Which of the following scatter plots does not have a zero correlation?

The solution of the questions is given as

a)

The alternative hypothesis is that mu is equal to 3.0. ( the average GPA at KSU is not equal to 3.0)

b)

This is a case with two separate arguments. (This is a two-tailed case)

c)

Test Statistic is given by:

z= 2.00

d)

Critical values of Z come from the table and equal [tex]\pm[/tex]1.96.

e)

, p-value = 0.0455

f)

CI= (3.006, 3.594)

g)

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

h)

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

i) Conclusion: The evidence does not lend credence to the assertion that KSU's typical grade point average is equivalent to 3.0.

j)

The choices you choose in Questions 7, 8, and 9 are identical.

a)

The null hypothesis states that mu equals 3.0. ( the average GPA at KSU is equal to 3.0) (Claim)

The alternative hypothesis is that mu is equal to 3.0. ( the average GPA at KSU is not equal to 3.0)

b)

This is a case with two separate arguments.

c)

Test Statistic is given by:

[tex]z=\frac{\barx-u^0}{σ/Vn}\\\\z=33 - 3.0/1.5/V100[/tex]

z= 2.00

d)

[tex]\alpha = 0.05[/tex]

Critical values of Z come from the table and equal [tex]\pm[/tex]1.96.

e)

Z Score = 2.00

reading from table, p-value = 0.0455

f)

Confidence Interval:

[tex]CI= \barx \pm \frac{ZX\sigma }{\sqrt((n)}[/tex]

[tex]CI =3.3 \pm 1.96 x 1.5 / \sqrt{100}[/tex]

CI= (3.006, 3.594)

g)

The disparity is noteworthy given that the computed value of Z = 2.00 is higher than the crucial value of z = 1.96. Cast doubt on the null hypothesis.

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

h)

The difference may be considered statistically significant since the p-value of 0.0455 is lower than the alpha threshold of 0.05. Cast doubt on the null hypothesis.

Conclusion: The evidence does not back up the assertion that KSU students have a cumulative grade point average of 3.0.

i)

The difference is statistically significant since the value of 3.0 is not included in the confidence range (3.006 - 3.594). Cast doubt on the null hypothesis.

Conclusion: The evidence does not lend credence to the assertion that KSU's typical grade point average is equivalent to 3.0.

j)

The choices you choose in Questions 7, 8, and 9 are identical.

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The answer is [tex]\boxed {\frac{3b - 8}{4b}}[/tex].

Answer:

C: 1 Quart

Step-by-step explanation:

2 pints are the equivalent of 1 quart

Answer:1 Quart

Step-by-step explanation: Because 2 pints equals=1 quart

Using proportions, it is found that the value of x is of 5.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

In this problem, since Figure A is a scale of Figure B, the side lengths are proportional, hence the following rule of three is established:

x/12.5 = 2/5

Applying cross multiplication:

5x = 25

x = 5.

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The statement that is true about the box plots is: D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

For a data distribution that is displayed by a box plot, the measure of variation that can be used to describe the data distribution is the interquartile range.

Interquartile range, which is a measure of variation can be determined from a box plot by finding the value of the third quartile and first quartile, the finding their difference. The difference is the interquartile range.

Interquartile range = third quartile - first quartile.

Interquartile range for crackers = 85 - 75

Interquartile range for crackers = 10

Interquartile range for trail mix = 105 - 90

Interquartile range for trail mix = 15.

The bigger the value of the interquartile range, the more variation there is that exist in a data distribution.

Interquartile range for trial mix is the largest, therefore, the statement that is true about the box plots is: D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

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Step-by-step explanation:

1. Entrance of Head of State and IOC President

2. Playing of the national anthem

3. The parade of the athletes

4. The symbolic release of doves

5. Olympic Laurel Award

6. Official Speeches

7. Opening of the Games

8. Raising the Olympic flag and playing the Olympic Anthem

9. Athletes, judges and coaches’ oath

10. Lighting of the Olympic flame

11. The artistic programme

The correct option regarding the data-sets represented by the box and whisker plots is:

B. The distribution for town A is positively skewed, but the distribution for town B is symmetric.

A box and whisker plot shows three things:

For data-set A, we have that:

Since 30 - 20 > 20 - 15, the distribution is positively skewed.

For data-set B, we have that:

Since 40 - 30 = 30 - 20, the distribution is symmetric.

Hence the correct option is:

B. The distribution for town A is positively skewed, but the distribution for town B is symmetric.

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Answer:3 3/4 meters

Step-by-step explanation:

6/(1 3/5)=3.75 meters or 3 3/4 meters

Answer:

Step-by-step explanation:

ggggggg

Step-by-step explanation:

we have

B = J + 2

as the number of laps Bill ran.

to get the number of laps Jill ran we need to transform this equation, so that we get

J = ...

or

... = J

J has to be on one side, and the rest on the other (just as it is with B = ...).

so,

B = J + 2

B - 2 = J

or

J = B - 2

hey ! we are finished already !

Any line through the points [tex](a,b)[/tex] and [tex](c,d)[/tex] has

[tex]\gcd(c-a,d-b)+1[/tex]

lattice points. In this case, the count is GCD(45, 264) + 1.

Using Euclid's algorithm, we have

264 = 5•45 + 39

45 = 1•39 + 6

39 = 6•6 + 3

6 = 2•3 + 0

so that GCD(45, 264) = 3. Then there are 3 + 1 = 4 lattice points.

Modern numbers are the ten numerical digits used in constructing other numbers.

Modern numbers are also called Arabic numerals. They are the ten numerical digits:

0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Uses of modern numbers;

Hence, modern numbers are the ten numerical digits used in constructing other numbers.

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a) The coordinates of point A that meet the linear equation y = 2 · x + 7 are A(x, y) = (0, 7).

b) The equation of the line parallel to the line y = 2 · x + 7 is the line y = 2 · x + 3.

c) The coordinates of the midpoint of line segment BC is M(x, y) = (1, 7).

Herein we find three cases related to the linear equation y = 2 · x + 7. a) We need to determine the coordinates of point A, the y-coordinate can be determined by evaluating the function:

y = 2 · 0 + 7

y = 0 + 7

y = 7

The coordinates of point A that meet the linear equation y = 2 · x + 7 are A(x, y) = (0, 7).

b) Two lines are parallel to each other whether both have the same slope but different intercept. Then, the new line is of the form:

y = 2 · x + b

Now we proceed to find the intercept of the line:

b = y - 2 · x

b = 3 - 2 · 0

b = 3

The equation of the line parallel to the line y = 2 · x + 7 is the line y = 2 · x + 3.

c) The coordinates of point C is:

y = 2 · 2 + 7

y = 11

If we know that B(x, y) = (0, 3) and C(x, y) = (2, 11), then the coordinates of the midpoint of BC is:

M(x, y) = 0.5 · B(x, y) + 0.5 · C(x, y)

M(x, y) = 0.5 · (0, 3) + 0.5 · (2, 11)

M(x, y) = (1, 7)

The coordinates of the midpoint of line segment BC is M(x, y) = (1, 7).

The statement presents typing mistakes and omissions, complete form is shown below:

a) The line of y = 2x + 7 meets the y-axis at A, where x = 0. Write down the co-ordinates of A.

b) A line parralel to y = 2x + 7 passes through B(x, y) = (0,3) Find the equation of this line.

c) C is the point on the line y = 2x + 7 where x = 2. Find the the co-ordinates of the midpoint of BC.

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Given

[tex]7 - px - x^2 = 16 - (q + x)^2[/tex]

expanding the right side gives

[tex]7 - px - x^2 = 16 - (q^2 + 2qx + x^2)[/tex]

[tex]7 - px - x^2 = 16 - q^2 - 2qx - x^2[/tex]

Two polynomials of equal degree are the same if their coefficients are identical. This means

[tex]\begin{cases}16 - q^2 = 7 \\ -2q = -p\end{cases}[/tex]

[tex]16 - q^2 = 7 \implies q^2 = 9 \implies q=\pm3[/tex]

[tex]-2q = -p \implies p = 2q = \pm6[/tex]

Both [tex]p[/tex] and [tex]q[/tex] are positive, so [tex]\boxed{p=6}[/tex] and [tex]\boxed{q=3}[/tex].

Answer:

nope

Step-by-step explanation:

simplify the first equation

Answer:

  yes

Step-by-step explanation:

The associative and commutative properties of multiplication allow you to rearrange the product to the form shown in the question.

Division by 2 is effectively multiplication by 1/2.

  [tex](n\times(n+1))/2\qquad\text{given}\\\\=\dfrac{n(n+1)}{2}=\dfrac{n}{2}(n+1)\\\\=\boxed{(n/2)(n+1)}[/tex]

__

Additional comment

Evaluation of the expressions according to the order of operations will proceed differently for the two expressions.

For the first expression, evaluation steps are ...

For the second expression, evaluation steps are ...

As we said above, the properties of multiplication ensure the results are the same either way.

As a practical matter, for integer values of n, one of n and (n+1) will be even. It is usually convenient to divide the even number by 2. This means the evaluation might be ...

For certain computer representations of the numbers, results may differ depending on the specific numbers and the order of evaluation.

Answer:

[tex]\displaystyle \int \dfrac{\ln x}{x^8}\:\:\text{d}x=-\dfrac{\ln x}{7x^7}- \dfrac{1}{49x^7}+\text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

[tex]\displaystyle \int \dfrac{\ln x}{x^8}\:\:\text{d}x[/tex]

[tex]\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}[/tex]

[tex]\implies \displaystyle \int x^{-8}\ln x\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{4.8 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}[/tex]

Using Integration by parts:

[tex]\textsf{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}[/tex]

[tex]\textsf{Let }\dfrac{\text{d}v}{\text{d}x}=x^{-8} \implies v=-\dfrac{1}{7}x^{-8+1}=-\dfrac{1}{7}x^{-7}=-\dfrac{1}{7x^7}[/tex]

Therefore:

[tex]\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x^{-8}\ln x\:\:\text{d}x & = -\dfrac{1}{7x^7}\ln x- \int -\dfrac{1}{7x^7} \cdot \dfrac{1}{x}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}+ \int \dfrac{1}{7x^8}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}+ \int \dfrac{1}{7}x^{-8}\:\:\text{d}x\\\\& = -\dfrac{\ln x}{7x^7}-\dfrac{1}{7 \cdot 7}x^{-8+1}+\text{C}\\\\& = -\dfrac{\ln x}{7x^7}- \dfrac{1}{49x^7}+\text{C}\end{aligned}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Differentiating $\ln x$}\\\\If $y=\ln x$, then $\dfrac{\text{d}y}{\text{d}x}=\dfrac{1}{x}$\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}[/tex]

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Substitute [tex]y=\ln(x) \iff e^y = x[/tex] and [tex]dy=\frac{dx}x[/tex] to get

[tex]\displaystyle \int \frac{\ln(x)}{x^8} \, dx = \int ye^{-7y} \, dy[/tex]

Integrate by parts with

[tex]u = y \implies du = dy[/tex]

[tex]dv = e^{-7y} \, dy \implies v = -\dfrac17 e^{-7y}[/tex]

Then

[tex]\displaystyle \int u\,dv = uv - \int v\,du \\\\ \implies \int y e^{-7y} \, dy = -\dfrac17 ye^{-7y} + \frac17 \int e^{-7y} \, dy + C \\\\ ~~~~~~~~~~~~ = -\frac17 ye^{-7y} - \frac1{49} e^{-7y} + C \\\\ ~~~~~~~~~~~~ = \boxed{-\frac{\ln(x)}{7x^7} - \frac1{49x^7} + C}[/tex]

19) The sum of the arithmetic series is - 397. (Correct choice: E)

33) The sum of the geometric series is 1700. (Correct choice: E)

19) Arithmetic series are sets of elements generated by a linear expression of the form:

aₙ = a₁ + (n - 1) · d      (1)

Where:

If we know that a₁ = 27, n = 20 and d = - 5, then sum of the first 20 terms of the series is:

x = 27 + 22 + 17 + 12 + 7 + 2 + (- 3) + (- 8) + (- 13) + (- 18) + (- 23) + (- 28) + (- 33) + (- 38) + (- 43) + (- 48) + (- 53) + (- 58) + (- 63) + (- 68)

x = - 397

The sum of the arithmetic series is - 397. (Correct choice: E)

33) Geometric series are sets of elements generated by a exponential expression of the form:

aₙ = a₁ · rⁿ     (2)

Where:

If we know that a₁ = 1458 and r = 1 / 3, then the sum of the first 6 terms of the geometric series is:

x = 1458 + 162 + 54 + 18 + 6 + 2

x = 1700

The sum of the geometric series is 1700. (Correct choice: E)

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Answer:

x= -4 + radical 13

Step-by-step explanation:

The speed of the wave is 0.83 m/s.

According to the statement

we have given that the equation and we have to find the speed of the wave.

So, For this purpose, we know that the

The given equation is y(x,t) = (9. 13 mm) sin [(6. 65 m^-1)x - (5. 52 s^-1)t]

Then we have to find the speed of the wave

So, Comparing y(x,t) = (9. 13 mm) sin[(6. 65 m-1)x - (5. 52 s-1)t]

to the general expression y(x,t)=y m sin(kx−ωt),

Because the general expression is y(x,t)=y m sin(kx−ωt)

we see that k= 6.65 m −1 and ω= 5.52 rad/s.

The speed of the wave is

v=ω/k=(5.52 rad/s)/(6.65 m )= 0.83 m/s

So, The speed of the wave is 0.83 m/s.

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Using limits, the correct option regarding the end behavior of the function is given by:

A. as x→∞, y→−∞ as x→−∞, y→−∞.

The end behavior is found calculating the limit of f(x) as x goes to infinity.

For this problem, the equation is given by:

[tex]f(x) = -x^6 - 6x^5 + 50x^3 + 45x^2 - 108x - 108[/tex]

Since x goes to infinity, we consider only the term with the highest exponent, hence the limits are given as follows:

[tex]\lim_{x \rightarrow -\infty} f(x) = \lim_{x \rightarrow -\infty} -x^6 - 6x^5 + 50x^3 + 45x^2 - 108x - 108 = \lim_{x \rightarrow -\infty} -x^6 = -(-\infty)^6 = -\infty[/tex]

[tex]\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} -x^6 - 6x^5 + 50x^3 + 45x^2 - 108x - 108 = \lim_{x \rightarrow \infty} -x^6 = -(\infty)^6 = -\infty[/tex]

Hence the correct option is:

A. as x→∞, y→−∞ as x→−∞, y→−∞.

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Answer: 24+12=36

Step-by-step explanation:

36= _+_

2x(2 times)

x

X+2x=36

3x=36

12

12+2(12)=36

12+24=36

Answer:

Step-by-step explanation:

From, the given word explanation try to create your own equation like

From, equation (1) solve for x

Substitute, equation (3) to equation (2).

Answer: [tex]\Large\boxed{f(x-1)=x}[/tex]

Step-by-step explanation:

Given function

f(x + 1) = x + 2

Isolate [ x + 1 ] on the right-hand side

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

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

Substitute [ x + 1 ] with x

f(x) = x + 1

Substitute [ x ] with [ x - 1 ]

f(x - 1) = (x - 1) + 1

[tex]\Large\boxed{f(x-1)=x}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Isolate y

Compare to slope intercept form y=mx+b

Slope=m=5/3

Answer:

[tex]\dfrac{5}{3}[/tex]

Step-by-step explanation:

To find the gradient (slope) of the given linear relation, use arithmetic operations to isolate y:

Given relation:

[tex]3y-5x=10[/tex]

Add 5x to both sides:

[tex]\implies 3y-5x+5x=10+5x[/tex]

[tex]\implies 3y=5x+10[/tex]

Divide both sides by 3:

[tex]\implies \dfrac{3y}{3}=\dfrac{5}{3}x+\dfrac{10}{3}[/tex]

[tex]\implies y=\dfrac{5}{3}x+\dfrac{10}{3}[/tex]

Slope-intercept form of a linear equation

[tex]y = mx+b[/tex]

where:

Comparing the rewritten equation with the slope-intercept formula, the gradient (slope) of the given linear relation is ⁵/₃.

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Answer:

540

Step-by-step explanation:

540 =18% of 97.2

Done please

it's quotient it's a synthax error..

Using it's concept, the probability of choosing a number that is a multiple of 9 is of 0.111.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

From 1 to 81, there are 81 numbers, of which 81/9 = 9 are multiples of 9, hence, considering that 81 is the number of total outcomes and that 9 is the number of desired outcomes, the probability is given the following division:

p = 9/81 = 1/9 = 0.111.

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The distance between the bird (B) and plane (P) exists at 10,779 feet.

Given the figure attached,

PT = OB = 7000 ft

PB = OT = x feet

and angle of elevation (∠POT) = 30°

By applying the Tan rule in the triangle POT,

tan 33° = Opposite side/Adjacent side = PT/OT

tan 33° = 7000/x

x = 7000/tan33°

x = 10779 feet

The distance between the bird (B) and plane (P) exists at 10,779 feet.

Therefore, the correct answer is option D. 10779 feet.

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Answer:

Solution:

here,

6 cans cost = 9 dollars

1 can costs = 9÷6 dollars

= 1.5 dollars

Now,

8 cans cost = 8×1.5 dollars

= 12 dollars

The assumption that is necessary before conducting a test of hypothesis is that there is a significant amount of complaints for use of e-tickets.

The assumption is that there is a significant amount of complaints for use of e-tickets.

H0: there is a significant amount of complaints for use of e-tickets.

H1: Number of complaints for use of e-tickets is not significant.

H1: parameter > value

Notice the inequality points to the right

Decision Rule: Reject H0 if test statistic > critical value

Decision Rule: Reject H0 if p -value = 0.05

The hypothetical mean is 15.00 and The actual mean is 13.50

The difference between these two values is -1.50.

The sample size is 20

Mean 13.50

SD 1.50

The test statistic will be -4.459.

Lastly z it's important to reject the null hypothesis since the test statistic is greater than the critical value of 1.7291.

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