A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is used to measure the glucose concentration in the blood sample five times. The measured glucose concentrations are 104.5 , 96.2 , 102.2 , 98.3 , and 101.8 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor. A. 104.5 mg/dL absolute error = 2.5 mg / dL relative error = 0.025 B. 96.2 mg/dL absolute error = −5.8 mg / dL relative error = 0.057 C. 102.2 mg/dL absolute error = 0.2 mg / dL relative error = 0.020 D. 98.3 mg/dL absolute error = −3.7 mg / dL relative error = 0.036 E. 101.8 mg/dL absolute error = −0.2 mg / dL relative error =

Answer:

a) h'(t)= (6cos3t,-9sin3t,3[tex]\sqrt[]{5}[/tex]cos3t)

b) t=0.93994736+πn/3

c) Magnitude of h(t) is 3 which is a constant, so h(t) lies on a sphere

Step-by-step explanation:

We are required to calculate the percentage increase in tax and determine who is right.

The percentage increase in tax is 25% and Linda is very correct

percentage increase = difference in tax /

percentage increase = difference in tax / old tax × 100

old tax = 2%

New tax = 2.5%

Difference = New tax - old tax

= 2.5% - 2%

= 0.5%

percentage increase = difference in tax /

percentage increase = difference in tax / old tax × 100

= 0.5% / 2% × 100

= 0.25 × 100

= 25%

Therefore, the percentage increase in tax is 25% and Linda is very correct

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Answer:

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X,Y), X and Y :

F(X,Y)(x,y)=P(X≤x,Y≤y)

Likewise for (s(X),t(Y)), s(X) and t(Y) :

F(s(X),t(Y))(u,v)=P(s(X)≤u

Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :

F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))

The last step is obtained by applying the functions s and t since s preserves order and t reverses it.

This can be further transformed into

F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))

Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.

Now, to transform this into a statement about copulas, note that

C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))

Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,

we get

F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))

The left hand side is the copula C(X,Y), the right hand side still needs some work.

Note that

Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a

and likewise

Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b

Combining all results we obtain for the relationship between the copulas

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

The difference between the volumes in cubic inches is option A) 72pi

Step-by-step explanation:

Volume = 4/3 π(6)³

⇒ 4/3(216)π

⇒ 4[tex]\times[/tex]72π

⇒ 288π cubic inches

Volume = π(6)²(6)

⇒ 6³π

⇒ 216π cubic inches

Difference between the volumes = 288π - 216π = 72π

The difference in volume between the two solids is 226.08in^3

Data;

The volume of a sphere is given as

[tex]v = \frac{4}{3} \pi r^3\\[/tex]

Let's substitute the values and find the volume

[tex]v = \frac{4}{3}*3.14*6^3\\v = 904.32in^3[/tex]

The formula of volume of a cylinder is given as

[tex]v = \pi r^2 h\\[/tex]

Let's substitute the values into the equation and solve

[tex]v = 3.14 * 6^2 * 6\\v = 678.24in^3[/tex]

The difference in volume between the two solids is

[tex]volume of sphere - volume of cylinder = 904.32 - 678.24 = 226.08in^3[/tex]

The difference in volume between the two solids is 226.08in^3

Learn more on volume of sphere and cylinder here;

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Answer:

0.0726 or 7.26%

Step-by-step explanation:

When choosing the first vial, there is a 16 in 58 chance that the vial does not have a hairline crack. When choosing the second vial, since on good vial was already picked, there is a 15 in 57 chance that the vial does not have a hairline crack. The probability that none of the 2 vials have hairline cracks is:

[tex]P = \frac{16}{58}*\frac{15}{57}\\P=0.0726[/tex]

There is a 0.0726 or 7.26% chance that none of the 2 vials have a hairline crack.

Answer:

a) Equation is

[tex]y = 0.1135x+1.590[/tex]

b) Slope = 0.1135 represents the change in y for a unit change in x

i.e. When nitrate content is increasedby 1, absorbence is increased by 0.1135

Step-by-step explanation:

Nitrates Absorbence

x y

50 7

50 7.6

100 12.7

200 24

400 47

800 93

1200 138

1600 183

2000 231

2000 226

SUMMARY OUTPUT        

Regression Statistics        

Multiple R 0.999911043        

R Square 0.999822094        

Adjusted R Square 0.999799856        

Standard Error 1.2890282        

Observations 10        

Coefficients

Intercept 1.589782721

x 0.113500259

we get regression line as

y = 0.1135x+1.590

a) Equation is

[tex]y = 0.1135x+1.590[/tex]

b) Slope = 0.1135 represents the change in y for a unit change in x

i.e. When nitrate content is increasedby 1, absorbence is increased by 0.1135

Answer:

[tex] ME = \frac{10.08}{2}= 5.04[/tex]

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=50.86[/tex] represent the sample mean  

[tex]\mu[/tex] population mean

s represent the sample standard deviation  

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

[tex] \bar X \pm ME[/tex]   (1)

Or equivalently:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (2)

Where the margin of error is given by:

[tex] ME=  t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

For this case we have the confidence interval limits given (45.82, 55.90)

We can find the width of the interval like this:

[tex] Width =55.90-45.82= 10.08[/tex]

And now the margin of error would be the half of the width since we assume that the confidence interval is symmetrical.

[tex] ME = \frac{10.08}{2}= 5.04[/tex]

Answer:

0.008 is the probability that a computer will take more than 42 seconds to boot up.                                  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 30 seconds

Standard Deviation, σ = 5 second

We are given that the distribution of time taken for a computer to boot up is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(computer will take more than 42 seconds to boot up)

P(x > 42)

[tex]P( x > 42) = P( z > \displaystyle\frac{42 - 30}{5}) = P(z > 2.4)[/tex]

[tex]= 1 - P(z \leq 2.4)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x > 42) = 1 - 0.992 = 0.008[/tex]

0.008 is the probability that a computer will take more than 42 seconds to boot up.

Answer:

[tex]E(X)=1*0.74 +2*0.1 +3*0.11+ 4*0.03 +5*0.02=1.49[/tex]  

[tex]Var(X)=E(X^2)-[E(X)]^2 =3.11-(1.49)^2 =0.8899[/tex]  

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{0.8899}=0.943[/tex]  

Step-by-step explanation:

For this case we have the following data given:

X      1    2    3    4    5

F     74   10  11    3    2

The total number of values are 100, so then we can find the empirical probability dividing the frequency by 100 and we got the followin distribution:

X          1          2        3         4          5

P(X)     0.74   0.10   0.11    0.03    0.02

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".  

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).  

And the standard deviation of a random variable X is just the square root of the variance.  

Solution to the problem

In order to calculate the expected value we can use the following formula:  

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]  

And if we use the values obtained we got:  

[tex]E(X)=1*0.74 +2*0.1 +3*0.11+ 4*0.03 +5*0.02=1.49[/tex]  

In order to find the standard deviation we need to find first the second moment, given by :  

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]  

And using the formula we got:  

[tex]E(X^2)=1^2 *0.74 +2^2 *0.1 +3^2 *0.11 +4^2 0.03 +5^2 *0.02=3.11[/tex]  

Then we can find the variance with the following formula:  

[tex]Var(X)=E(X^2)-[E(X)]^2 =3.11-(1.49)^2 =0.8899[/tex]  

And then the standard deviation would be given by:  

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{0.8899}=0.943[/tex]  

Answer: 13.04

Here are some consumer price indexes from the past 100+ years:

Year CPI

1909 9.1

1919 17.3

1929 17.1

1939 13.9

1949 23.8

1959 29.1

1969 36.7

1979 72.6

1989 118.3

1999 166.6

2009 214.5

2015 238.5

The admission price was $1.00 in 1909. How much would the Speedway have had to charge in 1989 to match the purchasing power of $1 in 1909? In other words, how much was that in 1989?

Answer:

(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French = 0.5 .

(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish = 0.6 .

Step-by-step explanation:

We are given that an elementary school is offering 2 language classes ;

 Spanish Language is denoted by S and French language is denoted by F.

Also we are given, P(S) = 0.5 {Probability of students taking Spanish language}

P(F) = 0.4 {Probability of students taking French language}

[tex]P(S\bigcup F)[/tex] = 0.7 {Probability of students taking Spanish or French Language}

We know that,  [tex]P(A\bigcup B)[/tex]  = [tex]P(A) + P(B) -[/tex] [tex]P(A\bigcap B)[/tex]

So, [tex]P(S\bigcap F)[/tex] = [tex]P(S) + P(F) - P(S\bigcup F)[/tex] = 0.5 + 0.4 - 0.7 = 0.2

[tex]P(S\bigcap F)[/tex] means Probability of students taking  both Spanish and French Language.

Also, P(S)' = 1 - P(S) = 1 - 0.5 = 0.5

         P(F)' = 1 - P(F) = 1 - 0.4 = 0.6

        [tex]P(S'\bigcap F')[/tex] = 1 -  [tex]P(S\bigcup F)[/tex] = 1 - 0.7 = 0.3

(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French is given by P(S/F);

  P(S/F) = [tex]\frac{P(S\bigcap F)}{P(F)}[/tex] = [tex]\frac{0.2}{0.4}[/tex] = 0.5

(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish is given by P(F'/S');

   P(F'/S') = [tex]\frac{P(S'\bigcap F')}{P(S')}[/tex] = [tex]\frac{1- P(S\bigcup F)}{1-P(S)}[/tex] = [tex]\frac{0.3}{0.5}[/tex] = 0.6 .

Note: 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears  ; This question is incomplete please provide with complete detail.

Answer:

the bridge has a height y₀ = 6.94 m

Step-by-step explanation:

The position y of the loose bolt is given by (0,y) where

y = y₀ - 1/2*g*t²

where

y₀ = initial position of the bolt (height of the bridge) , g= gravity , t=time

and the position x of the car is given by (x,0) where

x= x₀  + v*t

where

x₀= initial position of the car

v= car's velocity

then in order for the bolt to hit the windshield they should be at  x=0 and y=0 at the same time , then

0= x₀  + v*t

t= -x₀/v

replacing in the equation for y

0 = y₀ - 1/2*g*t²

0 = y₀ - 1/2*g*(-x₀/v)²

0 = y₀ - 1/2*g*x₀²/v²

y₀ =  1/2*g*x₀²/v²

replacing values

y₀ =  1/2*g*x₀²/v² = 1/2* 9.8m/s² * (-25 m)²/(21 m/s)² = 6.94 m

then the bridge has a height y₀ =6.94 m

We have assumed that

- The bolt has no horizontal velocity ( only vertical velocity) , starts from rest and neglected air friction

- Neglecting the height of the car , position of the windshield and size of the loose bolt

Answer:

0.749

Step-by-step explanation:

The probability that at least one of the three taken balls is blue is the inverse of the probability that none of the three taken balls is blue, aka all 3 of the taken balls are red. The probability of this to happen is

In the first pick: 13/20 chance of this happens

In the 2nd pick: 12/19 chance of this happens

In the 3rd pick: 11/18 chance of this happens

So the probability of picking up all 3 red balls is

[tex]\frac{13*12*11}{20*19*18} = \frac{1716}{6840} = 0.251[/tex]

So the probability of picking up at least 1 blue ball is

1 - 0.251 = 0.749

Answer:

Bit rate = 440 kBits/sec

Step-by-step explanation:

Band width = W= 22 kHz

Number of levels = L = 1024 levels

Bit per sample:

 [tex]m=log_2 L\\\\m =log_2(1024)\\\\m=10 bits/sample[/tex]

Ideal pulses are sent at the Nyquist rate then bit rate = 2 x W x m

[tex]bit\,\,rate= 2\times 22\times 10^3\times 10\\\\bit\,\,rate= 440\times 10^3 bits\,sec^{-1}[/tex]

bit rate = 440 kBits/sec

Answer:

Step-by-step explanation:

Given that A is a square matrix and A is idempotent

[tex]A^2 = A[/tex]

Consider I-A

i) [tex](I-A)^2 = (I-A).(I-A)\\= I^2 -2A.I+A^2\\= I-2A+A\\=I-A[/tex]

It follows that I-A is also idempotent

ii) Consider the matrix 2A-I

[tex](2A-I).(2A-I)=\\4A^2-4AI+I^2\\= 4A-4A+I\\=I[/tex]

So it follows that 2A-I matrix is its own inverse.

Answer:

a) 0.2

b) 0.28

c) 0.12

d) 0.7

Step-by-step explanation:

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

We have the following table:

                  At Least One Tattoo No Tattoo

Age 18-29 126                              324

Age 30-50 54                               396

So in total, there are

126 + 324 + 54 + 396 = 900 people

(A) P(tattoo)

This is the probability that a randomly selected person has a tattoo.

Desired outcomes:

126 + 54 = 180

180 people have at least one tattoo

Total outcomes:

There are 900 people.

P(tattoo) = 180/900 = 0.2

(B) P(tattoo | age 18-29)

This the probability that a person aged 18-29 has a tattoo

Desired outcomes:

126 people aged 18-29 have tattoos

Total outcomes:

126 + 324 = 450 people aged 18-29

P(tattoo | age 18-29) = 126/450 = 0.28

(C) P(tattoo | age 30-50)

This the probability that a person aged 30-50 has a tattoo

Desired outcomes:

54 people aged 18-29 have tattoos

Total outcomes:

54 + 396 = 450 people aged 30-50

P(tattoo | age 18-29) = 54/450 = 0.12

(D) P(age 18-29 | tattoo)

The probability that a tattoed person is 18-29.

Desired outcomes:

126 tattoed people are 18-29

Total outcomes

126 + 54 = 180 tattoed people

P(age 18-29 | tattoo) = 126/180 = 0.7

Answer:

The right answer is option 3.

lim x → −3 f(x) = [infinity] means the values of f(x) can be made arbitrarily large by taking x sufficiently close to (but not equal to) −3.

Step-by-step explanation:

The limit of a function is a fundamental concept concerning the behavior of that function near a particular input.

A function f assigns an output f(x) to every input x. We say the function has a limit L at an input a: this means f(x) gets closer and closer to L as x moves closer and closer to a. More specifically, when f is applied to any input sufficiently close to a, the output value is forced arbitrarily close to L.

That is,

lim x → a f(x) = L

Hope this helps!

The limit lim x → −3 f(x) = [infinity] means that as x values get closer to -3 (without becoming -3), the value of the function f(x) goes towards infinity i.e., it grows without bound. This is akin to certain function behaviors near a value at which an asymptote is present. However, the second part about f(x) values getting close to 0 seems contradiction to the first statement.

The statement lim x → −3 f(x) = [infinity] is related to a concept in Calculus known as a limit. When we say that the limit of f(x) as x approaches -3 is infinity, we mean that as we make x values closer and closer to -3 (without letting x actually be -3), the value of the function f(x) becomes larger and larger without bound, i.e., approaches infinity.

This is similar to some function behaviors near an asymptote. For example, the function y = 1/x has a vertical asymptote at x = 0, where y approaches infinity as x approaches zero from either direction. Here, as x gets arbitrarily close to 0, the value of y = 1/x gets arbitrarily large, or 'approaches infinity'.

On the other hand, when the question states, 'The values of f(x) can be made arbitrarily close to 0 by taking x sufficiently close to (but not equal to) -3', it signifies the tendency of the function values to get closer and closer to 0 as x gets closer to -3. This indicates a certain limit behavior, but it seems to be contradictory with the first part where the limit was stated to be infinity. It is important to scrutinize the function's properties and behavior around x = -3 carefully.

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Answer:

-1.28 ft/s

Step-by-step explanation:

We are given that

The height of tip of  fisherman's rod from the water surface=y=8 ft

[tex]\frac{dz}{dt}=-1ft/sec[/tex]

We have to find the rate at which the fish is approaching the base of the dock when x=10 ft

[tex]z=\sqrt{x^2+y^2}[/tex]

By Pythagoras theorem

[tex]Hypotenuse=\sqrt{base^2+(perpendicular\;side)^2}[/tex]

Substitute x=10 and y=8

[tex]z=\sqrt{(10)^2+8^2}=\sqrt{164}=2\sqrt{41}ft[/tex]

[tex]x^2+y^2=z^2[/tex]

Differentiate w.r.t t

[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}[/tex]

[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}[/tex]

Substitute the values

[tex]10\frac{dx}{dt}+8(0)=2\sqrt{41}\times (-1)[/tex]

[tex]\frac{dy}{dt}=0[/tex]

Because he never moves the rod.

[tex]\frac{dx}{dt}=\frac{-2\sqrt{41}}{10}=-1.28 ft/s[/tex]

Hence, the fish is approaching the base of the dock at the rate  of 1.28 ft/s

Final answer:

To find the probability of drawing two marbles without replacement, calculate the probability of each event separately and multiply them together.

Explanation:

To find the probability of drawing two marbles without replacement, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.

1. Both marbles are red:
First, we calculate the probability of drawing a red marble on the first draw (3 red marbles out of 6 total marbles). Then, we calculate the probability of drawing a red marble on the second draw, given that the first marble drawn was red (2 red marbles out of 5 total marbles remaining). The probability is (3/6) * (2/5) = 1/5.

2. Both marbles are yellow:
First, we calculate the probability of drawing a yellow marble on the first draw (2 yellow marbles out of 6 total marbles). Then, we calculate the probability of drawing a yellow marble on the second draw, given that the first marble drawn was yellow (1 yellow marble out of 5 total marbles remaining). The probability is (2/6) * (1/5) = 1/15.

Answer:

- 21 x + 24 y + 15 z =120

Step-by-step explanation:

Given that

Po(3,-2,5), Qo (-3,-1,-5), and Ro (0,-4,4) ,These are the point in the space.

We know that equation of a plane is given as

[tex]\begin{vmatrix}x-x_1 & y-y_1 &z-z_1 \\ x_2-x_1 & y_2-y_1 &z_2-z_1 \\ x_3-x_1 &y_3-y_1 & z_3-z_1\end{vmatrix}=0\\[/tex]

[tex]\begin{vmatrix}x-0 & y+4 &z-4 \\ 3-0 & -2+4 &5-4 \\ -3-0 &-1+4 & -5-4\end{vmatrix}=0.[/tex]

[tex]\begin{vmatrix}x & y+4 &z-4 \\ 3 & 2 &1 \\ -3 &3 & -9\end{vmatrix}=0.[/tex]

Now by solving above determinate we get

x( -18 -3 ) -(y+4 ) ( -27 +3 ) + ( z- 4) (9+6) = 0

-21 x +24 y -24 x 4 + 15 z - 24 = 0

- 21 x + 24 y + 15 z -120 = 0

- 21 x + 24 y + 15 z =120

Therefore the equation of the plane will be

- 21 x + 24 y + 15 z =120

Answer:

a) Probability of picking the first two answers wrong & the third answer correctly in that order, P(WWC) = 0.128

b) All possible outcomes = WWC, WCW, CWW

P(WWC) = 0.128; P(WCW) = 0.128; P(CWW) = 0.128

c) Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = 0.384

Step-by-step explanation:

P(Correct answer) = P(C) = number of correct options/total options available.

That is, P(C) = 1/5 = 0.2

P(Wrong answer) = P(W) = 1 - 0.2 = 0.8 or (number of wrong options)/(total options available) = 4/5 = 0.8

a) Using the multiple rule for independent events,

P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128

Probability of picking the first two answers wrong & the third answer correctly in that order = 0.128

b) All possible outcomes of picking two wrong answers and one right answer in whichever order for the first 3 questions are (WWC, WCW, CWW)

P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128

P(WCW) = P(W) × P(C) × P(W) = 0.8 × 0.2 × 0.8 = 0.128

P(CWW) = P(C) × P(W) × P(W) = 0.2 × 0.8 × 0.8 = 0.128

c) Using the addition rule for disjoint events,

Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = P(WWC) + P(WCW) + P(CWW) = 0.128 + 0.128 + 0.128 = 0.384

QED!

The probability of getting exactly one correct answer when guessing on three multiple-choice questions with five possible answers each is [tex]\(\frac{3}{25}\)[/tex].

a. To find the probability that the first two guesses are wrong and the third is correct, denoted as P(WWC), we use the multiplication rule for independent events. Since there are five possible answers for each question and only one correct answer, the probability of guessing wrong on one question is [tex]\(\frac{4}{5}\)[/tex] and the probability of guessing correctly is [tex]\(\frac{1}{5}\)[/tex]. Therefore, the probability of WWC is:

[tex]\[ P(WWC) = P(W) \times P(W) \times P(C) = \left(\frac{4}{5}\right) \times \left(\frac{4}{5}\right) \times \left(\frac{1}{5}\right) = \frac{16}{125} \][/tex]

b. The different possible arrangements of two wrong answers and one correct answer (WWC) are:

1. WWC

2. WWC

3. CWW

4. CW

5. WCW

6. WCW

For each of these arrangements, the probability is the same as calculated in part (a), which is \(\frac{16}{125}\).

c. Since there are six different ways to arrange two wrong answers and one correct answer, and each of these arrangements has the same probability, we multiply the probability of one such arrangement by the number of arrangements to find the total probability of getting exactly one correct

[tex]\[ P(\text{exactly one correct answer}) = 6 \times P(WWC) = 6 \times \frac{16}{125} = \frac{96}{125} \][/tex]

However, we must note that we have counted each arrangement twice because the order of the wrong answers (W) does not matter. Therefore, we need to divide the total by 2 to get the correct probability:

[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{2 \times 125} = \frac{48}{125} \][/tex]

Upon reviewing the calculations, it appears there was an error in the final step. We should not divide by 2 because the arrangements are distinct due to the position of the correct answer (C). The correct total probability is indeed [tex]\(\frac{96}{125}\)[/tex], but since we are looking for the probability of exactly one correct answer, we have to consider that there are three questions and the correct answer could be on any of them. Therefore, we have already accounted for all possible arrangements by multiplying by 6.

The correct probability of getting exactly one correct answer when guessing on three multiple-choice questions is:

[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{125} \][/tex]

However, this is not the final answer. We need to consider that there are three different ways to get exactly one correct answer out of three questions (CWW, WCW, WWC). Since these are mutually exclusive events, we add their probabilities:

[tex]\[ P(\text{exactly one correct answer}) = P(CWW) + P(WCW) + P(WWC) \][/tex]

[tex]\[ P(\text{exactly one correct answer}) = \frac{16}{125} + \frac{16}{125} + \frac{16}{125} \][/tex]

[tex]\[ P(\text{exactly one correct answer}) = 3 \times \frac{16}{125} \][/tex]

[tex]\[ P(\text{exactly one correct answer}) = \frac{48}{125} \][/tex]

Thus, the final correct probability is [tex]\(\frac{48}{125}\)[/tex], which simplifies to [tex]\(\frac{3}{25}\)[/tex].

Answer:

the mean is 82.75

Step-by-step explanation:

Amount                   Frequency             Mid Point           fx

0-19.99                             16                    9.995            159.92  

20.00-39.99                    13                   29.995          389.935

40.00-59.99                     21                    49.995        1049.895

60.00-79.99                      19                   69.995         1329.905

80.00-99.99                       12                89.995           1079.94

100.00-119.99                     10                109.995          1099.95

120.00-139.99                     7                 129.995              909.965

140.00-159.99                      8                  149.995            1199.96

160.00-179.99                       7                  169.995          1189.965

180,00-199.99                      1                    189.995         189.995

200.00-219.99                    3                    209.995        629. 985

220.00-239.99                    2                   229.995         459.99

240.00-259.99                    1                   249.995         240.995

∑                                         120                                           9930.4

Mean = ∑fx/∑x

Mean = 9930.4/120=82.7533 = 82.75

Answer: The mean amount spent on Valentine's day is;

$82.83

Step-by-step explanation: To find the mean amount we first arrange the numbers in a frequency table, then solve.

STEP1:

AMOUNT FREQUENCY

0-19.99 16

20.00-39.99 13

40.00-59.99 21

60.00-79.99 19

80.00-99.99 12

100.00-119.99 10

120.00-139.99 7

140.00-159.99 8

160.00-179.99 7

180,00-199.99 1

200.00-219.99 3

220.00-239.99 2

240.00-259.99 1

STEP 2: Find the center of each amount, to do this we have to find the average value of the amounts.

For the first amount is;

(0+19.99)/2 = 9.995

For the second amount is;

(20+39.99)/2 =29.995

Solving this for all the amount. Therefore the table comes

AMOUNT FREQUENCY

9.995 16

29.995 13

49.995 21

69.995 19

89.995 12

109.995 10

129.995 7

149.995 8

169.995 7

189.995 1

209.995 3

229.995 2

249.995 1

STEP 3: multiple each amount in step 2 with the frequency.

For the first amount;

9.995×16 = 159.92

For the second amount;

29.995×13= 389.935

For the third amount;

49.995×21= 1049.895

For the fourth amount;

69.99×19= 1329.81

For the fifth amount;

89.995×12=1079.94

For the six amount;

109.995×10= 1099.95

For the sixth amount;

129.995×7= 909.965

For the seventh amount;

149.995×8= 1199.96

For the eight amount;

169.995×7= 1189.965

For the ninth amount;

189.995×1= 189.995

For the tenth amount;

209.995×3= 629.985

For the eleventh amount;

229.995×2=459.99

For the twelveth amount;

249.995×1= 249.995

STEP 4: Sum up all the answers from the multiplication in step 3

Therefore;

159.92+389.935+1049.895+1329.81+1079.94+1099.95+909.965+1199.96+1189.965+189.995+629.985+459.99+629.985+459.99+249.995 = 9939.995

STEP 5: divide the sum of the value seen in step 4 with the total number of frequency to get the mean value.

The total number of frequency is 120

Therefore;

9939.305÷120=82.827541

Take the value to two decimal place, it becomes;

$82.83 this is the mean value of money spent on Valentine's day.

Answer:

(√6/√2)ft

Step-by-step explanation:

cos 45 = m / √6 ft

m = cos 45 x √6 ft

m = (1 / √2) x √6 ft = (√6/√2)ft

(a) The probability that a hand will contain exactly two wild cards is 0.076.

(b) The probability that a hand will contain two wild cards, two red cards, and three blue cards is 0.0007.

Let's solve these problems using the concept of combinations in probability.

Remember, [tex]C(n, k)[/tex] denotes the number of ways to choose k items from a set of n items, and is calculated as

[tex]C(n,k)=\frac{n!}{k!(n-k)!}[/tex]

where "!" denotes factorial. For example, [tex]5! = 5 \times 4 \times 3 \times 2 \times 1[/tex]

(a) The probability that a hand will contain exactly two wild cards:

The total number of ways to choose 7 cards out of 108 is [tex]C(108, 7)[/tex].

The number of ways to choose 2 wild cards out of 8 is [tex]C(8, 2)[/tex].

The number of ways to choose the remaining 5 cards out of the 100 non-wild cards is [tex]C(100, 5)[/tex].

So, the probability is [tex]\frac{C(8, 2) \times C(100, 5)}{C(108, 7)} \approx 0.076[/tex]

(b) The probability that a hand will contain two wild cards, two red cards, and three blue cards:

The number of ways to choose 2 wild cards out of 8 is [tex]C(8, 2)[/tex].

The number of ways to choose 2 red cards out of 25 is [tex]C(25, 2)[/tex].

The number of ways to choose 3 blue cards out of 25 is [tex]C(25, 3)[/tex].

So, the probability is [tex]\frac{C(8, 2) \times C(25, 2) \times C(25, 3)}{C(108, 7)} \approx 0.0007[/tex]

Answer:

[tex] -3*(3y+2) + 9y = -6[/tex]

[tex] -9y -6 + 9y = -6[/tex]

[tex]-6=-6[/tex]

So then as we can see we can have infinite solutions.

[tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]

Step-by-step explanation:

Assuming the following system of equations:

[tex] 2x-6y =4[/tex]   (1)

[tex] -3x+9y =-6[/tex]   (2)

For this case we can use the substitution method in order to find the possible solutions for the system.

If we solve for x from equation (1) we got:

[tex] 2x = 6y +4[/tex]

[tex] x = 3y +2 [/tex]   (3)

Now we can replace equation (3) into equation (2) and we got:

[tex] -3*(3y+2) + 9y = -6[/tex]

[tex] -9y -6 + 9y = -6[/tex]

[tex]-6=-6[/tex]

So then as we can see we can have infinite solutions.

And the possible solutions are for a fixed value of x, we can solve y from equation (3) and we got:

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

So the solution would be: [tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]

Answer:

Friend A

[tex]\hat p_A= \frac{100}{400}=0.25[/tex]

[tex]z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47[/tex]  

Friend B

[tex]\hat p_B= \frac{20}{120}=0.167[/tex]

[tex]z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80[/tex]  

Friend C

[tex]\hat p_C= \frac{65}{300}=0.217[/tex]

[tex]z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17[/tex]  

So then the best solution for this case would be:

-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)

Step-by-step explanation:

Data given and notation

n represent the random sample taken

X represent the number of scissors selected for each friend

[tex]\hat p=\frac{X}{n}[/tex] estimated proportion of  scissors selected for each friend

[tex]p_o=\frac{1}{3}=0.333[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion that the friend will pick scissors is less than 1/3 or 0.333, the system of hypothesis would be:  

Null hypothesis:[tex]p\geq 0.333[/tex]  

Alternative hypothesis:[tex]p < 0.333[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

Friend A

[tex]\hat p_A= \frac{100}{400}=0.25[/tex]

[tex]z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47[/tex]  

Friend B

[tex]\hat p_B= \frac{20}{120}=0.167[/tex]

[tex]z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80[/tex]  

Friend C

[tex]\hat p_C= \frac{65}{300}=0.217[/tex]

[tex]z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17[/tex]  

So then the best solution for this case would be:

-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)

Answer:

a) Null hypothesis: [tex] p \leq 0.113[/tex]

Alternative hypothesis: [tex] p >0.113[/tex]

b) [tex]p_v =P(z>1.07)=0.1423[/tex]  

c) So the p value obtained was a high low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of workers belonged to unions  is not significantly higher than 0.113.  

Step-by-step explanation:

Part a

For this case we want to check the following system of hypothesis:

Null hypothesis: [tex] p \leq 0.113[/tex]

Alternative hypothesis: [tex] p >0.113[/tex]

Part b

Data given and notation

n=400 represent the random sample taken

X=52 represent the workers belonged to unions

[tex]\hat p=\frac{52}{400}=0.13[/tex] estimated proportion of workers belonged to unions

[tex]p_o=0.113[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.13 -0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}}=1.07[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>1.07)=0.1423[/tex]  

Part c

So the p value obtained was a high low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of workers belonged to unions  is not significantly higher than 0.113.  

Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]p = 0.553, n = 10[/tex]

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003[/tex]

[tex]P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039[/tex]

[tex]P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219[/tex]

[tex]P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724[/tex]

[tex]P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567[/tex]

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552[/tex]

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Answer:

128

Step-by-step explanation:

Since the real track has a gauge of 3 feet but the model railroad track has a gauge of 3/4 inches, the scale factor must be 1/4, because you must follow the rule

"true measurement * scale factor = model measurement"

In fact, we have

[tex]3\cdot\dfrac{1}{4}=\dfrac{3}{4}[/tex]

This implies that the original driving wheels height, 44 inches, must be scaled down to

[tex]44\cdot\dfrac{1}{4}=11[/tex]

inches.