1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kipish [7]
3 years ago
8

Two coins, A and B, each have a side for heads and a side for tails. When coin A is tossed, the probability it will land tails-s

ide up is 0.5. When coin B is tossed, the probability it will land tails-side up is 0.8. Both coins will be tossed 20 times. Let A represent the proportion of times coin A lands tails-side up, and let PB represent the proportion of times coin B lands tails-side up. Is the number of fosses for each coin enough for the sampling distribution of the difference in sample proportions PA-PB to be approximately normal?
a. Yes, 20 tosses for each coin is enough.
b. No, 20 tosses for coin A is enough, but 20 tosses for coin B is not enough.
c. No, 20 tosses for coin A is not enough, but 20 tosses for coin B is enough.
d. No, 20 tosses is not enough for either coin.
e. There is not enough information given to determine it 20 tosses is enough.
Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Answer:

Is the number of tosses for each coin enough for the sampling distribution of the difference in sample proportions PA-PB to be approximately normal?

b. No, 20 tosses for coin A is enough, but 20 tosses for coin B is not enough.

Step-by-step explanation:

a) Data and Calculations:

The probability of coin A landing tails-side up = 0.5

The proportion of times coin A lands tails-side up (PA) = 20 * 0.5 = 10

Therefore, the probability of coin A landing heads-side up = 0.5 (1 - 0.5)  

And the proportion of times that coin A lands heads-side up = 20 * 0.5 = 10.  

The proportion on either side is equally distributed.

This is why 20 tosses for coin A is enough, since the sample proportions PA is approximately normal, symmetric, and equally distributed.  There will be equal amounts of 10 tosses (0.5 *20) for either heads-side up or tails-side up.

For coin B, the probability of landing tails-side up = 0.8

The proportion of times coin B lands tails-side up (PB) = 20 * 0.8 = 16

Therefore, the probability of coin B landing heads-side up = 0.2 (1 - 0-.8)

The proportion on either side is not equally distributed, but skewed.

This is why 20 tosses for coin B is not enough, since the sample proportions PB is not approximately normal, symmetric, and equally distributed.  There will be 16 tosses landing tails-side up (0.8*20) and only 4 tosses landing heads-side up (0.2*20).

You might be interested in
Write the equation of the line that passes through the points (-2,-5)(−2,−5) and (8,-8)(8,−8). Put your answer in fully reduced
Nadya [2.5K]

Answer:

y= -0.3x - 5.6

Step-by-step explanation:

m= (-5)-(-8)/(-2)-8

m= -0.3

y=mx+c

-8= -0.3(8)+c

-8= -2.4 +c

-8+2.4=c

c= -28/5

y= -0.3x - 5.6

3 0
2 years ago
(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla
natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}

    =\sqrt{1+1}\\\\=\sqrt{2}

calculating unit vector:

\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}

the point Q is at a distance h from P(6,6) Here, h=0.1  

a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071

the value of Q= (5.92928 ,6.07071  )

In point b:

Calculating the directional derivative of f (x, y) = \sqrt{x+3y} at P in the direction of \vec{v}

f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\

            =\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557

\vec{v} = 1.97651557

In point C:

Computing the directional derivative using the partial derivatives of f.

f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot   \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}

4 0
3 years ago
The price Jimmy paid for a video game is. 85x. What was the discount ?
rewona [7]
The discount was 85% off.
8 0
3 years ago
SOLVE 5 and 6 for me.
Elza [17]
Yes and yes are your two answer for the problems
7 0
3 years ago
Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a
NARA [144]

x^3y^2+\sin(x\ln y)+e^{xy}=0

Differentiate both sides, treating y as a function of x. Let's take it one term at a time.

Power, product and chain rules:

\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}

=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}

=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}

=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)

=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)

=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}

=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)

=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)

=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0

Isolate the derivative, and solve for it:

\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by y to get rid of that fraction in the denominator.

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}

3 0
2 years ago
Other questions:
  • Robert and his coworkers brought 11 work trucks to a service station for oil changes. While there they also put 6 gallons of gas
    15·1 answer
  • Please Help, Thank you!
    12·1 answer
  • 4.
    14·1 answer
  • Suppose that the functions p and q are defined as follows.<br> HELP PLEASE
    10·1 answer
  • Please hurry I need an answer
    12·2 answers
  • The copy machine at the library made a copy of Leslie's 3-page essay in just
    9·1 answer
  • Find m&lt;1<br><br> (Picture included)
    6·1 answer
  • A point f(–3, 1) is translated along the vector ⟨5, –1⟩.what are the coordinates of the image f′ ?
    14·1 answer
  • A) all real numbers; R: all real numbers.
    10·1 answer
  • Hey guy can you please explain this to me ​
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!