The table below shows the average SAT math scores from 1993-2002.

Year
SAT math scores
1993
503
1994
504
1995
506
1996
508
1997
511
1998
512
1999
511
2000
514

Using the data from the table determine if there is a linear trend between the year and the average SAT math scores and determine if there is an exact linear fit of the data. Describe the linear trend if there is one.
a.
Positive linear trend, an exact linear fit.
b.
Positive linear trend, not an exact linear fit.
c.
Negative linear trend, not an exact linear fit.
d.
Negative linear trend, an exact linear fit.


Please select the best answer from the choices provided

A
B
C
D
Mark this and return

The complete question is attached below.

The most correct phases that imply correlation without causation are 1, 5, and 6.

The correlation of two pairs of data values tells about the degree of movement(along or opposite) that can occur in one of the data values when another data value is increased or decreased respectively.

The most correct phases imply correlation without causation are;

The number of stuffed animals produced at a factory and the number of newborn babies.

The number of videos rented and the number of films in the theater.

The number of pets in the neighborhood and the amount of grass nearby.

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The equation for the red graph is f(x) = -x².

From the information given, g(x) = 4 - x².

The vertex of the point (0, 4) and (0, 0) for g(x) and f(x) respectively. The rule of the translation will be:

(x, y) = (x, y -4).

The equation of the function will be:

f(x) = g(x) - 4

f(x) = (4 - x²) - 4

f(x) = -x²

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The true statements are (A), (B) and (D)

Data is information that has been translated into a form that is efficient for movement or processing.

As, from the given table we can see that

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Answer:

p = 4

Step-by-step explanation:

[tex]m =\frac{p}{2} +3r[/tex]

[tex]\Longrightarrow m-3r=\frac{p}{2}[/tex]

[tex]\Longrightarrow 2m-6r=p[/tex]

[tex]\Longrightarrow p = 2m-6r[/tex]

If m = 5 and r = 1

Then

p = 2(5) - 6(1)

  = 10 - 6

  = 4

Step-by-step explanation:

To get the angle of QPR, we subtract 90-60=30. This proves that the triangle is a 30,60,90, as angle QRP must be 60°. To get the area, we must know the length and width. These can be derived from a known relationship within 30,60,90 triangles.

[tex]qp = \frac{\sqrt{3}}{2} n[/tex]

[tex]qr = \frac{1}{2} n[/tex]

[tex]pr = n[/tex]

because half of pr is 10, we can conclude that n is 20. So we plug 20 into N for QP and QR, and get values of

[tex]qp = 5 \sqrt{3} [/tex]

[tex]qr = 5[/tex]

Now we multiply (l*w)/2 and get:

[tex] \frac{(5 \sqrt{3} \times 5)}{2} = 21.65 \: units[/tex]

Answer: (1,4) and (1,3) because the have the same x-value

Step-by-step explanation:

A function is defined so that for each input (known as the domain), there is no more than one output (known as the range).

A one-to-one function has an inverse. The inverse is another function that undoes the action of the first one, so if we evaluate a function [tex]f[/tex] at some point [tex]x[/tex] to get the number [tex]f(x)[/tex], evaluating the inverse at [tex]f(x)[/tex] will recover the original input [tex]x[/tex]. In other words,

[tex]f^{-1}(f(x)) = x[/tex]

The process works in the opposite direction, too:

[tex]f\left(f^{-1}(x)\right) = x[/tex]

From the given definition of [tex]g[/tex], we have [tex]g(-4) = 3[/tex], so taking inverses on both sides, we find

[tex]g(-4) = 3 \implies g^{-1}(g(-4)) = g^{-1}(3) \implies \boxed{g^{-1}(3) = -4}[/tex]

Given [tex]h(x)=2x-13[/tex], evaluating [tex]h[/tex] at its inverse will recover [tex]x[/tex], so that

[tex]h\left(h^{-1}(x)\right) = x \implies 2h^{-1}(x) - 13 = x \implies \boxed{h^{-1}(x) = \dfrac{x+13}2}[/tex]

[tex](h\circ h^{-1})(x)[/tex] is another way of writing the compound function [tex]h\left(h^{-1}(x)\right)[/tex]. As already discussed, this reduces to [tex]x[/tex], so

[tex]\boxed{\left(h\circ h^{-1}\right)(-9) = -9}[/tex]

Answer:

x=12

Step-by-step explanation:

16x-30-(14x-6)=2x-24=0. add 24 on both sides to get 2x=24. divide by 2 on both sides to get x=12

The standard error will be equal to 0.75.

The standard deviation of a statistic's sample distribution, or an approximation of that standard deviation, is the statistic's standard error. The term "standard error of the mean" is used when referring to a statistic that is the sample mean.

The standard error will be calculated by using the formula:-

SE = [tex]\sigma[/tex] / √n

Here [tex]\sigma[/tex] is the standard deviation and n is the sample number.

SE = 6.5 / √75

SE = 0.75

Therefore the standard error will be equal to 0.75.

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Answer:

[tex]27[/tex]

Step-by-step explanation:

[tex]g(f(0))=g(2(0)+3)=g(3)=3^3=27[/tex]

Answer:

$809.25

Step-by-step explanation:

→ Find the sum of the sales amount

2400 + 3200 + 1900 + 800 = 8300

→ Find the amount that s commission

( 9.75 × 8300 ) ÷ 100 = $809.25

Step-by-step explanation:

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Answer:

See below

Step-by-step explanation:

Longest is the down tube.....it is opposite the largest angle (95°) angle in the triangle .    This is a SCALENE triangle ( all sides of different length)

The longest tube that we have here is the down tube. The triangle that is formed id the scalene triangle

A scalene triangle is a type of triangle where all three sides have different lengths. In other words, no two sides of a scalene triangle are equal in length. Additionally, the angles of a scalene triangle are also different from each other.

The term "scalene" comes from the Greek word "skalenos," which means "uneven" or "unequal." This name reflects the defining characteristic of the scalene triangle, which is the absence of any equal sides or angles.

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The solution of the logarithmic expression [tex]\log_2(17 -\sqrt{19}) + \log_2(17 +\sqrt{19})[/tex] is [tex]\log_2[270][/tex]

The expression is given as:

[tex]\log_2(17 -\sqrt{19}) + \log_2(17 +\sqrt{19})[/tex]

Apply the product law of logarithm

[tex]\log_2[(17 -\sqrt{19}) *(17 +\sqrt{19})]\\[/tex]

Apply the difference of two squares

[tex]\log_2[(17^2 -(\sqrt{19})^2][/tex]

Evaluate the squares

[tex]\log_2[(289 -19][/tex]

Evaluate the difference

[tex]\log_2[270][/tex]

Hence, the solution of the logarithmic expression [tex]\log_2(17 -\sqrt{19}) + \log_2(17 +\sqrt{19})[/tex] is [tex]\log_2[270][/tex]

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Bikash borrowed Rs150,000 from Anish at the rare 21% per year at the end of 9 month. How compound interest should he pay compound half yearly.

The probability that Jamie buys a telecaster or a Less Paul is 45% or forty-five percent.

1. Find the individual probabilities:

Probabilities to buy a telecaster:

Probabilities to buy a Les Paul:

2. Add the probabilities:

3. Multiply by 100:

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Answer:

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Step-by-step explanation:

i need it

Answer:

3^1 multiplied by 3^2 = 3^(1 + 2) = 3^3

162

Simplify ——————

(2•3^3)

Canceling out 2 as it appears on both sides of the fraction line

3^4 divided by 3^3 = 3^(4 - 3) = 3^1 = 3

Step-by-step explanation:

hope this helps you ♡

The function that represents the Scooter Company's total profit is given by:

P(x) = -2x² + 250x - 2,000.

The profit function is calculated by the revenue function subtracted by the cost function, hence:

P(x) = R(x) - C(x).

In this problem, we have that:

Hence the profit function is:

P(x) = R(x) - C(x).

P(x) = 400x - 2x² - (150x + 2000).

P(x) = -2x² + 250x - 2,000.

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Answer:

[tex]\frac{4}{3}[/tex] or [tex]1\frac{1}{3}[/tex]

Step-by-step explanation:

[tex]1 + \frac{1}{3} \\= \frac{3}{3} + \frac{1}{1}\\= \frac{4}{3}[/tex]

(if the question is asking for the simplified form , it is [tex]1\frac{1}{3}[/tex]

Answer:

C

Step-by-step explanation:

x = [tex]\frac{2}{3}[/tex] πr³ ( multiply both sides by 3 to clear the fraction )

3x = 2πr³ ( divide both sides by 2π )

[tex]\frac{3x}{2\pi }[/tex] = r³ ( take cube root of both sides )

[tex]\sqrt[3]{\frac{3x}{2\pi } }[/tex] = r

Answer:

[tex]r=\sqrt[3]{\frac{3x}{2\pi }}[/tex]

Step-by-step explanation:

This question make look scary, but don't judge its book by its cover.

[tex]x=\frac{2}{3} \pi r^{3} \\\frac{2}{3} \pi r^{3}=x\\r^{3} =x / \frac{2\pi }{3} \\r^{3} =x*\frac{3}{2\pi } \\r^{3} =\frac{3x}{2\pi } \\r=\sqrt[3]{\frac{3x}{2\pi }}[/tex]

The coordinate of C( a+c , b) ,  AC are ((a+c)/2 , b/2) , BD are ((a+c)/2 , b/2)

Coordinates is the value assigned to a point in a x-y graph to determine its location.

The coordinates of parallelogram is

A ( 0,0) B (a,0) , C ( __ , ___) , D ( c , b)

The y coordinate is same as D as b

and the x coordinate will be a+c

C( a+c , b)

The coordinate of the mid point of AC are ((a+c)/2 , b/2)

The coordinate of the mid point of BD are ((a+c)/2 , b/2)

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Answer:

No. greater than of equal to (-5) = -4,-3,-2,-1,0,1,2,3,4,5,6...........

No. less than 5 = 4,3,2,1 ,0,-1......

{ -4,-3,-2,-1,0,1,2,3,4}

The slope of line p is -5, so substitute into point-slope form, we get

[tex]\boxed{y-37=-5(x+2)}[/tex]

Answer:

x = 9 or B

Step-by-step explanation:

Formula

(x + 2)^2 = 121                        

Comment

The easiest way to do this is to take the square root immediately, which is usually possible.

Solution

√(x + 2)^2 = √121                     Take the square root

x + 2 = 11                                   Subtract 2 from both sides

x + 2 -  2 = 11 - 2

x = 9

Answer: B) 9 inches by 9 inches

Step-by-step explanation:

Edg 2022

Answer:

[tex]\text{C.} \ \ \ {\left(\textit{AB}\right)}^{2} \ = \ \left(\textit{AC}\right)\left(\textit{AD}\right)[/tex]

Step-by-step explanation:

This problem uses the concept of the tangent-secant theorem which describes the relationship of the segments a secant line and a tangent line with the associated circle. This theorem is found as Proposition 36 in Book 3 of Euclid's Elements.

As shown in the figure attached below, segment AB (in blue) forms a tangent with the circle BCD and segment AD (in orange) is the secant where it intersects the circle at point C.

Furthermore, let two segments (in green) be drawn one from point C and point D.

To show that [tex]\triangle ABC[/tex] is similar to [tex]\triangle ADB[/tex], notice that both triangles share a common angle [tex]\angle BAC[/tex]. Additionally, by the alternate segment theorem, [tex]\angle ABC[/tex] is equal to [tex]\angle ADB[/tex]. Therefore, [tex]\angle ACB[/tex] is also equal to [tex]\angle ABD[/tex].

Hence, [tex]\triangle ABC[/tex] is indeed similar to [tex]\triangle ADB[/tex]. This implies the ratio of the sides of both triangles is the same. Particularly,

                                                  [tex]\displaystyle{\frac{AB}{AD} \ \ = \ \ \frac{AC}{AB}}[/tex].

Then, performing cross multiplication yields

                                             [tex]{\left(AB\right)}^{2} \ \ = \ \ \left(AC\right)\left(AD\right)[/tex].

Therefore, the product of the lengths of the secant segment and its external segment is equal to the square of the length of the tangent segment.

Answer:

huh

Step-by-step explanation:

In a study of birth orders and intelligence, IQ tests were given to 18 and 19 years old men to estimate the size of the difference, we have that  [tex]|t|=0.618\leq t_c =2.101[/tex] and for this reason, we cannot reject the null hypothesis.

Null and Alternative Hypotheses

[tex]H_o: \mu_{1} = \mu_{2}\\\\Ha: \mu_1 \neq \mu_2 .[/tex]

Since the population standard deviations are unknown, we will apply a t-test on the means of the two populations, using two independent samples.

The degrees of freedom are df = 18, and the significance threshold is set at alpha = 0.05, based on the data presented. If we assume that the population variances are the same, we can calculate the degrees of freedom in this way:

The rejection region for this two-tailed test is R = \{t : |f| > 2.101}

Therefore the Test Statistics is

[tex]t= \frac{X 1 -X 2 }{\frac{\sqrt{ (n_1 -1)s_1 ^ 2 +(n_2 -1)s_2 ^2}} {n_1 +n_2 -2} (( 1/n_1 + 1 n_2 )}}[/tex]

[tex]t= \frac{X 1 -X 2 }{\frac{\sqrt{ (n_1 -1)s_1 ^ 2 +(n_2 -1)s_2 ^2}} {n_1 +n_2 -2} (( 1/n_1 + 1 n_2 )}}[/tex]

t=0.618

In conclusion. we have that  [tex]|t|=0.618\leq t_c =2.101[/tex]

For this reason, we cannot reject the null hypothesis.

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Answer: 49 in

Step-by-step explanation:

The value of the integration is

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C[/tex], where C is an integrating constant.

We know that finding the area of the curve's undersurface is the process of integration. To do this, cover the area with as many tiny rectangles as possible, then add up their areas. The sum gets closer to a limit that corresponds to the area under a function's curve. Finding an antiderivative of a function is the process of integration. If a function can be integrated and its integral over the domain is finite with the given bounds, then the integration is definite.

Here, we have to determine the value of the given integral

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx[/tex].

So,

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx = 7 \int {\frac{1-x^{3}-1 }{x^{3}+1} } \, dx[/tex][tex]= 7 \int [{1-\frac{1 }{x^{3}+1} } ]\, dx =7\int \, dx - 7 \int {\frac{1 }{x^{3}+1} } \, dx[/tex] ...(1)

Now,

[tex]\frac{1}{x^{3} +1} = \frac{1}{(x+1)(x^{2} -x+1)}[/tex]

We can write

[tex]\frac{1}{(x+1)(x^{2} -x+1)}=\frac{A}{(x+1)} + \frac{Bx+C}{(x^{2} -x+1)}[/tex]

i.e. [tex]1 = A(x^{2} -x+1)+(Bx+C)(x+1)[/tex]

i.e. [tex]1 = Ax^{2} -Ax + A+Bx^{2} +Bx+Cx+C[/tex]

i.e. [tex]1=(A+B)x^{2} +(-A+B+C)x +(A+C)[/tex]

Comparing both sides, we get

[tex]A+B=0 \implies B = -A[/tex] ...(2)

[tex]-A+B+C = 0 \implies -A+(-A)+C=0 \implies -2A+C=0[/tex] ...(3)

[tex]A+C=1[/tex] ...(4)

Subtracting (3) from (4),

[tex]A+C+2A-C=1-0 \implies 3A=1 \implies A=\frac{1}{3}[/tex]

From (4), [tex]C= 1-\frac{1}{3} =\frac{3-1}{3} =\frac{2}{3}[/tex]

From (2), [tex]B =-\frac{1}{3}[/tex]

We get

[tex]\frac{1}{x^{3}+1 } =\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{3} \frac{x-2}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-4}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1-3}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{3}{6} \frac{1}{x^{2} -x+1}[/tex]

[tex]=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{1}{2} \frac{1}{x^{2} -x+1}[/tex]

Integrate both sides,

[tex]\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \int \frac{1}{(x+1)} \, dx - \frac{1}{6} \int \frac{2x-1}{x^{2} -x+1}\, dx+\frac{1}{2} \int \frac{1}{x^{2} -x+1}\, dx[/tex]

i.e. [tex]\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \ln(x+1) - \frac{1}{6} ln (x^{2} -x+1)+\frac{\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })[/tex]

From (1),

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C[/tex]

Therefore, the value of the integration is

[tex]\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C[/tex], where C is an integrating constant.

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f(0) = -3(0) + 5 = 5

f(-4) = -3(-4) + 5 = 17

f(6) = ½(6) - 4 = -1

f(9) = 2

f(2) = ½(2) - 4 = -3

Using the restraints, find which equation the f(x) function validates and plug it in.