Answer:
A) P(Type I error) = 0.045
B) P(Type II) error = 0.1
Step-by-step explanation:
We are told that the reliability of the test is 90% reliable.
Also, we are told that the probability that the test erroneously detects a lie even when the individual is actually telling the truth is 0.045.
Thus;
A) To calculate the probability of type I error:
From statistics, in this question we can say that the probability of a type I error is the probability that the test will erroneously detect a lie even though the individual is actually telling the truth. Thus;
Probability of (type I error) = P(rejecting true null) = 0.045
B) For probability of type II error, it is defined as the error where we accept a null hypothesis that is false. We can say that it produces a false negative and the formula is;
P(Type II) error = 1 - reliability
Reliability in the question is 0.90
Thus;
P(Type II) error = 1 - 0.9
P(Type II) error = 0.1
Solve for x. Round your answer to the nearest tenth if necessary. Please look at the picture above
Answer:
veoba
Step-by-step explanation:
SOMEONE HELP PLEASE! I don’t know how to solve this problem nor where to start? Can some please help me out and explain how you got the answer please. Thank you for your time.
For a population with µ = 40 and σ = 8, what is the z-score corresponding to X = 34?
Answer:
Step-by-step explanation:
[tex]\frac{34-40}{8}= -.75[/tex]
Fill in the table using this function rule.
y=-10x+3
9514 1404 393
Answer:
see below
Step-by-step explanation:
Put the x-value in the equation and do the arithmetic.
For example, ...
for x = -5,
y = -10(-5) +3 = 50 +3 = 53
what is the H.C.F of 30 and 45
Answer:
15
Hope this helped ^^
2[30 5[45
3[15 3[9
5[5 3[3
[1 [1
30=2*3*5*1
45=5*3*3*1 HCF= 3*1=3
hope its helps you
keep smiling be happy stay safe
Can you please help me with this ☺️
Answer:
a=27.807
Step-by-step explanation:
Its simple, set it up for law of sine which is sinA/a = sinB/b
Sin108/a = Sin20/10
A bus started from Kathmandu and reached khanikhola,26km far from Kathmandu, in one hour. if the bus had uniform acceleration, calculate the final velocity of the bus and acceleration.
Answer:
a = 0.0040 m/s², v = 14.4 m/s.
Step-by-step explanation:
Given that,
The distance between Kathmandu and Khanikhola, d = 26 km = 26000 m
Time, t = 1 hour = 3600 seconds
Let a is the acceleration of the bus. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
Where
u is the initial speed of the bus, u = 0
So,
[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 26000}{(3600)^2}\\\\a=0.0040\ m/s^2[/tex]
Now using first equation of motion.
Final velocity, v = u +at
So,
v = 0+0.0040(3600)
v = 14.4 m/s
Hence, this is the required solution.
Which of the following best describes the relationship between angle a and angle bin the image below?
Which best describes what forms in nuclea fission?
O two smaller, more stable nuclei
O two larger, less stable nuclei
• one smaller, less stable nucleus
one larger, more stable nucleus
Answer:
One larger, more stable nucleus
Hannah would like to make an investment that will turn 8000 dollars into 33000 dollars in 7 years. What quarterly rate of interest, compounded four times per year, must she receive to reach her goal?
Answer:
20.76%
Step-by-step explanation:
[tex]33000=8000(1+\frac{i}{4})^{4*7}\\4.125=(1+\frac{i}{4})^{28}\\\sqrt[28]{4.125}=1+\frac{i}{4} \\i= .207648169[/tex]
which rounds to 20.76%
Answer:
About 0.2076 or 20.76%.
Step-by-step explanation:
Recall that compound interest is given by the formula:
[tex]\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}[/tex]
Where A is the final amount, P is the principal, r is the interest rate, n is the number of times the interest is applied per year, and t is the number of years.
Since Hannah wants to turn an $8,000 investment into $33,000 in seven years compounded quarterly, we want to solve for r given that P = 8000, A = 33000, n = 4, and t = 7. Substitute:
[tex]\displaystyle \left(33000\right)=\left(8000\right)\left(1+\frac{r}{4}\right)^{(4)(7)}[/tex]
Simplify and divide both sides by 8000:
[tex]\displaystyle \frac{33}{8}=\left(1+\frac{r}{4}\right)^{28}[/tex]
Raise both sides to the 1/28th power:
[tex]\displaystyle \left(\frac{33}{8}\right)^{{}^{1}\! / \! {}_{28}}= 1+\frac{r}{4}[/tex]
Solve for r. Hence:
[tex]\displaystyle r= 4\left(\left(\frac{33}{8}\right)^{{}^{1}\! / \! {}_{28}}-1\right)[/tex]
Use a calculator. Hence:
[tex]r=0.2076...\approx 0.2076[/tex]
So, the quarterly rate of interest must be 0.2076, or about 20.76%.
Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated samplingg distribution.
The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.5 years and a standard deviation of 2.1 years. Random samples of size 17 are drawn from the population and the mean of each sample is determined.
a. 1.33 years, 2.1 years
b. 5.5 years, 0.12 years
c. 5.5 years, 0.51 years
d. 1.33 years, 0.51 years
Answer:
c. 5.5 years, 0.51 years
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Mean of 5.5 years and a standard deviation of 2.1 years.
This means that, for the population, [tex]\mu = 5.5, \sigma = 2.1[/tex]
Random samples of size 17.
This means that [tex]n = 17[/tex]
Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution.
The mean is the same as the mean for the population, that is, 5.5 years.
The standard deviation is:
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{17}} = 0.51[/tex]
This means that the correct answer is given by option c.
The following integral requires a preliminary step such as long division or a change of variables before using the method of partial fractions. Evaluate the following integral. x^4 + 7/x^3 + 2x dx Find the partial fraction decomposition of the integrand. x^4 + 7/x^3 + 2x dx
Division yields
[tex]\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}[/tex]
Now for partial fractions: you're looking for constants a, b, and c such that
[tex]\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}[/tex]
[tex]\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a[/tex]
which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then
[tex]\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}[/tex]
Now, in the integral we get
[tex]\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx[/tex]
The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get
[tex]\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}[/tex]
Find the area of
1.Table
Length = 123cm
Width = 82cm
Height = 76cm
2.Living room
Length = 422cm
Width = 278cm
Height = 253cm
3. Door
Length = 87cm
Width = 2.3cm
Height = 208cm
Answer:
1. 766,536cm^3
2. 29,680,948cm^3
3. 41,620.8cm^3
Step-by-step explanation:
1. 123×82 = 10,086 10,086×76 = 766,536
2. 422×278 = 117,316 117,316×253 = 29,680,948
3. 87×2.3 = 200.1 200.1×208 = 41,620.8
Hope this helps! :)
on a piece of paper graph y=-2x^2+10x-12 and identify the zeros. Select the choice that matches the graph that you drew and correctly identifies the zeros
Answer:
C. 2 and 3 opens down
Step-by-step explanation:
see graph
If a projectile is fired with an initial speed of vo ft/s at an angle α above the horizontal, then its position after t seconds is given by the parametric equations x=(v0cos(α))t andy=(v0sin(α))t−16t2
(where x and y are measured in feet).
Suppose a gun fires a bullet into the air with an Initial speed of 2048 ft/s at an angle of 30 o to the horizontal.
(a) After how many seconds will the bullet hit the ground?
(b) How far from the gun will the bullet hit the ground? (Round your answer to one decimal place.)
(c) What is the maximum height attained by the bullet? (Round your answer to one decimal place.)
Answer:
a) The bullet hits the ground after 64 seconds.
b) The bullet hits the ground 113,511.7 feet away.
c) The maximum height attained by the bullet is of 16,384 feet.
Step-by-step explanation:
Equations of motion:
The equations of motion for the bullet are:
[tex]x(t) = (v_0\cos{\alpha})t[/tex]
[tex]y(t) = (v_0\sin{\alpha})t - 16t^2[/tex]
In which [tex]v_0[/tex] is the initial speed and [tex]\alpha[/tex] is the angle.
Initial speed of 2048 ft/s at an angle of 30o to the horizontal.
This means that [tex]v_0 = 2048, \alpha = 30[/tex].
So
[tex]x(t) = (v_0\cos{\alpha})t = (2048\cos{30})t = 1773.62t[/tex]
[tex]y(t) = (v_0\sin{\alpha})t - 16t^2 = (2048\sin{30})t - 16t^2 = 1024t - 16t^2[/tex]
(a) After how many seconds will the bullet hit the ground?
It hits the ground when [tex]y(t) = 0[/tex]. So
[tex]1024t - 16t^2 = 0[/tex]
[tex]16t^2 - 1024t = 0[/tex]
[tex]16t(t - 64) = 0[/tex]
16t = 0 -> t = 0 or t - 64 = 0 -> t = 64
The bullet hits the ground after 64 seconds.
(b) How far from the gun will the bullet hit the ground?
This is the horizontal distance, that is, the x value, x(64).
[tex]x(64) = 1773.62(64) = 113511.7[/tex]
The bullet hits the ground 113,511.7 feet away.
(c) What is the maximum height attained by the bullet?
This is the value of y when it's derivative is 0.
We have that:
[tex]y^{\prime}(t) = 1024 - 32t[/tex]
[tex]1024 - 32t = 0[/tex]
[tex]32t = 1024[/tex]
[tex]t = \frac{1024}{32} = 32[/tex]
At this instant, the height is:
[tex]y(32) = 1024(32) - 16(32)^2 = 16384[/tex]
The maximum height attained by the bullet is of 16,384 feet.
The weights for newborn babies is approximately normally distributed with a mean of 5.4 pounds and a standard deviation of 1.8 pounds. Consider a group of 1500 newborn babies: 1. How many would you expect to weigh between 3 and 6 pounds
Answer:
You would expect 807 babies to weigh between 3 and 6 pounds.
Step-by-step explanation:
We are given that
Mean,[tex]\mu=5.4[/tex]pounds
Standard deviation,[tex]\sigma=1.8[/tex]pounds
n=1500
We have to find how many would you expect to weigh between 3 and 6 pounds.
The weights for newborn babies is approximately normally distributed.
Now,
[tex]P(3<x<6)=P(\frac{3-5.4}{1.8}<\frac{x-\mu}{\sigma}<\frac{6-5.4}{1.8})[/tex]
[tex]=P(-1.33<Z<0.33)[/tex]
[tex]P(3<x<6)=P(Z<0.33)-P(Z<-1.33)[/tex]
[tex]P(3<x<6)=0.62930-0.09176[/tex]
[tex]P(3<x<6)=0.538[/tex]
Number of newborn babies expect to weigh between 3 and 6 pounds
=[tex]1500\times 0.538=807[/tex]
La señora Alcántara realiza una compra en el supermercado fortuna, ella solo tiene 12,400 pesos ,compra varios artículos y su compra es equivalente a 13,600 pesos. ¿Cuánto tiene que pagar si le realizan un descuento de un 15%? ¿Cuántos le quedaron de lo que tenía en efectivo?
Answer:
She spent = 11560 pesos
Amount left = 840 pesos
Step-by-step explanation:
Mrs. Alcántara makes a purchase at the fortuna supermarket, she only has 12,400 pesos, she buys several items and her purchase is equivalent to 13,600 pesos. How much do you have to pay if they give you a 15% discount? How many was left of what he had in cash?
Amount she has = 12400pesos
Item purchased = 13600 pesos
discount = 15 %
So, the total discount on the item purchased is
= 15 % of 13600
= 0.15 x 13600
= 2040 pesos
So, the amount spent = 13600 - 2040 = 11560 pesos
Amount she left = 12400 - 11560 = 840 pesos
a student takes two subjects A and B. Know that the probability of passing subjects A and B is 0.8 and 0.7 respectively. If you have passed subject A, the probability of passing subject B is 0.8. Find the probability that the student passes both subjects? Find the probability that the student passes at least one of the two subjects
Answer:
0.64 = 64% probability that the student passes both subjects.
0.86 = 86% probability that the student passes at least one of the two subjects
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Passing subject A
Event B: Passing subject B
The probability of passing subject A is 0.8.
This means that [tex]P(A) = 0.8[/tex]
If you have passed subject A, the probability of passing subject B is 0.8.
This means that [tex]P(B|A) = 0.8[/tex]
Find the probability that the student passes both subjects?
This is [tex]P(A \cap B)[/tex]. So
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
[tex]P(A \cap B) = P(B|A)P(A) = 0.8*0.8 = 0.64[/tex]
0.64 = 64% probability that the student passes both subjects.
Find the probability that the student passes at least one of the two subjects
This is:
[tex]p = P(A) + P(B) - P(A \cap B)[/tex]
Considering [tex]P(B) = 0.7[/tex], we have that:
[tex]p = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.64 = 0.86[/tex]
0.86 = 86% probability that the student passes at least one of the two subjects
More math sorry. But I honestly don’t know any of these
Answer: A
Step-by-step explanation:
The main parent functions are x, and x raised to the power of something (examples: [tex]x^2, x^3, x^4[/tex], etc)
Find the L. C. M in division method of the following
a) 18,27
b) 21,38
Answer:
hope it will be helpful to you.....
The table gives estimates of the world population, in millions, from 1750 to 2000. (Round your answers to the nearest million.)
Year Population
1750 790
1800 980
1850 1260
1900 1650
1950 2560
2000 6080
(a) Use the exponential model and the population figures for 1750 and 1800 to predict the world population in 1900 and 1950 1900 1950 million people million people
(b) Use the exponential model and the population figures for 1800 and 1850 to predict the world population in 1950 million people
(c) Use the exponential model and the population figures for 1900 and 1950 to predict the world population in 2000 million people
Answer:
A.) 1508 ; 1870
B.) 2083
C.) 3972
Step-by-step explanation:
General form of an exponential model :
A = A0e^rt
A0 = initial population
A = final population
r = growth rate ; t = time
1)
Using the year 1750 and 1800
Time, t = 1800 - 1750 = 50 years
Initial population = 790
Final population = 980
Let's obtain the growth rate :
980 = 790e^50r
980/790 = e^50r
Take the In of both sides
In(980/790) = 50r
0.2155196 = 50r
r = 0.2155196/50
r = 0.0043103
Using this rate, let predict the population in 1900
t = 1900 - 1750 = 150 years
A = 790e^150*0.0043103
A = 790e^0.6465588
A = 1508.0788 ; 1508 million people
In 1950;
t = 1950 - 1750 = 200
A = 790e^200*0.0043103
A = 790e^0.86206
A = 1870.7467 ; 1870 million people
2.)
Exponential model. For 1800 and 1850
Initial, 1800 = 980
Final, 1850 = 1260
t = 1850 - 1800 = 50
Using the exponential format ; we can obtain the rate :
1260 = 980e^50r
1260/980 = e^50r
Take the In of both sides
In(1260/980) = 50r
0.2513144 = 50r
r = 0.2513144/50
r = 0.0050262
Using the model ; The predicted population in 1950;
In 1950;
t = 1950 - 1800 = 150
A = 980e^150*0.0050262
A = 980e^0.7539432
A = 2082.8571 ; 2083 million people
3.)
1900 1650
1950 2560
t = 1900 - 1950 = 50
Using the exponential format ; we can obtain the rate :
2560 = 1650e^50r
2560/1650 = e^50r
Take the In of both sides
In(2560/1650) = 50r
0.4392319 = 50r
r = 0.4392319/50
r = 0.0087846
Using the model ; The predicted population in 2000;
In 2000;
t = 2000 - 1900 = 100
A = 1650e^100*0.0087846
A = 1650e^0.8784639
A = 3971.8787 ; 3972 million people
Precision manufacturing: A process manufactures ball bearings with diameters that are normally distributed with mean 25.0 millimeters and standard deviation 0.07 millimeter. Round the answers to at least four decimal places. (a) Find the 60th percentile of the diameters. (b) Find the 67th percentile of the diameters. (c) A hole is to be designed so that 2% of the ball bearings will fit through it. The bearings that fit through the hole will be melted down and remade. What should the diameter of the hole be
Answer:
a) The 60th percentile of the diameters is of 25.0177 millimeters.
b) The 67th percentile of the diameters is of 25.0308 millimeters.
c) The diameter of the hole should be of 24.8562 millimeters.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with mean 25.0 millimeters and standard deviation 0.07 millimeter.
This means that [tex]\mu = 25, \sigma = 0.07[/tex]
(a) Find the 60th percentile of the diameters.
This is X when Z has a p-value of 0.6, so X when Z = 0.253.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.253 = \frac{X - 25}{0.07}[/tex]
[tex]X - 25 = 0.253*0.07[/tex]
[tex]X = 25.0177[/tex]
The 60th percentile of the diameters is of 25.0177 millimeters.
(b) Find the 67th percentile of the diameters.
This is X when Z has a p-value of 0.67, so X when Z = 0.44.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.44 = \frac{X - 25}{0.07}[/tex]
[tex]X - 25 = 0.44*0.07[/tex]
[tex]X = 25.0308[/tex]
The 67th percentile of the diameters is of 25.0308 millimeters.
(c) A hole is to be designed so that 2% of the ball bearings will fit through it. The bearings that fit through the hole will be melted down and remade. What should the diameter of the hole be.
This is the 2nd percentile, which is X when Z has a p-value of 0.08, so X when Z = -2.054.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.054 = \frac{X - 25}{0.07}[/tex]
[tex]X - 25 = -2.054*0.07[/tex]
[tex]X = 24.8562[/tex]
The diameter of the hole should be of 24.8562 millimeters.
Someone please help thanks
Answer:
By similar triangles: BE/20 = 18/25 BE 14.4
Also, (ED + 26) / 26 = 18/14.4
ED = 6.5 and AD = 32.5
A cottage industry exists in the home-manufacture of 'country crafts'. Especially treasured are handmade quilts. If the fourth completed quilt took 40 hours to make, and the eighth quilt took 35 hours. What is the percentage learning
Answer:
6.7%
Step-by-step explanation:
if f(x)=-5^x-4 and g(x)=-3x-2,find (f+g) (x)
Answer: (f-g)(x) = - 5^x + 3x - 2
Step-by-step explanation:
if f(x) = -5^x - 4 and g(x)= - 3x - 2,find (f-g)(x)
(f-g)(x) = -5^x - 4 - (-3x - 2)
(f-g)(x) = -5^x - 4 + 3x + 2
(f-g)(x) = - 5^x + 3x - 2
Question 5 Multiple Choice Worth 1 points)
(01.03 MC)
Bunny Hill Ski Resort charges $35 for ski rental and $10 an hour to ski. Black Diamond Ski Resort charges $40 for ski rental and $5 an hour to ski. Create an equation to determine at what point
the cost of both ski slopes is the same.
Answer:
Bunny Hill Ski Resort:
y = 10x + 35
Diamond Ski Resort:
y = 5x + 40
Point where the cost is the same:
(1, 45)
Step-by-step explanation:
The question tells us that:
$35 and $40 are initial fees
$10 and $5 are hourly fees
This means that x and y will equal:
x = number of hours
y = total cost of ski rental after a number of hours
So we can form these 2 equations:
y = 10x + 35
y = 5x + 40
Now we are going to use System of Equations to find what point the cost of both ski slopes is the same.
Because they both equal y, we can set the equations equal to each other:
10x + 35 = 5x + 40
And we use basic algebra to solve for x:
10x + 35 = 5x + 40
(subtract 5x from both sides)
5x + 35 = 40
(subtract 35 from both sides)
5x = 5
(divide both sides by 5)
x = 1
Remember, x equals the number of hours.
That means when your rent out the skis for 1 hour, you will get the same price of $45 (you find the price by plugging in 1 into both of the equations)
Hope it helps (●'◡'●)
Which are correct representations of the inequality -3(2x-5) <5(2 - x)? Select two options.
Answer:
-6x+15 < 10-5x
x>5
third equation, first graph
Step-by-step explanation:
suppose you have a bank account earning 6% annual interest rate compounded monthly, and you want to put in enough money so that you can withdraw $100 at the end of each month over a time frame of ten years. calculate how much money you need to start with. show work.
Answer:
maybe 10000
Step-by-step explanation:
Answer:
9007.35
Step-by-step explanation:
First find the effective rate: .06/12= .005
let x= amount
[tex]x=100\frac{1-(1+.005)^{-12*10}}{.005}\\100*\frac{1-.549632733}{.005}\\9007.345333[/tex]
what is the correct answer to my question ?
Answer:
13/17
Step-by-step explanation:
Learning Task No. 1 Randy, Manny and Jan put 3 As, 4 Bs and 5 Cs in the box. They will take turns in getting a letter from the box. They are trying to test the probability of getting their favourite letter.
Randy - A
Manny-B
Jan-C
1. What is the probability of getting each boy's favourite letter? a. Randy b. Manny c. Jan
2. If you are next to Jan to pick up a letter and your favourite letter is A , What is the probability of getting your favourite letter?
3. Who is most unlikely to get his favourite letter.
Answer:
1. A = 3/12
B= 4/12
C = 5/12
2......
3. Randy
Step-by-step explanation:
3+4+5 = 12
therefore there are 12 letters in the box
we can say that there are 3/12 A's in the box and do the same for the remaining letters
question two does not make sense
3. the person who has the lowest fraction in value which is A
