Answer: Choice B
Game A is not fair, and game B is fair.
==============================================================
We'll start with game A.
The sides of the dice are {1,2,3,4,5,6}
There are 6*6 = 36 ways to roll two dice.
If we roll two odd numbers, say 3 and 5, the product is odd because 3*5 = 15.
If any of the numbers are even, then the product is even. Eg: 4*5 = 20 or 2*6 = 12.
To be an even number, 2 must be a factor.
So you can probably see that it's more likely to get an even number product considering that in order to get an odd product, both values must be odd (which narrows the selection pool). We can say that game A is not fair. It favors player 1 more. If you're sure about this, then I recommend looking at a multiplication table but only focus on the first 6 rows and 6 columns. Ignore the row and column that has 0s in it.
------------------------------
Now onto game B
The rules are nearly the same, except now instead of multiplying the dice values, we add them up.
We still have 6*6 = 36 outcomes.
Here are some handy rules we'll take into account
even+even = eveneven+odd = oddodd+odd = evenIf the first dice rolls to '1', then there are 3 ways to get an even number (rolling either 1,3,or 5 on the second dice) and there are 3 ways to get an odd number (rolling either 2,4,or 6 on the second dice). We can see that there are the same number of ways to get an even sum or an odd sum. Apply this in general to any result that the first dice shows, and this basically leads to the conclusion that half of the 36 sums are even, and the other half are odd. We have a fair game now. Refer to a sum of dice chart if you need more convincing.
Answer:
Game A is not fair, and game B is fair.
Step-by-step explanation:
test on iready
công thức đạo hàm của ln(u) = ?
Answer:
U'/U
Step-by-step explanation:
Wbich of the following statements describes the set of integers gieater than --3 but less than 6?
-4,-5,0,1,2,2,4,5
Hope it works perfectly...
5n+15 as an undistributed expression
Answer:
5(n + 3)
Step-by-step explanation:
Factor out 5
5(n + 3)
Answer:
5(n + 3)
should I also give u an explanation
A baseball is hit and its height at different one-second intervals is recorded (See attachment)
Answer:
[tex]h(t)[/tex] is likely a quadratic function.
Based on values in the table, domain of [tex]h(t)[/tex] : [tex]\lbrace 0,\, 1,\, 2,\, 3,\, 4,\, 5,\, 6,\, 7,\, 8\rbrace[/tex]; range of [tex]h(t)\![/tex]: [tex]\lbrace 0,\, 35.1,\, 60.1\, 75.2,\, 80.3,\, 75.3,\, 60.2,\, 35.0 \rbrace[/tex].
Step-by-step explanation:
By the power rule, [tex]h(t)[/tex] is a quadratic function if and only if its first derivative, [tex]h^\prime(t)[/tex], is linear.
In other words, [tex]h(t)[/tex] is quadratic if and only if [tex]h^\prime(t)[/tex] is of the form [tex]a\, x + b[/tex] for some constants [tex]a[/tex] and [tex]b[/tex]. Tables of differences of [tex]h(t)\![/tex] could help approximate whether [tex]h^\prime(t)\![/tex] is indeed linear.
Make sure that values of [tex]t[/tex] in the first row of the table are equally spaced. Calculate the change in [tex]h(t)[/tex] over each interval:
[tex]h(1) - h(0) = 35.1[/tex].[tex]h(2) - h(1) = 25.0[/tex].[tex]h(3) - h(2) = 15.1[/tex].[tex]h(4) - h(3) = 5.1[/tex].[tex]h(5) - h(4) = -5.0[/tex].[tex]h(6) - h(5) = -15.1[/tex].[tex]h(7) - h(6) = -25.2[/tex].[tex]h(8) - h(7) = -35.0[/tex].Consecutive changes to the value of [tex]h(t)[/tex] appears to resemble a line with slope [tex](-10)[/tex] within a margin of [tex]0.2[/tex]. Hence, it is likely that [tex]h(t)\![/tex] is indeed a quadratic function of [tex]t[/tex].
The domain of a function is the set of input values that it accepts. For the [tex]h(t)[/tex] of this question, the domain of [tex]h(t)\![/tex] is the set of values that [tex]t[/tex] could take. These are listed in the first row of this table.
On the other hand, the range of a function is the set of values that it outputs. For the [tex]h(t)[/tex] of this question, these are the values in the second row of the table.
Since both the domain and range of a function are sets, their members are supposed to be unique. For example, the number "[tex]0[/tex]" appears twice in the second row of this table: one for [tex]t = 0[/tex] and the other for [tex]t = 8[/tex]. However, since the range of [tex]h(t)[/tex] is a set, it should include the number [tex]0\![/tex] only once.
Line A passes through the points (10,6) qnd (2,15). Line B passes through the points (5,9) and (14,-1).
Answer:
Line A equation = y=-9/8x+69/4
Line B equation = y=-10/9x+131/9
Step-by-step explanation:
7)
8)
13x + 1/9x + 3
A) 6
C) -7
B) 8
D) 7
Α)
C)
I
Answer:
b) 8
Step-by-step explanation:
the angles are a linear pair meaning that they have a sum of 180. So 22x+4+180. Solve that and x=8
Answer:
B) 8
Step-by-step explanation:
(13x + 1) + (9x + 3) = 180
Combine like terms
22x + 4 = 180
Subtract 4 from both sides
22x = 176
Divide both sides by 22
x = 8
HW HELP ASAP PLZZZZZ
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \: (x - 5)(x - 4) }}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\blue{Step-by-step\:explanation}}{\blue{:}}}}}[/tex]
[tex] {x}^{2} - 9x + 20[/tex]
[tex] = {x}^{2} - 4x - 5x + 20[/tex]
Taking "[tex]x[/tex]" as common from first two terms and "[tex]5[/tex]" from last two terms, we have
[tex] = x(x - 4) - 5(x - 4)[/tex]
Taking the factor [tex](x-4)[/tex] as common,
[tex] = (x - 5)(x - 4)[/tex]
[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35♛}}}}}[/tex]
The ratio of copper to zinc in a certain alloy is 3 to 2. If 30 grams of copper are used, how many grams of zinc are needed to make this alloy?
Answer:
zinc is 20grams
Step-by-step explanation:
Given data
Ratio copper :zinc = 3:2
Copper =30 grams
Applying the ratio
3/2=30/x
Cross multiply
3x=30*2
3x=60
Divide both sides by 3
x=60/3
x=20
Hence zinc is 20grams
The elevation of Sports Authority Field in Denver, Colorado is 5280ft. If this is 275ft higher than seven times the elevation of Lucas Oil Field in Indinapolis, what is the elevation of Lucas Oil Field?
Answer:
The elevation of Lucas Oil Field is of 715 feet.
Step-by-step explanation:
Elevation of Sports Authority Field:
Of 5280 ft.
Compared to Lucas Oil Field:
275ft than seven times the Lucas Oil Field elevation of x.
Equation:
[tex]275 + 7x = 5280[/tex]
What is the elevation of Lucas Oil Field?
Solving the above equation:
[tex]7x = 5280 - 275[/tex]
[tex]7x = 5005[/tex]
[tex]x = \frac{5005}{7}[/tex]
[tex]x = 715[/tex]
The elevation of Lucas Oil Field is of 715 feet.
course
Look at the following number line:
- 10
-5
0
5
10
What are two ways to write the inequality graphed?
x>-1 and -1
XS-1 and -12X
x < -1 and -1 > X
x2-1 and -1 5x
first and last one i think
The square below represents one whole.
What percent is represented by the shaded area?
%
The anwser is 6%
Answer:
the answer is 6%
hdhxbxcbxbxszznzj
> Next question Get a similar question You this question below Find the distance between the points (-3,-3) and (-5,-1) Distance vo Х Submit Question
Answer:
The answer is the square root of 8, or 2root2.
Step-by-step explanation:
Use the distance formula.
d = sqrt((-3-(-5))^2 + (-3-(-1)))
d = sqrt(2^2 + 2^2)
d = sqrt(4 + 4)
d = sqrt8
d = sqrt4 * sqrt2
d = 2 * sqrt2
help what's the answer??
Which equation is equivalent to 4 x = t + 2
s = t-2
s=4/t+2
s=t+2/4
s=t+6
Which expression is equivalent to ?
x ^-5/3
Answer:
[tex] {x}^{ - \frac{5}{3} } \\ \frac{1}{ {x}^{ \frac{5}{3} } } [/tex]
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. The profit contribution and production time in hours per unit are as follows:
Category Product 1 Product 2 Product 3
Profit/unit $30 $50 $20
Machine 1 time/unit 0.5 2.0 0.75
Machine 2 time/unit 1.0 1.0 0.5
Two operators are required for machine 1; thus, 2 hours of labor must be scheduled for each hour of machine 1 time. Only one operator is required for machine 2. A maximum of 100 labor-hours is available for assignment to the machines during the coming week. Other production requirements are that product 1 cannot account for more than 50% of the units produced and that product 3 must account for at least 20% of the units produced.
How many units of each product should be produced to maximize the total profit contribution?
Product # of units
1
2
3
What is the projected weekly profit associated with your solution?
Profit = $
How many hours of production time will be scheduled on each machine? If required, round your answers to two decimal places.
Machine Hours Schedule:
Machine 1 Hours
Machine 2 Hours
What is the value of an additional hour of labor? If required, round your answers to two decimal places.
$
Assume that labor capacity can be increased to 120 hours. Develop the optimal product mix, assuming that the extra hours are made available.
Product # of units
1
2
3
Profit = $
Would you be interested in using the additional 20 hours available for this resource?
Answer:
z (max) = 1250 $
x₁ = 25 x₂ = 0 x₃ = 25
Step-by-step explanation:
Profit $ mach. 1 mach. 2
Product 1 ( x₁ ) 30 0.5 1
Product 2 ( x₂ ) 50 2 1
Product 3 ( x₃ ) 20 0.75 0.5
Machinne 1 require 2 operators
Machine 2 require 1 operator
Amaximum of 100 hours of labor available
Then Objective Function:
z = 30*x₁ + 50*x₂ + 20*x₃ to maximize
Constraints:
1.-Machine 1 hours available 40
In machine 1 L-H we will need
0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40
2.-Machine 2 hours available 40
1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40
3.-Labor-hours available 100
Machine 1 2*( 0.5*x₁ + 2*x₂ + 0.75*x₃ )
Machine 2 x₁ + x₂ + 0.5*x₃
Total labor-hours :
2*x₁ + 5*x₂ + 2*x₃ ≤ 100
4.- Production requirement:
x₁ ≤ 0.5 *( x₁ + x₂ + x₃ ) or 0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0
5.-Production requirement:
x₃ ≥ 0,2 * ( x₁ + x₂ + x₃ ) or -0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0
General constraints:
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers
The model is:
z = 30*x₁ + 50*x₂ + 20*x₃ to maximize
Subject to:
0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40
1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40
2*x₁ + 5*x₂ + 2*x₃ ≤ 100
0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0
-0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers
After 6 iterations with the help of the on-line solver AtomZmaths we find
z (max) = 1250 $
x₁ = 25 x₂ = 0 x₃ = 25
In how many ways can a committee of 3 men and 2 women be formed from a group of 9 men and 10 women?
Answer:
first you have to find the number of ways 3 men can be chosen, then the number of ways 2 women can be chosen, and then you need to multiply these numbers together to get the number of ways because multiplication will show the total arrangements possibilities. use combination since order does not matter.
number ways for 2 out of 10 women total: 10 choose 2= 45
number of ways for 3 out of 9 men total: 9 choose 3= 84
84x45= 3780
3780 total ways
number of ways
PLEASE HELP WKLL MARK IF YOU HELP ME !!!
Answer :
Let in Triangle abc,
a=90,
b = (we have to find)
c = (we have to find)
c + 144=180 (linear pair)
c = 180 - 144
c = 36
a + b + c = 180 (Angle sum property)
90 + b + 36 = 180
126 + b = 180
b = 180-126
b = 54
8. A bag contains 72 toffees. How many toffees can be stored in 122 such bags? Estimate the number of toffees to nearest tens.
9.The working of a metro station is controlled by 28 persons. Estimate the number of persons required to control the working of 88 such stations to the nearest tens.
Answer:
8.) 8,780 toffees can be stored in 122 such bags.
9.) 2,460 people are required to control the working of 88 such stations.
Step-by-step explanation:
QUESTION 8:
We know that there are 72 toffees per bag
So if we want to know how many toffees can be stored in 122 such bags:
72 x 122 = 8,784
ROUND IT TO THE NEAREST TENS:
Look at the digit to the right of the tens place: 8,784–>4. 4 and below means that the digit will be repkaced by 0 and the tens place remains the same.
So 8,784 rounded to tens is 8,780
Tye answer for question 9 is
2,460 (do the same thing)
simplify (1+3i) + (2-5i)
Answer:
3 - 2i
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightAlgebra I
Terms/CoefficientsAlgebra II
Imaginary number i = √-1Step-by-step explanation:
Step 1: Define
Identify
(1 + 3i) + (2 - 5i)
Step 2: Simplify
[Addition] Combine like terms: 3 + 3i - 5i[Subtraction] Combine like terms (i): 3 - 2iA checker board is a square board that is divided into smaller squares, with eight squares along each side. Describe how to find the number of small squares on a checker board without counting.
I really need your help. I am having a lot of difficulty in solving this question, so please help me..
Answer:
Step-by-step explanation:
[tex]log\ \frac{1}{x} = - log \ x \\\\log_a \ a = 1\\\\log \ a^x = x \ log \ a[/tex]
(iii)
[tex]m = log_2 \ \frac{1}{32}\\\\m = log_2 \ \ 32^{-1}\\\\m = - 1 \ log_2 32\\\\m = - 1 \ log_2 \ 2^5\\\\m = -1 \times 5 \ log_2 \ 2\\\\m = -5 \ log_2 \ 2\\\\m = -5 \times 1 = -5[/tex]
(iv)
[tex]log_a \ m = 0\\\\m = a^0 = 1[/tex] [tex][ \ log _a \ 1 = 0 \ ][/tex]
(vii)
[tex]log_{32} \ 8 = m\\\\8 = 32^m\\\\8^{ \frac{1}{m}} = 32 \\\\(2^{3 })^ { \frac{1}{m}} = 2^{5}\\\\2^{\frac{3}{m}} = 2^5 \\\\\frac{3}{m} = 5\\\\3 = 5 \times m \\\\m = \frac{3}{5}[/tex]
(viii)
[tex]log_5 \ ( 2m + 5 ) = 3\\\\(2m +5 ) = 5^3\\\\2m + 5 = 125\\\\2m = 125 - 5 \\\\2m = 120 \\\\m = 60[/tex]
(x)
[tex]log_{m^3} \ 64 = \frac{2}{3}\\\\64 = (m^3)^{ \frac{2}{3}}\\\\64 = m^{ 3 \times \frac{2}{3}}\\\\64 = m ^2\\\\\sqrt{64} = m \\\\8 = m[/tex]
The side of an equilateral triangle is 12. Its area is
Select one:
a. 144
b. 72
c. 36 √3
d. 48
Answer:
area is 1/2bh
in an euqilateral triangle, all sides are equal so base and height are 12
1/2 x 12 x 12
1/2 x 144= 72 (B)
Answer: Choice C. [tex]36\sqrt{3}[/tex]
This is the same as writing 36*sqrt(3)
===================================================
Work Shown:
x = 12 = side length of equilateral triangle
A = area of equilateral triangle
A = 0.25*sqrt(3)*x^2
A = 0.25*sqrt(3)*12^2
A = 0.25*sqrt(3)*144
A = 0.25*144*sqrt(3)
A = 36*sqrt(3) .... answer is choice C
Side note: the area approximates to 62.354 square units.
The volume of a rectangular prism is given by 24x3+78x2+49x+10. The height of the prism is given by 2x+5. Find an expression for the area of the base of the prism
Answer:
?
Step-by-step explanation:
i cant not explian that
The temperature at 2 a.m. was -10°C.
At an earlier time the temperature was
0°C. It changed by -2°C each hour
until 2 a.m. At what earlier time was
the temperature 0°C?
Answer:
-1
Step-by-step explanation:
this question is much too hard would anyone please help me
Answer:
B and C are the same angles so if B is 60 so is C
Answer:
b= 60
c= 60
Step-by-step explanation:
<b and 120 form a straight line so the add to 180
b+120 =180
b = 180-120
b = 60
angles b and c are alternate interior angles so they are equal
b = c= 60
How is the graph of y = 8x2 − 1 different from the graph of y = 8x2?
It is shifted 1 unit down.
It is shifted 1 unit to the right.
It is shifted 1 unit to the left.
It is shifted 1 unit up.
Answer:
since it's the lhs we are concerned about, i. e., Y axis, so it must be either up or down. now look at the question, it says 8x2 - (1), it means one lower value of y i. e., 1 unit down
Answer:It is shifted 1 unit down.
Step-by-step explanation:
Two points on the line are (0, 4) and (7, 18). Use the points to first determine the slope and y-intercept. Then write the equation of the line. a. m = 2; y intercept = (0, 4); y = 2 x + 4 b. m = negative 2; y intercept = (0, 4); y = negative 2 x + 4 c. m = negative 2; y intercept = (4, 0); y = negative 2 x + 4 d. m = 2; y intercept = (0, 4); y = 4 x + 2
9514 1404 393
Answer:
a. m = 2; y intercept = (0, 4); y = 2 x + 4
Step-by-step explanation:
The slope can be found using the formula ...
m = (y2 -y1)/(x2 -x1)
m = (18 -4)/(7 -0) = 14/7
m = 2
The y-intercept is the first given point: (0, 4).
The equation for the line in slope-intercept form is
y = mx + b
Here, we have m=2 and b=4, so the equation is ...
y = 2x +4
Find the value of k, if (x - 2) is a factor of the polynomial p(x) = 2x2 + 3x - k
Answer:
The value of k is 14.
Step-by-step explanation:
(x - 2) is a factor of the polynomial
[tex]x - 2 = 0 \rightarrow x = 2[/tex]
This means that [tex]p(2) = 0[/tex]
p(x) = 2x² + 3x - k
[tex]p(2) = 2(2)^2 + 3(2) - k[/tex]
[tex]0 = 8 + 6 - k[/tex]
[tex]14 - k = 0[/tex]
[tex]k = 14[/tex]
The value of k is 14.
Please help me with solving these. Thank you very much. Have a great day!
Answer:
Problem 20)
[tex]\displaystyle \frac{dy}{dx}=(\cos x)^x\left(\ln \cos x-x\tan x\right)[/tex]
Problem 21)
A)
The velocity function is:
[tex]\displaystyle v(t) =2\pi(\cos(2\pi t)-\sin(\pi t))[/tex]
The acceleration function is:
[tex]\displaystyle a(t)=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))[/tex]
B)
[tex]s(0)=2\text{, }v(0) = 2\pi \text{ m/s}\text{, and } a(0) = -2\pi^2\text{ m/s$^2$}[/tex]
Step-by-step explanation:
Problem 20)
We want to differentiate the equation:
[tex]\displaystyle y=\left(\cos x\right)^x[/tex]
We can take the natural log of both sides. This yields:
[tex]\displaystyle \ln y = \ln((\cos x)^x)[/tex]
Since ln(aᵇ) = bln(a):
[tex]\displaystyle \ln y =x\ln \cos x[/tex]
Take the derivative of both sides with respect to x:
[tex]\displaystyle \frac{d}{dx}\left[\ln y \right]=\frac{d}{dx}\left[x \ln \cos x\right][/tex]
Implicitly differentiate the left and use the product rule on the right. Therefore:
[tex]\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x+x\left(\frac{1}{\cos x}\cdot -\sin(x)\right)[/tex]
Simplify:
[tex]\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x-\frac{x\sin x}{\cos x}[/tex]
Simplify and multiply both sides by y:
[tex]\displaystyle \frac{dy}{dx}=y\left(\ln \cos x-x \tan x\right)[/tex]
Since y = (cos x)ˣ:
[tex]\displaystyle \frac{dy}{dx}=(\cos x)^x\left(\ln \cos x-x\tan x\right)[/tex]
Problem 21)
We are given the position function of a particle:
[tex]\displaystyle s(t)= \sin (2\pi t)+2\cos(\pi t)[/tex]
A)
Recall that the velocity function is the derivative of the position function. Hence:
[tex]\displaystyle v(t)=s'(t)=\frac{d}{dt}[\sin(2\pi t)+2\cos(\pi t)][/tex]
Differentiate:
[tex]\displaystyle \begin{aligned} v(t) &= 2\pi \cos(2\pi t)-2\pi \sin(\pi t)\\&=2\pi(\cos(2\pi t)-\sin(\pi t))\end{aligned}[/tex]
The acceleration function is the derivative of the velocity function. Hence:
[tex]\displaystyle a(t)=v'(t)=\frac{d}{dt}[2\pi(\cos(2\pi t)-\sin(\pi t))][/tex]
Differentiate:
[tex]\displaystyle \begin{aligned} a(t)&=2\pi[-2\pi\sin(2\pi t)-\pi\cos(\pi t)]\\&=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))\end{aligned}[/tex]
B)
The position at t = 0 will be:
[tex]\displaystyle \begin{aligned} s(0)&=\sin(2\pi(0))+2\cos(\pi(0))\\&=\sin(0)+2\cos(0)\\&=(1)+2(1)\\&=2\end{aligned}[/tex]
The velocity at t = 0 will be:
[tex]\displaystyle \begin{aligned} v(0)&=2\pi(\cos(2\pi (0)-\sin(\pi(0))\\&=2\pi(\cos(0)-\sin(0))\\&=2\pi((1)-(0))\\&=2\pi \text{ m/s}\end{aligned}[/tex]
And the acceleration at t = 0 will be:
[tex]\displaystyle \begin{aligned} a(0) &= -2\pi ^2(2\sin(2\pi(0))+\cos(\pi(0)) \\ & = -2\pi ^2(2\sin(0)+\cos(0)) \\ &= -2\pi ^2(2(0)+(1)) \\ &= -2\pi^2(1) \\ &= -2\pi^2\text{ m/s$^2$} \end{aligned}[/tex]