Kelsey has a list of possible functions. pick one of the g(x) functions below and then describe to kelsey the key features of g(x), including the end behavior, y-intercept, and zeros. g(x) = (x 2)(x − 1)(x − 2) g(x) = (x 3)(x 2)(x − 3) g(x) = (x 2)(x − 2)(x − 3) g(x) = (x 5)(x 2)(x − 5) g(x) = (x 7)(x 1)(x − 1)

The standard error of the sample mean to the nearest hundredth  is  mathematically given as

S.E=0.784

The amount by which the population means deviates from a sample mean is represented by the standard error of the mean, which is more often referred to as simply the standard error.

It informs you how much the sample means would change if you were to perform an experiment using fresh samples from the same population but this time uses different samples.

The standard error is obtained by calculating the standard deviation and then dividing that number by the square root of the sample size. It determines the accuracy of a sample mean by factoring in the variation in sample means that exists from one set of data to the next.

Generally, The population's mean height is 62.95 inches, so let's use that.

Taking into account that the population has a standard deviation of 5.65 inches

The sample mean is used to calculate the standard error of the mean.

[tex]SE=\frac{SD}{\sqrt{n}}[/tex]

Therefore

[tex]SE=\frac{5.65}{ \sqrt{52}}[/tex]

S.E=0.7835

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The total time in seconds the ball will take to travel before returning the ground is 2.25 seconds.

The breaking or breakdown of an entity (such as an integer, a matrices, or a polynomials) into a products of another unit, or factors, whose multiplication results in the original number, matrix, etc., is known as factorisation or factoring in mathematics. 

Calculation for the time;

Let 't' be the time in second the ball will travel.

Let H(t) be the total height travelled by the ball.

The equation of the height is given as

[tex]H(t)=-16 t^{2}+36 t[/tex]

As soon as the ball will return to the ground the total height will become zero.

So, put equation of height equal to zero.

[tex]-16 t^{2}+36 t=0[/tex]

Factorise the above equation;

[tex]-16 t\left(t-\frac{9}{4}\right)=0[/tex]

Put each value equals to zero to get the value of time.

[tex]\begin{aligned}&t=0 \mathrm{sec} \\&t=9 / 4=2.25 \mathrm{sec}\end{aligned}[/tex]

As, time can not be zero

Therefore, the total time taken by the ball to return to the ground is 2.25 sec.

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Answer:

The answer is B.

[tex] \frac{1025}{4} [/tex]

Step-by-step explanation:

[tex] \frac{x {}^{6} - x}{4y} ...substitute \: x = - 4 \: and \: y = 4 \\ \frac{( - 4) {}^{6} - (- 4) }{4(4)} ...apply \: exponent \: rule \\ \frac{4096 + 4}{16} = \frac{4100}{16} = \frac{1025}{4} [/tex]

(-4)^6 - - 4 / 4 x 4

4096 + 4 / 16

4100/16

= 1025/4

So, answer is B. 1025/4

Hope this helps!

Ellipses that are represented by equations, whose eccentricities are less than 0.5 are:

To identify the ellipses:

For the standard-form equation of an ellipse: Ax^2+Bx+Cy^2+Dy+E=0

We can define:

Then the eccentricity can be shown to be:

For Eccentricity<0.5, we want:

Checking the values of p/q for the given equations, we have p/q= equations (A), (B), and (F) are of ellipses with eccentricity < 0.5.

Therefore, ellipses that are represented by equations, whose eccentricities are less than 0.5 are:

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The complete question is given below:

Identify the ellipses, represented by equations, whose eccentricities are less than 0.5.

(A) 49x2 -98x + 64yz +256y -2,831=0            

(B) 81x2 -648x +100yz +200y -6,704 =0                                      

(C) 6x2 -12x +54y2 +108y -426 =0                  

(D) 49x2 + 196x +36y2 +216y -1,244 =0                                        

(E) 4x +32x +25y2 - 250y +589 =0                

(F) 64x2 +512x +81y2 -324y -3,836 =0

[tex]\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{lllllllll} \log_{\underline{10}}(10)=1&\implies &\underline{10}^1=10 ~~ \checkmark\\\\ \log_{10}(100)=2&\implies &10^2=100~~ \checkmark\\\\ \log_{10}(55)=1.5&\implies &10^{1.5}=55\implies 10^{\frac{3}{2}}=55\implies \sqrt[2]{10^3}=55\\\\ &&\sqrt{1000}\ne 55 ~~ \bigotimes \end{array}[/tex]

increments over the exponent, are not directly proportional to increments on the resulting values, namely the half-way from 10² and 10⁴ is not 10³, because 10⁴ is 100 times 10², and 10³ is simply 10 times 10².

There would not be enough time to reach the half marathon distance.

The distance run in a week can be represented with a linear equation. A linear equation is an equation that has a single variable raised to the power of 1.

The linear equation that would be used to represent the distance run would be in the form:

Miles run in a week = miles run the previous week - increase in miles

Increase in miles = 6 - 4.5 =  1.5 miles

Miles run in the fourth week = 1.5 + 6  = 7.5 miles

Miles run in the fifth week = 7.5 miles + 1.5 miles = 9 miles

Miles run in the sixth week = 9 + 1.5 = 10.5 miles

There would not be enough time to reach the half marathon distance because in the sixth week she would be able to run 10.50 miles

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Subtracting the volume of the cylinder from the volume of the prism, the volume of metal in the hex nut to the nearest tenth exists [tex]$$23.6 cm^3[/tex]

Diameter of the cylinder be d = 1.6 cm

Apothem of the hexagon be a = 2 cm

Thickness of the steel hex nut be t = 2 cm

Volume of the prism be [tex]V_p[/tex]

Volume of the cylinder be [tex]V_c[/tex]

Volume of metal in the hex nut,

[tex]$$V = V_p - V_c[/tex]

To estimate the volume of a prism,

[tex]$$V_p = A_b h[/tex]

Ab = n L a / 2

Number of the sides, n = 6

The side of the hexagon be L

Height of the prism, h = t = 2 cm

Central angle in the hexagon, A = 360°/n

A = 360°/6 = 60°

[tex]$tan (\frac{A}{2} )=(\frac{L/2}{a})[/tex]

simplifying the value of L, we get

[tex]$tan (\frac{60}{2} )=(\frac{L/2}{2})[/tex]

[tex]$tan 30}=(\frac{L/2}{2})[/tex]

[tex]$tan (\frac{\sqrt{3}}{3} )=(\frac{L/2}{2})[/tex]

Solving for L/2:

[tex]$\frac{2 \sqrt{3}}{3} =\frac{L}{2}[/tex]

Solving the value of L, we get

[tex]$2\frac{2 \sqrt{3}}{3} =L[/tex]

[tex]$\frac{4 \sqrt{3}}{3} =L[/tex]

[tex]$L=4 \sqrt{3}/3 cm[/tex]

Ab = n L a / 2

Substitute the values in the above equation, we get

[tex]$A_b=\frac{6 (4 \sqrt{3}/3)(2)}{2}[/tex]

[tex]$$A_b=24 \sqrt{3}/3 $$cm^2[/tex]

[tex]$A_b=8 \sqrt{3} cm^2[/tex]

[tex]V_p = A_b h[/tex]

substitute the values in the above equation, we get

[tex]$V_p=(8 \sqrt{3})(2)[/tex]

[tex]$V_p=16 \sqrt{3} cm^3[/tex]

[tex]$$V_p=16 (1.732) cm^3[/tex]

[tex]$$V_p=27.712 cm^3[/tex]

To estimate the volume of cylinder,

[tex]$V_c[/tex] = (π[tex]d^2[/tex]/4) h

Here, π = 3.14 and d = 1.6 cm

Height of the cylinder, h = t = 2 cm

substitute the values in the above equation, we get

[tex]$V_c=[3.14 (1.6) / 4] (2)[/tex]

[tex]$V_c=[3.14 (2.56) / 4] (2)[/tex]

[tex]$V_c=(2.0096) (2)[/tex]

[tex]$$V_c=4.019 cm^3[/tex]

Substitute the values in the equation, we get

[tex]$$V=V_p-V_c[/tex]

[tex]$$V=27.712 - 4.019[/tex]

[tex]$$V=23.693 cm^3[/tex]

[tex]$$V=23.6 cm^3[/tex]

Therefore, the volume of metal in the hex nut, to the nearest tenth exists [tex]$$23.6 cm^3[/tex].

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[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

let's solve ~

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4}{y - 6} + \cfrac{5}{y + 3} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4(y + 3) + 5(y - 6)}{(y - 6)(y + 3)} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4y + 12 + 5y - 30}{ {y}^{2} + 3y - 6y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{9y - 18}{ {y}^{2} - 3y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 9y - 18 = 7y - 4[/tex]

[ denominator is same, so numerator must have same value to be equal ]

[tex]\qquad \sf  \dashrightarrow \: 9y - 7y = - 4 + 18[/tex]

[tex]\qquad \sf  \dashrightarrow \: 2y = 14[/tex]

[tex]\qquad \sf  \dashrightarrow \: y = 7[/tex]

Answer:

Step-by-step explanation:

The most basic quadratic function is f(x) = x2, whose graph appears below. ... us to graph an entire family of quadratic functions using transformations.

Answer:

20.0

Step-by-step explanation:

We can use the distance formula, d = [tex]\sqrt{(x_{1}-x_{2}) ^{2} +(y_{1}-y_{2}) ^{2}}[/tex]  to find the length of three sides separately.

AB =  [tex]\sqrt{(-4+2) ^{2} +(1-3}) ^{2}}[/tex] = [tex]\sqrt{8}[/tex]

BC =  [tex]\sqrt{(-2-3) ^{2} +(3+4) ^{2}}[/tex] = [tex]\sqrt{74}[/tex]

CA =  [tex]\sqrt{(3+4) ^{2} +(-4-1) ^{2}}[/tex] = [tex]\sqrt{74}[/tex]

Perimeter = [tex]\sqrt{74}[/tex] + [tex]\sqrt{8}[/tex] + [tex]\sqrt{74}[/tex] = 20.0330776588 units

Answer:

Step-by-step explanation:

Answer:

144π or 452.16 units³

Step-by-step explanation:

The formula for volume of a cylinder is [tex]\pi r^2 h[/tex]. We can use the given values for radius and height to find the volume of the cylinder.

[tex]V=\pi r^2 h\\V=\pi6^2(4)\\V=\pi36(4)\\V=144\pi \ OR\ 452.16[/tex]

The volume of the cylinder is 144π or 452.16 units³.

The rational expression that results into x-4/2(x+4) is x²-16/2x+8 * x+4/x²+8x+16

Rational expressions are written in the form a/b where a and b are integers.

According to the question, we need to determine the expression that is equal to x-4/2(x+4)

From the fourth option;

x²-16/2x+8 * x+4/x²+8x+16

Factorize the quadratic part to have;

x²-16/2x+8 * x²+8x+16/x+4

(x+4)(x-4)/2(x+4) * x+4/(x+4)^2

Cancel out the common expression to have;

x-4 * 1/2(x+4)

x-4/2(x+4)

Hence the rational expression that results into x-4/2(x+4) is x²-16/2x+8 * x+4/x²+8x+16

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For the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80

Minimum = 48

Lower quartile = (54+56)/2

Lower quartile = 110/2

Lower quartile = 55

Median = (63+64)/2

Median = 127/2

Median = 63.5

Upper quartile = (74+74)/2

Upper quartile = 148/2

Upper quartile = 74

Maximum = 80

Interquartile range = Upper quartile - Lower quartile

Interquartile range = 74 - 55

Interquartile range = 19

Hence, for the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

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When two coins are flipped simultaneously 24 times, on average 12 times we will get a head and a tail.

For given question,

When two coins are tossed simultaneously, the sample space is as follows:

S = { HH, HT, TH, TT} where H denotes Head and T denotes tail.

Using the formula of probability,

P = Number of favorable outcomes/total number of outcomes,

we get,

P(HT) = 1/4

and

P(TH) = 1/4

We know that the occurrence of two mutually exclusive events is the sum of their individual probabilities.

So, P(Head on one and Tail on other)

= P(HT) + P(TH)

= 1/4 + 1/4

= 2/4

= 1/2

= 0.5

So, when two coins flipped simultaneously, the probability of flipping 2 coins and getting a head and a tail is 0.5

In this question, we need to find the number of times getting a head and a tail when two coins are flipped simultaneously 24 times

= 0.5 × 24

= 12

Therefore, When two coins are flipped simultaneously 24 times, on average 12 times we will get a head and a tail.

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Angle <QAB is =15° because the opposite angles of an isosceles triangle are equal.

The length of the straight line AB = 80cm

The angle at a point = 360°

Angle AQB= 360 - 210° = 150

But the angle that makes up a triangle= 180°

180-150= 30°

But <QAB = <QBA because triangle AQB is an isosceles triangle.

30/2 = 15°

To calculate the length of the straight line the following is carried out using the sine laws.

a/ sina, = b sinb

a= 8cm, sin a { sin 15)

b= ? , sin B = 150

make b the subject formula;

8/sin15= b/sin 150

b= 8 × sin 150/sin 15

b= 80cm

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Answer:

9.224

Step-by-step explanation:

The answer is.........

25.99

Answer:

x < -19/7

Step-by-step explanation:

Solve for x

2x+5  < (x+1)/4

Multiply each side by 4

4( 2x+5) <  (x+1)/4 *4

8x + 20  < x+1

Subtract x from each side

8x+20-x < x+1-x

7x + 20 < 1

Subtract 20 from each side

7x+20-20 < 1-20

7x< -19

Divide by 7

7x/7 < -19/7

x < -19/7

Answer:

[tex] \red{x < \frac{ - 19}{7} }[/tex]

Step-by-step explanation:

[tex]2x + 5 < \frac{x + 1}{4} \\ \text{use \: cross \: muliplication} \\ 4(2x + 5) < x + 1 \\ 8x + 20 < x + 1 \\ 8x - x < - 20 + 1 \\ 7x < - 19 \\ \frac{7x}{7} < \frac{ - 19}{7} \\ x < \frac{ - 19}{7} [/tex]

From the given similar triangles we can conclude that the length of Side SQ is; 4

We are given that;

PQ is parallel and congruent to MT.

Now, from the concept of similar triangles , we can say that;

PS/TS = QS/MS

From the given triangle we see that;

PS = 3 - 2x

QS = 6x - 1

MS = 30

TS = 10

Thus;

(3 - 2x)/10 = (6x - 1)/30

Cross multiply to get;

3(3 - 2x) = 6x - 1

9 - 6x = 6x - 1

Rearranging gives;

6x + 6x = 9 + 1

12x = 10

x = 10/12

x = 5/6

Thus;

SQ = 6(5/6) - 1

SQ = 4

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Multiply 9 with 9, and we get:

9 × 9 = 81

Multiply 10 with 10, and we get:

10 × 10 = 100

We get the final result as:

Hope this helps :)

Answer:

Slope (m) = infinity

Step-by-step explanation:

Given points:

⇒ (9, 2) and (9, 27)

In order to find the slope of these points, we must use the slope formula.

So the formula is ⇒ y₂ - y₁ / x₂ - x₁

Plug in the given points:

⇒ 27 - 2 / 9 - 9

Solve both the top and bottom (numerator and denominator):

⇒ 25 / 0

Divide:

⇒ Undefined.

(Anything divided by zero is undefined).

The approximate solution to the given system of equation is (-2.71, 2.14)

From the question, we are to determine the approximate solution to the given system of linear equations.

The given equations are

y = 0.5x + 3.5

y = -2/3 x+ 1/3

From the given information, we are to show the solutions on a graph

The graph that shows the solution to the given system of equation is shown below.

The solution to the given system is the point of intersection of the lines. The coordinate of the point of intersection of the lines is (-2.71, 2.14). That is, x = -2.71, y = 2.14.

Hence, the approximate solution to the given system of equation is (-2.71, 2.14)

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From the two functions function 2 has the highest maximum value.

Given two functions,one is y=-[tex]x^{2} -2x-2[/tex] and f(-2)=1,f(-1)=6,f(0)=9,f(1)=10,f(2)=9,f(3)=6.

We ae required to choose a function which is having highest maximum possible value.

Function is like a relationship between two or more variables expressed in equal to form. Each value of x of a function has some value of y.

The highest value in the second function is 10 which is at x=1. Put the value of x=1 in y=-[tex]x^{2} -2x-2[/tex].

y=-1-2-2

y=-5

So it might not be highest value of this function.So the second function has highest maximum value of 10 at x=1.

Hence the second function has highest value.

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Coplanar circles that have the same center, but not necessarily the congruent radii are called concentric circles.

From the question, we have the following statements:

As a general rule, circles that have the above features are referred to as concentric circles.

This is so because concentric circles have the same center, and they do not intersect

Hence, coplanar circles that have the same center, but not necessarily the congruent radii are called concentric circles.

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The perimeter of the figure to the nearest tenth is 44.6 units

The perimeter of a figure is the sum of the whole sides.

Therefore,

A parallelogram has opposite sides equal to each other.

Therefore, the perimeter of the figure is as follows;

The figure combines a parallelogram and a semi circle.

Therefore,

perimeter of the figure = 4 + 14 + 14 + πr

perimeter of the figure = 32 + 3.14 × 4

perimeter of the figure = 32 + 12.56

perimeter of the figure = 44.56

perimeter of the figure ≈ 44.6 units

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Using the normal distribution, it is found that the standard deviation is of 26.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

For this problem, we have that the mean is of [tex]\mu = 95[/tex], while X = 110 has a p-value of 0.75, hence X = 110, Z = 0.575, and then we solve for [tex]\sigma[/tex] to find the standard deviation.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.575 = \frac{110 - 95}{\sigma}[/tex]

[tex]0.575\sigma = 15[/tex]

[tex]\sigma = \frac{15}{0.575}[/tex]

[tex]\sigma = 26[/tex]

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For given power series [tex]\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}[/tex] the radius of convergence is 2.

For given question,

We have been given a power series [tex]\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}[/tex]

We need to find the radius of convergence of the power series.

We use ratio test to find the radius of convergence of the power series.

Let [tex]a_n=\frac{(-1)^nx^n}{2^n}[/tex]

[tex]\Rightarrow a_{n+1}=\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}[/tex]

Consider,

[tex]\lim_{n \to \infty}|\frac{a_{n+1}}{a_n} |\\\\= \lim_{n \to \infty} |\frac{\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}}{ \frac{(-1)^nx^n}{2^n} } |\\\\=\lim_{n \to \infty} |\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}\times \frac{2^n}{(-1)^nx^n} |\\\\=\lim_{n \to \infty} |\frac{(-1)x}{2} |\\\\=\lim_{n \to \infty}|\frac{-x}{2} |\\\\=\frac{x}{2}[/tex]

By Ratio test, given power series converges at [tex]|\frac{x}{2} | < 1[/tex]

⇒ |x| < 2

So, the radius of convergence is 2.

Therefore, for given power series [tex]\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}[/tex] the radius of convergence is 2.

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By algebraic handling, the value of k of the system of equations is equal to 2.

Herein we find a system formed by two equations, a linear function and a quadratic equation with the following characteristics:

y = x² + b · x + 5        (1)

y = 2 · x + b               (2)

If we eliminate y in (1) and (2), then we have this expression:

x² + b · 3 + 5  = 2 · x + b

3² + b · x + 5 = 2 · 3 + b

3 · b + 14 = 6 + b

14 = 6 - 2 · b

8 = - 2 · b

b = - 4

By (2), y = k, x = 3 and b = - 4, we find the value of k:

k = 2 · 3 - 4

k = 6 - 4

k = 2

By algebraic handling, the value of k of the system of equations is equal to 2.

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Step-by-step explanation:

[tex] \sf \: T_n = a+(n-1)d[/tex]

[tex] \sf \: T_n = \frac{1}{4} +(n-1) \frac{1}{4} [/tex]

[tex]\sf \: T_n = \frac{1}{4} + \frac{n}{4} - \frac{1}{4} [/tex]

[tex]\sf \: T_n = \frac{n}{4}[/tex]

an equation for the nth term of the arithmetic sequence. is n/4

Step-by-step explanation:

TN=n/4 is the nth term of arithmetic sequence.