A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain using the visual analog scale. Assume that both samples are independent simple random samples from populations having normal distributions. Use a significance level to test the claim that those given a sham treatment have pain reductions that vary more than the pain reductions for those treated with magnets.
n xbar s
Sham 20 0.41 1.26
Magnet 20 0.46 0.93

Answer:

each ice coffee is $2.25

Step-by-step explanation:

13.50 - 9 = 4.50

4.50 / 2 = 2.25

Answer:

[tex]x> \frac{8}{51} [/tex]

Step-by-step explanation:

[tex]7x - \frac{5}{8} x + 3>4[/tex]

Bring constants to one side, simplify:

[tex] \frac{51}{8} x>4 - 3 \\ \frac{51}{8} x>1 \\ x>1 \div \frac{51}{8} \\ x>1 \times \frac{8}{51} \\ x> \frac{8}{51} [/tex]

*Note that the inequality sign only changes when you divide the whole inequality by a negative number.

Answer:

[tex]x>\frac{8}{51}[/tex]

Step-by-step explanation:

[tex]7x-\frac{5}{8}x+3>4\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\7x-\frac{5}{8}x+3-3>4-3\\\mathrm{Simplify}\\7x-\frac{5}{8}x>1\\\mathrm{Multiply\:both\:sides\:by\:}8\\7x\times \:8-\frac{5}{8}x\times \:8>1\times \:8\\\mathrm{Simplify}\\56x-5x>8\\51x>8\\\mathrm{Divide\:both\:sides\:by\:}51\\\frac{51x}{51}>\frac{8}{51}\\\\x>\frac{8}{51}[/tex]

I hope it helps :)

Answer:

choice 1,2,4,5 from top to bottom

Step-by-step explanation:

1:the points given are in the line where both planes intersect

2:point H is not on any plane

3:in the diagram point F is on plane R so false

4:if you connect the points given they will intersect so not collinear

5:the points F and G are on the plane R

6:so F is on plane R but H is not on any do false

Answer:

B

Step-by-step explanation:

The weight of the jar depends on the number of marbles in it, therefore the weight is the dependent variable (y) and the number of marbles is the independent variable (x). The y-intercept is when x = 0 and since the number of marbles is x, the answer is that the y-intercept represents the weight of the jar without the marbles.

Answer:

total number(n)=10

divided the total number into two equal parts so

in one group of number =5

ln first phase group of number middle value =44

In second phase group of number middle value=50

then using formula,q1=44 q3=50

inter quartile range =q3_q1

=50-44

=6

therefore interquartile range =6

Answer: The answer is 6

Answer:

The probability that the diagnosis is correct is 0.95249.

Step-by-step explanation:

We are given that the American Diabetes Association estimates that 8.3% of people in the United States have diabetes.

Suppose that a medical lab has developed a simple diagnostic test for diabetes that is 98% accurate for people who have the disease and 95% accurate for people who do not have it.

Let the probability that people in the United States have diabetes = P(D) = 0.083.

So, the probability that people in the United States do not have diabetes = P(D') = 1 - P(D) = 1 - 0.083 = 0.917

Also, let A = event that the diagnostic test is accurate

So, the probability that a simple diagnostic test for diabetes is accurate for people who have the disease = P(A/D) = 0.98

And the probability that a simple diagnostic test for diabetes is accurate for people who do not have the disease = P(A/D') = 0.95

Now, the probability that the diagnosis is correct is given by;

    Probability = P(D) [tex]\times[/tex] P(A/D) + P(D') [tex]\times[/tex] P(A/D')

                      = (0.083 [tex]\times[/tex] 0.98) + (0.917 [tex]\times[/tex]0.95)

                      = 0.08134 + 0.87115

                      = 0.95249

Hence, the probability that the diagnosis is correct is 0.95249.

Answer:

The sample size is  [tex]n = 2401[/tex]

Step-by-step explanation:

From the question we are told that

    The margin of  error is  [tex]E= 2\% = 0.02[/tex]

Given that the  confidence level is  95% then the level of significance is mathematically evaluated as

          [tex]\alpha = (100 - 95)\%[/tex]

            [tex]\alpha = 5\% = 0.05[/tex]

The  critical value of [tex]\frac{\alpha }{2}[/tex]  obtained from the normal distribution table is   [tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]  

Let assume that the sample proportion is  [tex]\r p = 0.5[/tex]

 Generally the sample size is evaluated as

            [tex]n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \r p ( 1- \r p )[/tex]

            [tex]n = [\frac{1.96}{0.02} ]^2 * 0.5 ( 1- 0.5 )[/tex]

            [tex]n = 2401[/tex]

Answer:

The fraction that represents the heart in the diagram shown is 7/3

Step-by-step explanation:

For this problem, we have to find the fraction expressed by the number line in the diagram shown.

First off, we know that the fraction will be between 2 and 3. Second, we know that each little dash between 2 and 3 represents 1/6.So, let's use this information to find the fraction.

Since the heart is two dashes away from 2, then this part of the fraction is 2/6 which can also be simplified to 1/3.

2 1/3

Since we can not have a mixed fraction, then we are going to turn this mixed number into an improper fraction. We do this by multiplying 2 with the denominator (which is 3) and adding the numerator (which is 1) to that product. Our denominator will stay the same in the final fraction.

2 1/3 = 7/3

So, the fraction represented by the heart is 7/3

Answer:

16/7

Step-by-step explanation:

There are 7 divisions between the numbers 2 and 3

So the denominator is 7

The heart is at the second mark

We are past the 2 mark so it is

2 2/7

Changing this from a mixed number to an improper fraction

(7*2+2) /7

16/7

Answer:

The least possible price is  p  =  £110

Step-by-step explanation:

From the question we are told that

     The number of bracelets to be made is  [tex]n = 150[/tex]

      The length of silver require for on bracelet is  [tex]x = 26 \ cm = 0.26 \ m[/tex]

      The option of silver length packs that she buys is  a =  10.5 m  packs

                                                                                           b  =  3.7 m packs

    Generally

                   1 bracelet   [tex]\to[/tex]  0.26 m  

                   150 bracelet [tex]\to[/tex]  z

=>        [tex]z = \frac{150 * 0.26}{1}[/tex]

=>        [tex]z = 39 \ m[/tex]

Now for option a  i.e  10.5 m  per pack

    The number of packs require is

        [tex]v = \frac{z}{a}[/tex]

=>     [tex]v = \frac{39}{ 10.5}[/tex]

=>      [tex]v = 3.7 1[/tex]

given that the number of packs cannot be a fraction but an integer hence she needs to purchase v =  4

and that 4 packs would equal  t  =  4 * 10.5 =  42 meters of  silver

 Now for option d  i.e  3.7 meters per pack  

 The number of packs requires is  

           [tex]w = \frac{z}{b}[/tex]

=>        [tex]w = \frac{39}{3.7}[/tex]

=>        [tex]w = 10.54[/tex]

given that the number of packs cannot be a fraction but an integer hence she needs to purchase w= 11

and that 11 packs would equal  t  = 11 * 3.7 =  40.7 meters of  silver

So the comparing the option and option b we see that for her to pay as little as possible she needs to go for option b since option be will produce the 150  bracelet with a little excess while option a will produce the 150 bracelet with much excess

    Assuming the price for the 3.7 m  pack is    £10

     And the price for the 10.7 pack is  £30

The least possible amount she would pay is  

                 [tex]p = 10 * 11[/tex]

                 p  =  £110  

Answer:

Option A.

Step-by-step explanation:

The given equation of ellipse is

[tex]4x^2+9y^2=36[/tex]

Divide both sides by 36.

[tex]\dfrac{4x^2}{36}+\dfrac{9y^2}{36}=1[/tex]

[tex]\dfrac{x^2}{9}+\dfrac{y^2}{4}=1[/tex]

[tex]\dfrac{x^2}{3^2}+\dfrac{y^2}{2^2}=1[/tex]    ...(1)

The standard form of an ellipse is

[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex]    ...(2)

where, (h,k) is center, (h±a,k) are vertices and (h±c,k) are foci.

On comparing (1) and (2), we get

[tex]h=0,k=0,a=3,b=2[/tex]

Now,

Center [tex]=(h,k)=(0,0)[/tex]

Vertices [tex]=(h\pm a,k)=(0\pm 3,0)=(3,0),(-3,0)[/tex]  

We know that

[tex]c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}=\sqrt{5}[/tex]

Foci [tex]=(h\pm c,k)=(0\pm \sqrt{5},0)=(\sqrt{5},0),(-\sqrt{5},0)[/tex]

Therefore, the correct option is A.

Answer:

B

Step-by-step explanation:

Calculate the ratio of corresponding sides, image to preimage, that is

scale factor = [tex]\frac{DE}{AC}[/tex] = [tex]\frac{12.09}{16.12}[/tex] = 0.75 → B

The scale factor of the dilation will be 1.33. Then the correct option is A.

Dilation is the process of increasing the size of an item without affecting its form. Depending on the scale factor, the object's size can be raised or lowered.

There is no effect of dilation on the angle.

In the diagram, ∆ABC and ∆DBE are similar.

Then the scale factor of the dilation that will map the preimage ΔABC onto the image ΔDBE will be

⇒ 16.12 / 12.09

⇒ 1.33

Then the correct option is A.

More about the dilation link is given below.

https://brainly.com/question/2856466

#SPJ2

Answer:

(0,9.8) and (10, 1.2)

Step-by-step explanation:

These are the only points that are the best fit for the garph correlation.

Answer:

(0, 9.8) and (10, 1.2)

Step-by-step explanation:

:) hope this helped

Answer:

A) 0.12. Yes. Choosing a green and blue M&M is possible

B) 0.43. Yes. Choosing a yellow and red M&M is possible

C) 0.78

Step-by-step explanation:

First of all, the summation of the distribution of all colours is;

Σ(all colors ) = 22% + 20% + 23% + 10% + 6% + 6% + 13% = 100%, or 1.

Thus;

a) P(green candy or blue candy) is;

P(GREEN ∪ BLUE) = P(G) + P(BL)

P(GREEN ∪ BLUE) = 6%+6%

P(GREEN ∪ BLUE) = 12% or 0.12

Now, due to the fact that we have to choose ONE candy and only ONE candy at random, then they are mutually exclusive: Yes. Choosing a green and blue M&M is possible

b)P(yellow candy or red candy is;

P(YELLOW ∪ RED) = P(Y) + P(R)

P(YELLOW ∪ RED) = 20% + 23% = 43% or 0.43

Yes. Choosing a yellow and red M&M is possible

c) P(NOT PURPLE)

the probability of having a purple is;

P(PURPLE) = 22% or 0.22

So, the Probability of NOT having a PURPLE is 1 - 0.22 = 0.78

Answer:

800700

Step-by-step explanation:

800000 + 00000 + 0000 + 000 + 00 + 0

000000 + 00000 + 0000 + 700 + 00 + 0

------------------------------------------------------------

= 800700

Answer:

Hey there!

800000+700=800700

Hope this helps :)

Answer: Vu is 15 years old now.

Step-by-step explanation:

Let present age of WU be x.

Then, the present age of Vu = 3x

Also, After 25 years

Age of Wu = x+25

According to the question:

[tex](x+25)=2(3x)\\\\\Rightarrow\ x+25=6x\\\\\Rightarrow\5x=25\\\\\Rightarrow\ x=5[/tex]

Present age of Vu = 3(5) = 15

Hence, Vu is 15 years old now.

Answer:

j

Step-by-step explanation:j

Answer:

a.

[tex]\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}[/tex]

[tex]\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}[/tex]

[tex]\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}[/tex]

b.

[tex]\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}[/tex]

Step-by-step explanation:

Given that:

C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

a. Find a piecewise smooth parametrization of the path C.

r(t) = { 0

If C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1),

Then:

[tex]C_1 = (0,0) \\ \\ C_2 = (1,0) \\ \\ C_3 = (0,1)[/tex]

Also:

[tex]\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }[/tex]

[tex]\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}[/tex]

[tex]\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }[/tex]

[tex]\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}[/tex]

[tex]\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }[/tex]

[tex]\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}[/tex]

b Evaluate :

Integral of (x+2y^1/2)ds

[tex]\mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \int \limits ^1_{0} \ (t + 0) \sqrt{1} } \\ \\ \mathtt{ \int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \begin {pmatrix} \dfrac{t^2}{2} \end {pmatrix} }^1_0 \\ \\ \mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \dfrac{1}{2}}[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits (x+2 \sqrt{y} \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2 \ dt } }[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits 2- t + 2\sqrt{t-1} \ \sqrt{1+1} }[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} \int \limits^2_1 2- t + 2\sqrt{t-1} \ dt }[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2t - \dfrac{t^2}{2}+ \dfrac{2(t-1)^{3/2}}{3} (2) \end {pmatrix} ^2_1}[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{1}{2} (4-1)+\dfrac{4}{3} (1)^{3/2} -0 \end {pmatrix} }[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{3}{2} + \dfrac{4}{3} \end {pmatrix} }[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{12-9+8}{6} \end {pmatrix} }[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{11}{6} \end {pmatrix} }[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ \sqrt{2} }{6} \ (11 )}[/tex]

[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ 11 \sqrt{2} }{6}}[/tex]

[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 0+2 \sqrt{3-t} \ \sqrt{0+1} }[/tex]

[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 2 \sqrt{3-t} \ dt}[/tex]

[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits^3_2 \begin {pmatrix} \dfrac{-2(3-t)^{3/2}}{3} (2) \end {pmatrix}^3_2 }[/tex]

[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [(0)-(1)]}[/tex]

[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [-(1)]}[/tex]

[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \dfrac{4}{3}}[/tex]

[tex]\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}}{6}+\dfrac{1}{2}+ \dfrac{4}{3}}[/tex]

[tex]\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+3+8}{6}}[/tex]

[tex]\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}[/tex]

Answer:

(-5, 0) and (0,4)

Step-by-step explanation:

Given equation: -4x + 5y = 20

Sub. in the values

When x = -5 and y = 0 (-5,0),

[tex] - 4( - 5) + 5(0) = 20 [/tex]

When x = 0 and y = 4 (0,4),

[tex] - 4(0) + 5(4) = 20[/tex]

That's how I would do it, not sure if your school has another method. Hope this helps :)

Answer: x = -5 and y = 4

Step-by-step explanation: its the first option do you need me to exlain how cuz its multiple choice  

Answer:

6 sandwiches to choose from because

3×2 = 6

Nope, i dont think so

Answer:

no the numbers are infinite

Step-by-step explanation:

Answer:

x² + 7x + 49/4 = (x + 7/2)²

Step-by-step explanation:

x² + bx + c is a perfect square when c = (b/2)².

b = 7, c = (7/2)² = 49/4.

x² + 7x + 49/4 = (x + 7/2)²

Answer:

x² + 7x + 12.25 = (x+3.5)²

Step-by-step explanation:

(a+b)² = a²+ 2ab + b²

then:

7x = 2ab

7/2 = 3.5

then:

(x+3.5)² = x² + 2*3.5*a + 3.5²

x² + 7x + 12.25 = (x+3.5)²

Answer:

[tex]x = 3.6[/tex]

Step-by-step explanation:

To find the area of a rectangle, you multiply its length by its width. The formula is [tex]lw = a[/tex].

We already know the length, 5, and the area, 18, so we can plug it into the equation.

[tex]5\cdot w=18[/tex]

We can simplify this equation by dividing both sides by 5.

[tex]5\cdot w \div5 = 18\div5\\\\w = 3.6[/tex]

Hope this helped!

Answer: x= 13

Step-by-step explanation:

Answer:

4 units

Step-by-step explanation:

A transformation is the movement of a point from one position to another position. If a shape is transformed all its points are also transformed. Types of transformations are translation, rotation, reflection and dilation.

If a shape is transformed, the length of its sides and shape remains the same, only the position changes.

If Line segment AB has a length of 4 units. It is translated 1 unit to the right on a coordinate plane to obtain line segment A'B, the length of A'B' remains the same which is 4 unit. To prove this:

Let A be at ([tex]x_1,y_1[/tex]) and B be at ([tex]x_2,y_2[/tex]). The length of AB is:

[tex]AB=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]

If AB is translated to the right by 1 unit the new points are A' at ([tex]x_1+1,y_1[/tex]) and B' at ([tex]x_2+1,y_2[/tex]). The length of A'B' is:

[tex]A'B'=\sqrt{(y_2-y_1)^2+(x_2+1-(x_1+1))^2}=\sqrt{(y_2-y_1)^2+(x_2+1-x_1-1)^2}\\\\A'B'=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]

AB = A'B' = 4 units

Answer:

2

Step-by-step explanation:

The term 3x² has 2 as an exponent, the correct option is C.

Exponents are the base raised by power, it is written in the superscript of a number.

The expression is  3x² +y-5

The term 3x² has 2 as an exponent.

Therefore, the correct option is C.

The missing options are

A.3x2

B. y

A 2

C. -5

D. None of these choices are correct.

To know more about Exponents

https://brainly.com/question/5497425

#SPJ2

Answer:36

Step-by-step explanation:

Answer:

the answer will be 2 to 12

Answer:

2 to 12

Step-by-step explanation:

just did a quiz got it right

Answer:

13

Step-by-step explanation:

A)5+5=10

B)5+7=12

C) impossible

D)7+7=14

E)5+5+5=15

Answer:

Hey there!

We can write the equation:

0.65x=39

x=60

The dress originally sold for 60 dollars.

Hope this helps :)

Answer:

[tex]A.\ 16\sqrt3\ cm^2[/tex] is the correct answer.

Step-by-step explanation:

Given that:

Side of an equilateral triangle = 8 cm

To find:

Area of the triangle will be:

[tex]A.\ 16\sqrt3\ cm^2[/tex]

[tex]B.\ \dfrac{32}{3} cm^2[/tex]

[tex]C.\ 48\ cm^2[/tex]

[tex]D.\ 36\sqrt3\ cm^2[/tex]

Solution:

First of all, let us have a look at the formula for area of an equilateral triangle:

[tex]A =\dfrac{\sqrt3}{4}a^2[/tex]

Where [tex]a[/tex] is the side of equilateral triangle and an equilateral triangle is a closed 3 sided structure in 2 dimensions which has all 3 sides equal to each other.

Here, we are given that side, [tex]a=8\ cm[/tex]

Putting the value in formula:

[tex]A =\dfrac{\sqrt3}{4}\times 8^2\\\Rightarrow A =\dfrac{\sqrt3}{4}\times 64\\\Rightarrow A =\sqrt3\times 16\\OR\\\Rightarrow \bold{A =16\sqrt3\ cm^2}[/tex]

Hence, [tex]A.\ 16\sqrt3\ cm^2[/tex] is the correct answer.