the squared deviations from the sample mean is 304.

This expression is a quadratic equation of the form ax² + bx + c = 0, where the constants are a = 45, b = 27, and c = 18. By substituting these values into the quadratic formula, we find that the equation has no real solutions.

This expression is a quadratic equation of the form ax² + bx + c = 0, where the constants are a = 45, b = 27, and c = 18. To complete the statement, we need to find the values of x for which the equation is satisfied:

45ax² + 27ax + 18a = 9a

Simplifying the equation:

45ax² + 27ax + 18a - 9a = 0

45ax² + 27ax + 9a = 0

Dividing the equation by 9a:

5ax² + 3ax + 1 = 0

Now, we have a quadratic equation that we can solve using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Substituting the values of a = 5, b = 3, and c = 1 into the formula:

x = (-3 ± √(3² - 4(5)(1))) / (2(5))

x = (-3 ± √(9 - 20)) / (10)

x = (-3 ± √(-11)) / (10)

Since the discriminant (b² - 4ac) is negative, the equation has no real solutions. Therefore, the statement cannot be completed with real values of x.

Proof by contradiction.

Let assume that when [tex]x[/tex] is an irrational number, then [tex]\dfrac{1}{x}[/tex] is a rational number.

If [tex]\dfrac{1}{x}[/tex] is a rational number, then it can be expressed as a fraction [tex]\dfrac{a}{b}[/tex] where [tex]a,b\in\mathbb{Z}[/tex].

[tex]\dfrac{1}{x}=\dfrac{a}{b} \Rightarrow x=\dfrac{b}{a}[/tex]

Since [tex]a,b\in\mathbb{Z}[/tex], the number [tex]x=\dfrac{b}{a}[/tex] is also a rational number. But this contardicts our initial assumption that [tex]x[/tex] is an irrational number. Therefore [tex]\dfrac{1}{x}[/tex] must be an irrational number.

Answer: Second option is correct.

Step-by-step explanation:

Since we have given that

61 ones times [tex]\frac{1}{10}[/tex]

which will be equal to

[tex]61\times \frac{1}{10}\\\\=6.1[/tex]

So, 6.1 will read as six point one

and

after decimal it is read as tenths.

So, it becomes,

61 tenths.

Hence, Second option is correct.

The expression for 18+12 is  6(3+2)

A mathematical operation such as subtraction, addition, multiplication, or division is used to combine terms into an expression. In a mathematical expression, the following terms are used:

Given:

18+12

Now, prime factorize

18 = 2 x 3 x 3

12= 2 x 2 x 3

So,  18+12 = 2 x 3 x 3 + 2 x 2 x 3

= 2 x 3( 3 + 2)

=6 (3+ 2)

Hence, the expression is 6 (3+ 2).

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Answer:

1/36

Step-by-step explanation:

Each dice can take 6 values and if we roll two dices the values are independent from each another. All the possible combinations are 6 × 6 = 36.

There is only one way of rolling a sum of 12, which is that each dice takes the value 6.

The probability of rolling a sum of 12 is:

[tex]P(12)=\frac{favorable\ cases }{possible\ cases} =\frac{1}{36}[/tex]

The probability of rolling a sum of 12 on two six-sided dice is [tex]\( \frac{1}{36} \) or approximately 0.028.[/tex]

To find the probability of rolling a sum of 12 on a standard pair of six-sided dice, we first need to determine the number of favorable outcomes (rolling a sum of 12) and the total number of possible outcomes when rolling two six-sided dice.

**Total Number of Possible Outcomes**:

  When rolling two six-sided dice, each die has 6 faces, so the total number of possible outcomes is [tex]\(6 \times 6 = 36\)[/tex].

**Number of Favorable Outcomes**:

  To get a sum of 12, we need one die to show a 6 and the other die to show a 6 as well. There is only one way to get each die to show a 6.

**Probability**:

  Probability is the ratio of favorable outcomes to total outcomes.

  [tex]\[ \text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \][/tex]

  [tex]\[ \text{Probability} = \frac{1}{36} \][/tex]

So, the probability of rolling a sum of 12 on a standard pair of six-sided dice is [tex]\( \frac{1}{36} \)[/tex].

This can also be expressed as a decimal rounded to three decimal places: [tex]\[ \text{Probability} \approx 0.028 \][/tex]

To eliminate the parameter t in the equations x=t^2 and y=2+10t, solve [tex]t = \sqrt(x)[/tex] and substitute into the second equation to get [tex]y = 2 + 10\sqrt(x)[/tex]. This results in the cartesian equation [tex]y = 2 + 10\sqrt(x).[/tex]

To eliminate the parameter t and find the cartesian equation, follow these steps:

Start with the given parametric equations: x=t^2 and y=2+10t

Solve the first equation for t:

[tex]t = \sqrt(x)[/tex]

Substitute this expression for t into the second equation:

[tex]y = 2 + 10(\sqrt(x))[/tex]

Thus, the cartesian equation is [tex]y = 2 + 10 \sqrt(x).[/tex]

The percentage of students who passed the class with a C or better is 82.1%. This is found by adding the number of students who earned A, B, or C grades, dividing by the total number of students, and multiplying by 100.

The subject is Mathematics, specifically a problem in percentages. First, we need to find the total number of students in the class. We add the number of students that earned each grade: 5 (A) + 8 (B) + 10 (C) + 3 (D) + 2 (F) = 28 students in total.  

Next, we need to identify the number of students who passed the class with a C or better. This includes the students who earned an A, B, or C. Adding these numbers together gets us 5 (A) + 8 (B) + 10 (C) = 23 students.  

Then, to find the percentage of students who passed the class, we divide the number of students who passed (23) by the total number of students (28) and then multiply by 100 to convert the result to a percentage. So the calculation is as follows: (23/28)*100 = 82.14. Rounded to one decimal place, that is 82.1%. Therefore, 82.1% of students passed the class with a grade of C or better.

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The percent of students who passed the course with a C or better is: 82.14%

To find the percent of students who passed the course with a C or better, we need to divide the number of students who passed by the total number of students and multiply by 100%.

The number of students who passed is 5 + 8 + 10 = 23.

The total number of students is 5 + 8 + 10 + 3 + 2 = 28.

Therefore, the percent of students who passed the course with a C or better is:

(23 / 28) * 100% = 82.14%

Rounding to one decimal place, the answer is 82.1%.

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Answer: y=(yx−16x)x+16

Step-by-step explanation:

Answer:

answer is C.g(2)>h(2)

Step-by-step explanation:

The algebraic rule for rotating a point [tex]\( 270^\circ \)[/tex] clockwise about the origin is:

[tex]\[ (x, y) \rightarrow (y, -x) \][/tex]

Rotations in the coordinate plane can be performed using rotation matrices or by following specific rules based on the degree of rotation and the direction (clockwise or counterclockwise). For a rotation of [tex]\( 270^\circ \) or \( -90^\circ \) (since \( 270^\circ \)[/tex]  clockwise is equivalent to [tex]\( -90^\circ \)[/tex] counterclockwise), the rule can be derived as follows:

1.Starting Point:Consider a point [tex]\( (x, y) \)[/tex] in the coordinate plane.

2. 90° Rotation Counterclockwise (or 270° Clockwise):

  When you rotate a point [tex]\( 90^\circ \)[/tex] counterclockwise about the origin, the new position is [tex]\( (-y, x) \)[/tex].

3. 80° Rotation (second 90° Counterclockwise):**

  If you rotate [tex]\( (-y, x) \)[/tex] another [tex]\( 90^\circ \)[/tex] counterclockwise (making a total of 180°), the new position is [tex]\( (-x, -y) \)[/tex].

4.270° Rotation (third 90° Counterclockwise or 90° Clockwise):

Finally, rotating [tex]\( (-x, -y) \)[/tex] another [tex]\( 90^\circ \)[/tex] counterclockwise (or the original point [tex]\( (x, y) \) \( 90^\circ \)[/tex] clockwise), the new position is [tex]\( (y, -x) \).[/tex]

Therefore, the algebraic rule for rotating a point [tex]\( 270^\circ \)[/tex] clockwise about the origin is:

[tex]\[ (x, y) \rightarrow (y, -x) \][/tex]

This rule effectively swaps the x and y coordinates and negates the new x-coordinate, which corresponds to a [tex]\( 270^\circ \)[/tex] clockwise rotation.