Jennifer had a headache and was told she was given a name brand medicine but it was really a sugar pill. Three hours later she was asked if she feels better, and she agreed she did.
What does the placebo effect mean in this specific situation?
a.) The benefit is due to the fact that Jennifer was given the treatment (pill, although only a sugar pill) and not an effect of the "active ingredient" (headache medicine).
b.) Jennifer received the benefit from the treatment since she also took a nap.
c.) The benefit is the fact that Jennifer received the treatment (pill, although only a sugar pill) and it contains another active ingredient.
d.) The benefit is that Jennifer was given the control (pill, although only a sugar pill) and not the active ingredient (cough medicine).

Step-by-step explanation:

here,

-8=32

-8/2=32/2

-4=16

-(2)^2=(4)^2

-2=4

[tex]\frac{442}{x} = \frac{85}{100} \\85x = 44200\\x = \frac{44200}{85} = 520[/tex]

Original price is 520

Step-by-step explanation:

3(x + 6)

3 × x + 3×6

3x + 18

-2 (x - 1)

-2 × x -2 × -1

-2x + 2

-2 (x + 2)

-2 × x + -2 × 2

-2x - 4

Answer:

9514 1404 393

Answer:

  (f) Plane CEA or Plane CAD

Step-by-step explanation:

The name must include three points that are in the plane and not all on the same line. That is, the name will not include either of points B or F, and must include point C. The only names listed that are appropriate are ...

  Plane CEA or Plane CAD

Answer:

6 + 4

Step-by-step explanation:

Using the keywords:

Sum of, we know that it means for us to add.

Answer:

The given ratios 3: 5 and 15: 25 are equal. Because when you divide the ratio 15: 25 by 5 on both numerator and denominator, the first ratio 3: 5 can be obtained. Similarly, when you multiply the first ratio 3: 5 by 5, the ratio 15: 25 can be obtained.

Please mark as brainliest thanks!

Step-by-step explanation:

The angle between vectors is given by

[tex]\cos{\theta} = \dfrac{\vec{\textbf{a}}\cdot \vec{\textbf{b}}}{|\vec{\textbf{a}}||\vec{\textbf{b}}|}[/tex]

The magnitudes for the vectors are as follows:

[tex]|\vec{\textbf{a}}| = \sqrt{a_x^2 + a_y^2 + a_z^2}[/tex]

[tex]\:\:\:\:\:\:\:=\sqrt{(2)^2 +(-5)^2 + (3)^2} = 6.16[/tex]

[tex]|\vec{\textbf{b}}| = \sqrt{b_x^2 + b_y^2 + b_z^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(3)^2 + (1)^2 + (4)^2} = 5.10[/tex]

The dot product between the vectors is

[tex]\vec{\textbf{a}}\cdot \vec{\textbf{b}} = (2)(3) + (-5)(1) + (3)(4) = 13[/tex]

Therefore, the angle between the two vectors is

[tex]\cos{\theta} = \dfrac{\vec{\textbf{a}}\cdot \vec{\textbf{b}}}{|\vec{\textbf{a}}||\vec{\textbf{b}}|} = \dfrac{13}{(6.16)(5.10)}[/tex]

or

[tex]\theta = 65.56°[/tex]

It looks like the differential equation is

[tex] \left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0[/tex]

Check for exactness:

[tex]\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x[/tex]

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

[tex] \mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0[/tex]

*is* exact. If this modified DE is exact, then

[tex]\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}[/tex]

We have

[tex]\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu[/tex]

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

[tex]x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}[/tex]

The modified DE,

[tex]\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0[/tex]

is now exact:

[tex]\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}[/tex]

So we look for a solution of the form F(x, y) = C. This solution is such that

[tex]\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}[/tex]

Integrate both sides of the first condition with respect to x :

[tex]F(x,y) = -e^{-x}y - \dfrac1x + g(y)[/tex]

Differentiate both sides of this with respect to y :

[tex]\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C[/tex]

Then the general solution to the DE is

[tex]F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}[/tex]

Answer:

57

Step-by-step explanation:

9514 1404 393

Answer:

  x = 50°

Step-by-step explanation:

The angle marked 73° is the average of the arcs marked 96° and x.

  (x +96°)/2 = 73°

  x = 2(73°) -96° . . . . solve for x

  x = 50°

__

Additional comment

Then z = 360° -50° -114° -96° = 100°.

Answer:

23 and 66 are the dimensions

Step-by-step explanation:

178/2=89

89+43=132

178-132=46

132/2=66

46/2=23

(I think this is the answer I'm so sorry if I am wrong)

Answer:

The answer is 705,000

Step-by-step explanation:

pls mark brainlest

The value of (7 ten thousand 5 hundreds) x 10 is 705,000.

Here,

The written expression is,

(7 ten thousand 5 hundreds) x 10

We have to find the value of given written expression.

What is Mathematical expression?

Mathematical expression is a finite combination of symbols that is well-formed according to the rules.

Now,

The written expression is,

⇒ (7 ten thousand 5 hundreds) x 10 = 70, 500 x 10 = 705,000

Hence,

The value of (7 ten thousand 5 hundreds) x 10 is 705,000.

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Answer:

158.733 that's the answer

9514 1404 393

Answer:

  about 159 pounds

Step-by-step explanation:

The exact conversion factor from kilograms to pounds is ...

  1 lb = 0.45359237 kg

So, to find the number of pounds, you multiply:

  [tex]72\text{ kg}\times\dfrac{1\text{ lb}}{0.45359237\text{ kg}}\approx\boxed{158.7328\text{ lb}}[/tex]

The conversion is often approximated by 1 kg = 2.2 lb, which gives an answer that is about 0.21% low.

Answer:

what are we finding

Are we to derive an equation of a straight line

Answer:

i).2x2(5+9)

4(14)

4x14=56

ii) 2x5x9=90

Answer:

s

RS = 24

Step-by-step explanation:

A bisector, in mathematics, is a two-dimensional line or line segment that divides another two-dimensional shape in half. In this case, as per the given drawing line (s) divides the line (RS) in half. This is indicated by the two small red lines on either side of the line (s) on the line (RS), signifying that the two segments are congruent. Thus, proving that line (s) bisects line (RS).

The length of line (RS) is is twice the length of line (RM), thus line (RS) is (24).

Answer:

2:2 3:3

Step-by-step explanation:

9514 1404 393

Answer:

  B

Step-by-step explanation:

The exact conversion is ...

  [tex]\dfrac{10000\text{ ft}}{1\text{ s}}\times\dfrac{3600\text{ s}}{1\text{ h}}\times\dfrac{3.048\times10^{-4}\text{ km}}{1\text{ ft}}=10972.8\text{ km/h}[/tex]

The closest answer choice is B, 10975.6 km/h.

Answer:

She made 70% of the free throws.

Step-by-step explanation:

All I did was multiply 270 by percents like 40%, 50%, 60%, 70%, etc. In this case, it would be 270*70% or 270*.70

Answer:

1vbrbdhrgdvsvdbbd

Step-by-step explanation:

bhegsbdvsvdvdbrbdbr

ddghehrhebfhfhrhd

ehebrnrjrjrnr

dvdbdbfbfbfbtnyn

9514 1404 393

Answer:

  I and III only

Step-by-step explanation:

The graph is symmetrical about the y-axis, so is an even function.

The function is decreasing in the 1st quadrant, so is not increasing everywhere.

The values of the function approach y=0 for extreme values of x, so the graph shows that as a horizontal asymptote.

  I and III only

Answer:

1

Step-by-step explanation:

The GCF of 24 and 97 can be obtained like this:

Answer:

y = |x| – 4

Step-by-step explanation:

If we substitute x as 0, we get -4 therefore this is the answer.

Answer:

x≥ -43

Step-by-step explanation:

at least means greater than or equal to

Let x be the unknown number

x≥ -43

Answer:

Step-by-step explanation:

4 Section on the spinner and three spins:

4^3=64

The number of individual outcomes when spinning the wheel three times is 64 times

In mathematics, a base number raised to an exponent is referred to as a power. The base number is the factor that is multiplied by itself, and the exponent indicates how many times the base number has been multiplied.

A power exists as the product of multiplying a number by itself. Usually, power is illustrated with a base number and an exponent. The base number tells what number exists being multiplied. The exponent, a small number written beyond and to the right of the base number, tells how many times the base number exists being multiplied.

Given

Sections = 4

no. of spinners per player = 3

no. of spinners per player per game = 4 ³ = 4 * 4 * 4 = 64

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Step-by-step explanation:

everything can be found in the picture

Answer:

It is 1. A(n) =3(3^n-1)

Step-by-step explanation:

It is a geometric progression

General recursive formular:

[tex]a_{n} = a( {r}^{n-1} )[/tex]

a » first term, a = 3

r » common difference, r = 9 ÷ 3 = 3

n » number of terms

[tex]A(n) = 3( {3}^{n-1} )[/tex]

______________end_____________________

further more:

[tex]A(n) = {3}^{(n)} [/tex]

Answer:

97.6 + 77.8

Step-by-step explanation:

97.6-(-77.8)

= 97.6+77.8

97.6+77.8= 175.4

97.6-(-77.8) = 175.4

Answer:

There are 5 red ✎ pencils

Step-by-step explanation:

fraction of green pencils:

[tex] = 1 - \frac{1}{3} \\ \\ = \frac{2}{3} [/tex]

let total pencils be x :

[tex] \frac{2}{3} \: of \: x = 10 \: green \: pencils \\ \\ \frac{2}{3} \times x = 10 \\ \\ 2x = 3 \times 10 \\ x = \frac{3 \times 10}{2} \\ \\ x = 15 \: pencils[/tex]

Total pencils = 15

Red pencils:

[tex] = 15 - 10 \\ = 5 \: pencils[/tex]