is 5 yards and 2 feet greater than 200 in

Given the Figure shown below is, ΔABC~ΔAXY if so, 1.67 is the scale factor between the two triangles.

The ratio between the scale of an original thing and a new object that is a representation of it but has a different size is known as a scale factor (bigger or smaller).

Consider two comparable squares as an illustration. One has a side length of 2 inches, while the other has a side length of 4 inches. This results in a scale factor of 1: 2 from the small square to the huge square. The scale factor between these two comparable squares is 1: 2, or tiny square to huge square.

Given that,

12 / 20 = 17 / x

or, 12 x = 340

or, x = 28.33

on the other hand,

25 / 15 = 1.66

20 / 12 = 1.66

28.33 / 17 = 1.66

now, scale factor = 1.67

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The mean age of the 12 students is approximately 17.02 years, the variance is 37.41, and the standard deviation is approximately 6.117 years.

The statistic is the study of mathematics that deals with relations between comprehensive data.

Here,
To calculate the mean, variance, and standard deviation of the ages of the 12 students, we first need to find the sum of the ages:

Sum = 16 + 17 + 18 + (16/0.5) + 17 + 18 + 19 + 17 + 17 + 13 + (17/2/2) + 16

= 16 + 17 + 18 + 32 + 17 + 18 + 19 + 17 + 17 + 13 + 4.25 + 16

= 204.25

Mean = Sum / Number of students = 204.25 / 12 ≈ 17.02 years

To calculate the variance, we need to subtract the mean from each age, square the differences, add up the squares, and divide by the number of students minus one:

Variance = [ (16 - 17.02)² + (17 - 17.02)² + ... + (16 - 17.02)² ] / (12 - 1)

= [ 0.16 + 1.96 + ... + 0.16 ] / 11

= 37.41

Finally, we can calculate the standard deviation by taking the square root of the variance:

Standard deviation = √(Variance) ≈ 6.117

Therefore, the mean age of the 12 students is approximately 17.02 years, the variance is 37.41, and the standard deviation is approximately 6.117 years.

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We have shown that the product of two rational numbers is rational.

A number system is defined as a system of writing to express numbers.

Let a and b be two rational numbers, such that:

a = p/q

b = r/s

where p, q, r, and s are integers and q, s are non-zero.

We need to show that the product ab is also a rational number.

We have:

ab = (p/q) x (r/s)

= (p x r) / (q x s)

Since p, q, r, and s are all integers,

ab = m/n

where m = p x r and n = q x s.

Since both m and n are integers and n is non-zero ,we can conclude that ab is a rational number.

Therefore, we have shown that the product of two rational numbers is rational.

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The difference quotient expression for the given function is

[tex]\frac{f(x+h)-f(x)}{h} =\frac{\sqrt{(x+h+1)(x+h-1)}-\sqrt{(x+1)(x-1)} }{h}[/tex]

The expression in single-variable calculus is usually referred to as the difference quotient.

[tex]\frac{f(x+h)-f(x)}{h}[/tex]

When taken to the limit as h gets closer to zero, h frac f(x+h)-f(x)h, which gives the derivative of the function f.

The slope of a secant line passing through the curve of f(x) is measured by the difference quotient.

Consider the difference quotient formula,

[tex]\frac{f(x+h)-f(x)}{h}[/tex]

Evaluate the function at x = x + h

replace the variable x with (x + h) in the given expression

[tex]f(x+h)=\sqrt{(x+h)^2-1}[/tex]

simplify the result ,

[tex]f(x+h)=\sqrt{(x+h+1)(x+h-1)}[/tex]

find the components of the definition,

[tex]f(x+h)=\sqrt{(x+h+1)(x+h-1)}[/tex]

[tex]f(x)=\sqrt{(x+1)(x-1)}[/tex]

plug in the components,

[tex]\frac{f(x+h)-f(x)}{h} =\frac{\sqrt{(x+h+1)(x+h-1)}-\sqrt{(x+1)(x-1)} }{h}[/tex]

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Answer:

Therefore, the total value of Mary's investment after two years is $700.

Step-by-step explanation:

To calculate the total value of Mary's investment after two years, we need to calculate the value of the investment after the first year, and then use that value to calculate the value after the second year.

In the first year, the value of the investment increases by 8%, so the value after the first year is:

Value after first year = $600 + 8% of $600

Value after first year = $600 + 0.08 * $600

Value after first year = $648

In the second year, the value of the investment increases by another 8%, so the value after the second year is:

Value after second year = Value after first year + 8% of Value after first year

Value after second year = $648 + 0.08 * $648

Value after second year = $699.84

Rounding to the nearest dollar, the total value of Mary's investment after two years is:

Total value after two years = $700

Answer:

3

Step-by-step explanation:

9-7=2

2*3=6

6*2=12

75/5=15

15-12=3

Answer:

Therefore, the value of the expression 75÷5−[(9−7)×3]×2 is 138.

Step-by-step explanation:

Using the order of operations (also known as PEMDAS), we simplify the expression as follows:

First, we solve the parentheses: (9-7) = 2.

Next, we multiply 2 by 3: 2 x 3 = 6.

Then, we subtract 6 from 75: 75 - 6 = 69.

Finally, we multiply 69 by 2: 69 x 2 = 138.

Therefore, the value of the expression 75÷5−[(9−7)×3]×2 is 138.

True. bar graphs (aka. column charts) with error bars are used to show the comparison between two variables, not the relationship.

A confidence intervals bar chart (shown as red lines)

In graphs, error bars are used to graphically show data variability and to denote error or uncertainty in reported measurements. They provide a basic indication of a measurement's accuracy or, alternatively, how far the true (error-free) value may deviate from the reported value. The standard deviation of uncertainty, the standard error, or a specific confidence range (such a 95% interval) are frequently represented as error bars. Due to the fact that these values are not the same, the measure chosen needs to be specified clearly in the graph or supporting text.

According to the definition the statement is true.

True. bar graphs (aka. column charts) with error bars are used to show the comparison between two variables, not the relationship.

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Answer:

x = 54

Step-by-step explanation:

See picture below :)

[-1.5, 3.5] and [-2, infty) are the range and domain of the equation.

Quadratic function are functions that have a leading degree of 2.

The domain of the function are values that lies along the x-axis of the graph. Hence the domain of the curve will be [-1.5, 3.5]

The range of the function are values that lies along the y-axis of the graph. Hence the domain of the curve will be [-2, infty)

Hence the range and domain of the graph are [-1.5, 3.5] and [-2, infty)

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The value of the transformed function after a vertical stretch in the y direction is h ( x ) = k ( | x | - 3 ) , where k is the scale factor

The transformation of a function may involve any change.

Usually, these can be shifted horizontally (by transforming inputs) or vertically (by transforming output), stretched (multiplying outputs or inputs), etc.

If the original function is y = f(x), assuming the horizontal axis is the input axis and the vertical is for outputs, then:

Horizontal shift (also called phase shift):

Left shift by c units: y=f(x+c) (same output, but c units earlier)

Right shift by c units: y=f(x-c)(same output, but c units late)

Vertical shift:

Up by d units: y = f(x) + d

Down by d units: y = f(x) - d

Stretching:

Vertical stretch by a factor k: y = k × f(x)

Horizontal stretch by a factor k: y = f(x/k)

Given data ,

Let the parent function be represented as f ( x )

Now , the value of f ( x ) is

f ( x ) = ( | x | - 3 )   be equation (1)

Now , when the function is stretched in the y direction , we get

Vertical stretch by a factor k: y = k × f(x)

On simplifying , we get

Let the transformed function be represented as h ( x )

where the value of h ( x ) is

Let the scale factor of stretching be k

h ( x ) = k ( | x | - 3 )

Hence , the transformed function is h ( x ) = k ( | x | - 3 )

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f(x) is discontinuous.

Since LHS ≠ RHS ≠ f(x).

A function is a relationship between inputs where each input is related to exactly one output.

Example:

f(x) = 2x + 1

f(1) = 2 + 1 = 3

f(2) = 2 x 2 + 1 = 4 + 1 = 5

The outputs of the functions are 3 and 5

The inputs of the function are 1 and 2.

here, we have,

We have,

f(x) = x + 1, for x ≤ 2

f(x) = 2x - 1, for 1 < x < 2

f(x) = x - 1, for x < 1

Now,

A x = 1

LHS of f(x) = lim f((h - 1)) =  (h - 1 - 1) = 0 -2 = -2

RHS of f(x) = lim  f(h + 1) =  (2(h + 1) - 1) = 2h + 2 - 1 = 0 + 1

f(1) = x + 1 = 1 + 1 = 2

We see that,

LHS ≠ RHS ≠ f(x)

So,

f(x) is not continuous, it is discontinuous.

Thus,

f(x) is discontinuous.

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The complete question.

Show that the following functions are continuous or discontinuous at x = 1.

f(x) = x + 1, for x ≤ 2

f(x) = 2x - 1, for 1 < x < 2

f(x) = x - 1, for x < 1

Answer:

the efficiency of the generator is 81.82%.

Step-by-step explanation:

The efficiency of a generator is defined as the ratio of the electrical power output to the mechanical power input. In this problem, the electrical power output is given as 90 hp, and the mechanical power input is the power supplied by the engine, which is 110 hp. Therefore, the efficiency of the generator can be calculated as follows:

Efficiency = Electrical power output / Mechanical power input

Efficiency = 90 hp / 110 hp

Simplifying the fraction, we get:

Efficiency = 0.8182

To convert this decimal to a percentage, we multiply by 100:

Efficiency = 0.8182 * 100

Rounding to two decimal places, we get:

Efficiency = 81.82%

Therefore, the efficiency of the generator is 81.82%.

The cost of the burger and the cost of an order of fries to Lynn would be :

From the problem, we know that Lynn spent $20.00 and purchased 4 hamburgers and 5 orders of fries. Similarly, we know that Ricardo spent $41.20 and purchased 10 hamburgers and 7 orders of fries, which gives us the equations:

4h + 5f = 20

10h + 7f = 41.20

Using substitution, we can express h in form of f to be :

4h + 5f = 20

4 [ ( 3 f + 1. 20 ) / 2] + 5 f = 20

6 f + 2 . 40 + 5 f = 20

11 f = 17. 60

f = $ 1. 60

We can then find the cost of hamburgers using the first equation :

4h + 5f = 20

4h + 5 ( 1. 60 ) = 20

4h + 8 = 20

4h = 12

h = $ 3

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Answer:

No solution

Step-by-step explanation:

2x - 4y= -6 ----- eq(1)

-x + 2y = -2 ----- eq(2)

Simplifying eq(1) by taking 2 as common,

2(x-2y) = 2(-3)

x - 2y= -3 -----eq(1)

from eq(1)

x - 2y = -3

x = -3 + 2y ---eq(3)

substituting eq(3) into eq(2)

-(-3 + 2y) + 2y = -2

3 - 2y + 2y = -2

3 = -2

Therefore This pair of linear Equations Has No Solution

Answer:

true

Step-by-step explanation:

cause I’ve done this before

The percentage error of the student's estimated mass is 2.15%.

The percentage or percent error is the difference between the estimated value and the actual value divided by the actual value, multiplied by 100.

The estimated mass of the student = 82.0 kg

The actual mass of the student = 83.8 kg

Absolute error = Actual Value - Estimated Value

= 1.8 kg (83.8 - 82.0)

Percentage error = Actual Value - Estimated Value /Actual Value x 100

= 2.15% (1.8/83.8 x 100)

Thus, we can conclude that the student's estimate contains a 2.15% error.

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The standard error is of 0.2246, which means that the mean scores for samples of 30 vary around 0.2246 from the mean.

The confidence interval is the range of values that you expect your estimate to fall between a certain percentage of the time if you run your experiment again or re-sample the population in the same way.

The confidence interval formula is [tex]\bar x\pm t\frac{s}{\sqrt{n}}[/tex].

Where, [tex]\bar x[/tex] is the sample mean, t is the critical value, n is the sample size and s is the standard deviation for the sample.

Here,

[tex]\bar x[/tex] =1.267, s=1.23, n=30

The standard error is Se= 1.23/√30

= 0.2246

Item a:

The standard error is of 0.2246, which means that the mean scores for samples of 30 vary around 0.2246 from the mean.

Item b:

Using a t-distribution calculator, considering a confidence level of 0.99 with 30 - 1 = 29 df, the critical value is t = 2.7564.

[tex]\bar x\pm tS_c[/tex]

[tex]\bar x+ tS_c[/tex] =1.267-2.7564(0.2246) =0.6479

[tex]\bar x- tS_c[/tex] =1.267+2.7564(0.2246) =1.8861

Therefore, the 99% confidence interval to estimate the true mean score on this question is (0.6479, 1.8861). It means that we are 99% that the true mean score of all students in this question is between 0.6479 and 1.8861.

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Answer:

(0.0301, 0.1733)

Step-by-step explanation:

[tex]\displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\\\\\\CI_{98\%}=\biggr(\frac{287}{522}-\frac{233}{520}\biggr)\pm 2.326\sqrt{\frac{\frac{287}{522}(1-\frac{287}{522})}{522}+\frac{\frac{233}{520}(1-\frac{233}{520})}{520}}\\\\\\CI_{98\%}\approx\{0.0300,0.1734\}[/tex]

Hence, the best choice is (0.0301, 0.1733) which tells us that we are 98% confident that the difference between the two proportions of men and women who successfully found Egypt on the map when highlighted is between 0.0301 and 0.1733

Answer A: Construct a two-sample z-interval for the difference between population proportions.

What is fraction?

A fraction is a mathematical expression that represents a part of a whole or a ratio between two quantities. It is written as a/b, where a is called the numerator and b is called the denominator. The numerator represents the number of parts or units being considered, while the denominator represents the total number of parts or units in the whole.

According to given conditions:

The best method for the researcher to use to estimate the true difference between the population proportions would be to:

Answer A: Construct a two-sample z-interval for the difference between population proportions.

This is the most appropriate method for estimating the difference between population proportions, as it takes into account the variability of the sample proportions and provides a confidence interval for the true difference between the two populations. The formula for calculating the two-sample z-interval for the difference between population proportions is:

(p₁−p₂) ± z x √[(p₁(1−p₁))/₁ + (p₂(1−p₂))/₂]

Therefore, Answer A: Construct a two-sample z-interval for the difference between population proportions.

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Answer:

Step-by-step explanation:

(3x + 7y)(3x - 7y)  FOIL

9x² -49y²   yes, this is difference of two squares

(2x - 7y)(2x - 7y)   FOIL

4x² -14x -14x +49y²

4x² -28x + 49y²   not difference of two squares

9x²(5x - 7)    Distribute the 9x²

45x³ - 63x²   not difference of two squares because one term is cubed

(6x - 4y)²    Distribute the square term

36x² - 16y²   yes, difference of two squares

Answer:

The ordered pair that is a solution is (2, 10)

Step-by-step explanation:

2x + 6 = 4x + 2  Subtract 2x from both sides

2x - 2x + 6 = 4x - 2x + 2

6 = 2x + 2  Subtract 2 from both sides

6 - 2 = 2x + 2 -2

4 = 2x  Divide both sides by 2

2 = x

Substitute 2 for x in either of the two original equations.

y = 2x + 6

y = 2(2) + 6

y = 4 + 6

y = 10

The ordered pair that is a solution is (2, 10)

Answer:

x = 40

Step-by-step explanation:

Alternate exterior angles are congruent (equal).

2x + 16 = 96   Subtract 16 from both sides

2x + 16 -16 = 96 - 16

2x = 80  Divide both sides by 2

x = 40

Answer:

x=40

Step-by-step explanation:

The question you have indicated appears incomplete. However, here is a very similar example that uses the same logic and principles of probability and proportion. In this case, the proof is given below.

a.

Given: f(x) = 2(x+2)/5 , 0 <x< 1

Calculating P(0 <X< 1)

P(0 <X< 1) = ∫ f(x) dx {0.1}

Substitute 2(x+2)/5 for f(x)

P(0 <X< 1) = ∫ 2(x+2)/5 dx {0.1}

P(0 <X< 1) = 2/5 ∫(x+2) dx {0.1}

Integrate with respect to x

P(0 <X< 1) = 2/5 (x²/2+2x) {0.1}

P(0 <X< 1) = 2/5 (1²/2+2(1))

P(0 <X< 1) = 2/5 (½+2)

P(0 <X< 1) = 2/5 * 5/2

P(0 <X< 1) = 1 ----- Proved

b.

Here we're to solve for P(¼<X< ½)

P(¼<X< ½) = ∫ f(x) dx {¼,½}

Substitute 2(x+2)/5 for f(x)

P(¼<X< ½) = ∫ 2(x+2)/5 dx {¼,½}

P(¼<X< ½) = 2/5 ∫(x+2) dx {¼,½}

P(¼<X< ½) = 2/5 (x²/2+2x) {¼,½}

P(¼<X< ½) = 2/5 [ (½²/2+2(½)) - (¼²/2+2(¼))]

P(¼<X< ½) = 2/5 [(⅛+1)-(1/32 + ½)]

P(¼<X< ½) = 0.2375

The probability that more than ¼ but fewer than ½ of the people contacted will respond to this type of solicitation is 0.2375

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Full Question:

The proportion of people who respond to a certain mail-order solicitation is a continuous random variable X that has the density function f(x) = 2(x+2)/5 , 0 <x< 1, 0, elsewhere.

(a) Show that P(0 <X< 1) = 1.

(b) Find the probability that more than ¼ but fewer than ½ of the people contacted will respond to this type of solicitation.

Answer:

11° 33'

Step-by-step explanation:

[tex]21' = \left(\dfrac{21}{60}\right)^\circ = \left(\dfrac{7}{20}\right)^\circ \\50^\circ 21' = 50 \dfrac{21}{60}\; degrees\\\\38^\circ 48' = 38 \dfrac{48}{60} \;degrees[/tex]

Break
[tex]50 \dfrac{21}{60} = 49 + 1 \dfrac{21}{60}\\[/tex]

[tex]1 \dfrac{21}{60} = \dfrac{1 \times 60 + 21}{60} = \dfrac{81}{60}[/tex]

Therefore
[tex]50^\circ21' - 38^\circ48' = 49\dfrac{81}{60} - 38\dfrac{48}{60}[/tex]

Subtract the whole numbers:
49 - 38 = 11

Subtract the fractions:
[tex]\dfrac{81}{60} - \dfrac{48}{60} = \dfrac{33}{60} = 33'[/tex]

So

50°21' - 38°48' = 11° 33'

We expect that, on average, 1 out of 25 patients calling in claiming to have the flu will actually have the flu.

According to the nurse's observation, there is a 4% chance that a patient who calls the medical advice line and claims to have the flu truly has.

As a result, we anticipate that 4 out of every 25 people who phone in will truly have the flu.

To figure this out, we can use the formula below:

Probability of having the flu multiplied by the total number of patients equals the anticipated number of flu patients.

Estimated number of influenza patients = 0.04 times 25

Expected number of influenza patients is one.

Hence, on average, we anticipate that 1 out of every 25 individuals who call in and claim to have the flu will truly have it.

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[tex]\textit{Partial Variation} \\\\ y = k_ox+k_1\hspace{5em}\textit{"y" varies partly with a constant and directly with "x"} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{"D" varies partly with a constant and directly with "V"}}{D=k_oV+k_1}\qquad \textit{we know that} \begin{cases} V=40\\ D=150\\[-0.5em] \hrulefill\\ V=54\\ D=192 \end{cases}[/tex]

[tex]\boxed{\begin{array}{llll} 150=40k_o+k_1\\\\ 192=54k_o+k_1 \end{array}}\qquad \stackrel{ \textit{using elimination method} }{\begin{array}{llll} -150=-40k_o-k_1\\\\ ~~ 192= ~~ 54k_o+k_1\\\cline{1-1} ~~ 42~= ~~ 14k_o+0 \end{array} }\qquad \implies 42=14k_o \\\\\\ \cfrac{42}{14}=k_o\implies \boxed{3=k_o}\hspace{5em}\stackrel{\textit{substituting on the 1st equation}}{150=40(3)+k_1}[/tex]

[tex]\cfrac{150}{40(3)}=k_1\implies \boxed{\cfrac{5}{4}=k_1}\hspace{5em} {\Large \begin{array}{llll} D=3V+\cfrac{5}{4} \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \textit{when V=73, what is "D"?}\qquad D=3(73)+\cfrac{5}{4}\implies D=\cfrac{881}{4}\implies D=220\frac{1}{4}[/tex]

We kindly invite to see the image attached below to know the representation of the parabola according to its features.

A parabola is the graphic representation of any quadratic equation, that is, a polynomial of the form y = a · x² + b · x + c, where a, b, c are real coefficients. An alternative presentation is the vertex form:

y - k = C · (x - h)²

Where:

The procedure to sketch the parabola according to its features is described below:

Step 1 - Determine the y-intercept:

y = 6 · 0² - 17 · 0 + 12

y = 12

Step 2 - Determine the x-intercepts.

6 · x² - 17 · x + 12 = 0

6 · [x² - (17 / 6) · x + 2] = 0

6 · (x - 3 / 2) · (x - 4 / 3) = 0

x = 3 / 2 or x = 4 / 3

Step 3 - Determine the coordinates of the vertex:

y = 6 · x² - 17 · x + 12

y = 6 · [x² - (17 / 6) · x + 2]

y + 6 · (1 / 144) = 6 · [x² - (17 / 6) · x + 2 + 1 / 144]

y + 1 / 24 = 6 · [x² - (17 / 6) · x + 289 / 144]

y + 1 / 24 = 6 · (x - 17 / 12)²

(h, k) = (17 / 12, - 1 / 24)

Steps 4 & 5 - Add the points on Cartesian plane and match the resulting curve.

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The set of data that is skewed-right, there is at least 99.7% of the data within 3 standard deviations.

A statistical principle known as the empirical rule, also known as the three-sigma rule or 68-95-99.7 rule, holds that with a normal distribution, almost all observed data will lie within three standard deviations (denoted by ) of the mean or average (denoted by ).

Some points about standard deviation using Empirical Formula:

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the probability of selecting a card that is not prime from the set {4, 5, 6, 7, 8} is 0.6 or 60%.

How we solved?

To determine the probability of selecting a card that is not prime from the set {4, 5, 6, 7, 8}, we first need to identify which numbers in the set are prime.

The prime numbers in the set are 5 and 7, so the non-prime numbers are 4, 6, and 8.

The total number of cards in the set is 5.

Therefore, the probability of selecting a non-prime card is:

P(not prime) = number of non-prime cards / total number of cards

P(not prime) = 3 / 5

P(not prime) = 0.6 or 60%

So the probability of selecting a card that is not prime from the set {4, 5, 6, 7, 8} is 0.6 or 60%.

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