The probability of choosing a coin that is worth at leas 10 cents is:
[tex]P=\frac{22}{40}=\frac{11}{20}[/tex]then the probability of obtaining a penny is:
[tex]P=\frac{8}{39}[/tex]Then the probability to obtain a coin of at least 10 cents and then a penny is:
[tex]P=\frac{11}{20}\cdot\frac{8}{39}=\frac{22}{195}[/tex]For each linear system of equations, identify the
solution; if there is no solution, put an asterisk (*) in
the boxes for the coordinate values.
System A has a solution at
System B has a solution at(
System C has a solution at
A. y=x+5
V = 3x -3
B.
-2V
-1+7
8V = 4x + 10
C. y=
-3x
x+ y = 16
EXPLANATION
First equation:
y = x + 5
y = 3x - 3
[tex]\mathrm{Substitute\:}y=3x-3[/tex][tex]\begin{bmatrix}3x-3=x+5\end{bmatrix}[/tex][tex]\mathrm{Add\:}3\mathrm{\:to\:both\:sides}[/tex][tex]x-3+3=x+5+3[/tex]Simplify:
[tex]3x=x+8[/tex][tex]\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}[/tex][tex]3x-x=x+8-x[/tex]Simplify:
[tex]2x=8[/tex][tex]\mathrm{Divide\:both\:sides\:by\:}2[/tex][tex]\frac{2x}{2}=\frac{8}{2}[/tex]Simplify:
[tex]x=4[/tex][tex]\mathrm{For\:}y=3x-3[/tex][tex]\mathrm{Substitute\:}x=4[/tex][tex]y=3\cdot \:4-3[/tex][tex]\mathrm{Simplify}[/tex][tex]y=9[/tex]Therefore, System A has a solution at (4,9)
Second equation:
[tex]\begin{bmatrix}-2y=-x+7 \\ 8y=4x+10\end{bmatrix}[/tex][tex]\mathrm{Divide\:both\:sides\:by\:}-2[/tex][tex]\frac{-2y}{-2}=-\frac{x}{-2}+\frac{7}{-2}[/tex]Simplifying:
[tex]y=-\frac{-x+7}{2}[/tex][tex]\mathrm{Substitute\:}y=-\frac{-x+7}{2}[/tex][tex]\begin{bmatrix}8\left(-\frac{-x+7}{2}\right)=4x+10\end{bmatrix}[/tex]Removing the parentheses:
[tex]=-8\cdot \frac{-x+7}{2}[/tex]Multiplying fractions:
[tex]=-\frac{\left(-x+7\right)\cdot \:8}{2}[/tex][tex]\mathrm{Divide\:the\:numbers:}\:\frac{8}{2}=4[/tex][tex]-4\left(-x+7\right)=4x+10[/tex][tex]\mathrm{Refine\:}-4\left(-x+7\right)=4x+10[/tex][tex]-38=0[/tex][tex]-38=0\:\mathrm{\:is\:false,\:therefore\:the\:system\:of\:equations\:has\:no\:solution}[/tex]System B has a solution at (*,*)
Third equation:
[tex]\begin{bmatrix}y=-3x\\ x+y=16\end{bmatrix}[/tex][tex]\mathrm{Substitute\:}y=-3x[/tex][tex]\begin{bmatrix}x-3x=16\end{bmatrix}[/tex][tex]\begin{bmatrix}-2x=16\end{bmatrix}[/tex][tex]\mathrm{Divide\:both\:sides\:by\:}-2[/tex][tex]\frac{-2x}{-2}=\frac{16}{-2}[/tex]Simplify the expression:
[tex]x=-8[/tex][tex]\mathrm{For\:}y=-3x[/tex][tex]\mathrm{Substitute\:}x=-8[/tex][tex]y=-3\left(-8\right)[/tex][tex]\mathrm{Simplify}[/tex][tex]y=24[/tex][tex]\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}[/tex][tex]y=24,\:x=-8[/tex]Therefore, System C has a solution at (-8,24)
In a two hour TV broadcast , the commercials show 18% of the time. How many minutes of commercials are shown during the 2 hour duration of the broadcast ? Round to the nearest 10th of a hundreth.
Given:
In a two hour TV broadcast , the commercials show 18% of the time.
The number of hours of broadcast is 2 hours.
Required:
We need to find the number of minutes of a commercial show for 2 hours broadcast.
Explanation:
Convert the hours into minutes.
Multiply 2 by 60 to convert hours to minutes.
[tex]2hours\text{ =2}\times60\text{ minutes =120 minutes.}[/tex]18 % of 120 minutes is the commercials.
[tex]\frac{18}{100}\times120=21.6minutes[/tex]21.6 minutes of commercials are shown during the 2 hours duration of the broadcast
Final answer:
21.6 minutes of commercials are shown during the 2 hours duration of the broadcast
Range of 49,49,51,47,49
The range of a set of numbers is the difference between the highest and lowest value.
To find the range:
1) Find the highest and lowest number.
Highest = 51
Lowest = 47
2) Substract the lowest from the highest value.
Range = 51 - 47
Range = 4
Answer: Range = 4.
Question 20 - 5 PointsJon 14, 10:17:14 PM?In APRQ shown below, point S is on QR, and point T is on PR so thatZPQR = ZSTR. If PQ = 104, ST = 40, and RP = 130, find the length ofRS. Figures are not necessarily drawn to scale.TQSRoo
Since we are told that angles PQR and STR are congruent, we can draw each triangle apart from each other, as follows
Since they have one congruent angle, we can see that one triangle is a drawn as scale from the other. Having the congruent angle in the same position (left most corner of each triangle) allows us to relate the sides of each triangle regarding their positions with respect to the congruent angle. In the picture we have draw colored lines for each pair of related sides. Since they are similar triangles, we can establish the following equation
[tex]\frac{ST}{PQ}=\frac{SR}{PR}[/tex]We have PQ=104, ST =40 and PR = 130. Thus
[tex]\frac{SR}{130}=\frac{40}{104}[/tex]So, by multiplying both sides by 130 we get
[tex]SR=\frac{130\cdot40}{104}=\frac{5200}{104}=50[/tex]find the sum of the angle measures of each polygon 50 - gon
To be able to find the sum of all angles in a 50 - gon, we will be using the following formula:
[tex]\text{ (n - 2) x 180}[/tex]Where,
n = number of sides = 50
We get,
[tex]\text{ (n - 2) x 180}[/tex][tex]\text{ = (50 - 2) x 180}[/tex][tex]\text{ = 48 x 180}[/tex][tex]\text{ = 8,640}^{\circ}[/tex]Therefore, the sum of all angles inside a polygon is 8,640
Answer:
8640°
Step-by-step explanation:
Formula: 180(n-2) = 180(50-2) = 8640
I need help with the steps to solve the attach problem in the photo.
• We are told that there were 3/8 in the begining , where 3 is the number of advaced maths and 8 is the number of regular maths .
,• 4 /7 were the final results . meaning 4 advanced and 7 regular maths
,• 4/7 : 92/x ( where x is final advanced, and x is the regular maths )
so, 3+8 = 11 and 4+7 = 11
if 92 students are equal to 4 in the final ratio;
92 / 4 = 23 in the ratio of 1
therefore :
if 4 : 92
then 7 : 23
7 x 23 = 161 students in regular maths
161 + 92 = 253 students in total
(extra notes :if you want to find the original class numbers
253 /11 = 23
23 X 3 = 69
23 X8 = 184
this means that there were 69 advanced and 184 regular students)
M R 3 5 N 6 10. B S 6 L 7 ALMN А E 6 D B 8 20 20
8) To see if the triangles are similar que hace to calculate the ratio beetwen the side so:
[tex]\begin{gathered} \frac{5}{6}=\frac{6}{7}=\frac{3}{4} \\ 0.83=0.88=0.75 \end{gathered}[/tex]So the ratios are nor constant so the triangles are not similar
You are given rectangle ABCD on the grid shown and are told that the figure is reflected, but you are not told over which axis this has occurred. Which COULD be the new coordinates of point D? Responses A (-6, -4)(-6, -4) B (-1, -4)(-1, -4) C (-2, -2)(-2, -2) D (4, -4)(4, -4)
The new coordinate of point D is Option A which is ( -6 , -4 )
The coordinates of point D from the figure is (-4,-4)
The all possible cases to move the figure 2 units is as following:
If the figure is moved two units right point D will be (-2,-4)
If the figure is moved two units left point D will be (-6,-4)
And if it is moved two units up point D will be (-4,-2)
also if it is moved two units down point D will be (-4,-6)
By checking the given options, we will find that:
The options b , c and d is not at the previous possible cases
So, the answer is option a. (-6,-4)
Which mean the figure is moved two units left
To know more about coordinates you may visit the link which is mentioned below:
https://brainly.com/question/27749090
#SPJ1
f(x) = 3x2 - 4x + 2; f(x - 1)
Answer:
They are parallel lines.
Positive lines
Step-by-step explanation:
You can plug both equations into desmos and you can see for yourself.
Also if it was f(-1), it would equal 0. If it was f(1) it would equal 4. But those are not what you are asking for. The second equation is a y-intercept on (0,0)
Hope this helps :)
What is 12 = 4 + 7 x 8
Answer:
Step-by-step explanation:
PEMDAS
1. Choose the correct verbal expression for thealgebraic expression 2(y2-9) + 4.*
All algebraix expression follow rules of ( PEMDAS ) where,
[tex]\begin{gathered} P\colon\text{ Paranthesis} \\ E\colon\text{Exponents} \\ M\colon\text{ Multiplication} \\ D\colon\text{ Division} \\ A\colon\text{ Addition} \\ S\colon\text{ Subtraction} \end{gathered}[/tex]The acronyms for algebraic manipulation rule ( PEMDAS ) is given above. The same rule is also applied when you are verbally describing any algebraic expressions or equations.
The priority of algebraic manipulation decreases down the list given above.
We are given an algebraic expression as follows:
find three points that solve the equation X - 3y = -6
Given the linear equation:
[tex]x-3y=-6[/tex]The single linear equation above contains two variables. Hence, there is no unique solution but infinitely many pairs of solutions. For every value of x, there is a value for y.
Three points that solve the equation would be:
Solution 1
Put x = -6 into the equation
[tex]\begin{gathered} -6-3y=-6 \\ -3y=-6+6 \\ \frac{-3y}{-3}=\frac{0}{-3} \\ y=0 \end{gathered}[/tex]Solution 2
Put x = 0 into the equation
[tex]\begin{gathered} 0-3y=-6 \\ -3y=-6 \\ \frac{-3y}{-3}=\frac{-6}{-3} \\ y=2 \end{gathered}[/tex]Solution 3
Put x = 12 into the equation
[tex]\begin{gathered} 12-3y=-6 \\ -3y=-6-12 \\ -3y=-18 \\ \frac{-3y}{-3}=\frac{-18}{-3} \\ y=6 \end{gathered}[/tex]Therefore, three points that solve the linear equation as shown above are:
[tex]\begin{gathered} (x,y)=(-6,0) \\ (x,y)=(0,2) \\ (x,y)=(12,6) \end{gathered}[/tex]
What is the equation of the circle shown?785432.А1-B-8-5-4-3-2-1 02.4-1-2-3-4-5(x2 + (y)2 =IIWord Bank:10 -2 5 + 2 - 1+ 1 25Blank 1:Blank 2:Blank 3:
Consider that the equation of a circle with center (h,k) and radius 'r' units, is given by,
[tex](x-h)^2-(y-k)^2=r^2[/tex]Observe the given diagram carefully.
The center of the circle lies at (-2,1),
[tex](h,k)=(-2,1)[/tex]The radius of the circle is the distance of any point on the circle from the center. Since the point (3,0) lies on the circle, the radius is calculated as,
[tex]r=3-(-2)=3+2=5[/tex]Substitute the values in the standard equation of circle,
[tex]\begin{gathered} (x-(-2))^2-(y-1)^2=5^2 \\ (x+2)^2-(y-1)^2=25 \end{gathered}[/tex]Thus, the equation of the given circle is,
[tex](x+2)^2-(y-1)^2=25[/tex]what number could replace c below 4/8=c/4
The correct answer is 2
Let f(x) = x + 3 and g(x) = . The graph of (fog)(x) is shown below.83-6 +5+432->4--2-2343*What is the range of (fºg)(x)?
we can notice the range from the graph, then we see the possible values of y
the graph can take all values of y except y=3 because the graph has an asymptote, never touch 3
the green line is the asymptote
then the range is
[tex](-\infty,3)U(3,\infty)[/tex]or Real numbers except 3
The table shows the height, in centimeters, of the water in a swimming pool at different times since the pool started to be filled.Does the height of the water increase by the same amount each minute? Explain how you know.Does the height of the water increase by the same factor each minute? Explain how you know.
The table shows us the next information:
0 mins - 150 height
1 min - 150.5, the height increase 0.5 because 150+0.5 = 150.5
2 mins - 151, the height increase 0.5 because 150.5 +0.5 = 151
3 min - 151, the height increase 0.5 becuase 151.5
Question 1
Therefore, we can deduce that we have a linear equation because we have a common difference. Hence, the height of water increase by the same amount each time.
Question 2
Now, the table doesnt increase by the same factor because the tables doesnt represents an exponential equation.The factors are close but not the same.
Louise Tammy Dolores and Cheryl score 78 points in the tennis matches they play. A score consecutive number of points from smallest to greatest in respect to the order of the name is mentioned. How many points does Lewis score? Options:A. 19 pB. 20 pC. 21 pD. 18 p
Given that:
• Louise Tammy Dolores and Cheryl score 78 points in the tennis matches they play.
• A score consecutive number of points from smallest to greatest in respect to the order of the names: Louise Tammy Dolores and Cheryl.
Let be:
- The number of points that Louise scored in the tennis matches:
[tex]n[/tex]- The number of points that Tammy scored in the tennis matches:
[tex]n+1[/tex]-The number of points that Dolores scored in the tennis matches:
[tex]n+2[/tex]- The number of points that Cheryl scored in the tennis matches:
[tex]n+3[/tex]You can set up the following equation to represent the total number of points they scored in the tennis matches:
[tex]n+(n+1)+(n+2)+(n+3)=78[/tex]Now you can solve for "n":
[tex]n+n+1+n+2+n+3=78[/tex][tex]4n+6=78[/tex][tex]4n=78-6[/tex][tex]\begin{gathered} n=\frac{72}{4} \\ \\ n=18 \end{gathered}[/tex]Hence, the answer is: Option D.
Find the equation of the perpendicular to 5x+3y=7 and passing through (2,3)
The first step to solve this problem is arrange the equation in the standard line equation form, which is given below:
[tex]y\text{ = m}\cdot x\text{ + }b[/tex]y = m*x + b
Where "m" is the slope of the line and "b" is the constant shift. To find the perpendicular line we need to find a line with the slope as shown below:
[tex]m_1\text{ =-}\frac{1}{m}\text{ }[/tex]m1 = -1/m
We need to arrange the given expression.
[tex]\begin{gathered} 5x\text{ + 3y = 7} \\ 3y\text{ = 7 - 5x} \\ y\text{ = -}\frac{5}{3}x\text{ + }\frac{7}{3} \end{gathered}[/tex]5x + 3y = 7
3y = 7 - 5x
y = (-5/3)x + 7/3
The slope of the given line is -5/3 to find the perpendicular line we need to find the slope -1/m. Which is done below:
m1 = -1/(-5/3) = -1*(-3/5) = 3/5
The perpendicular line we want has the a slope of 3/5. We can find the constant shift by applying the given point.
Perpendicular:
y = (3/5)*x + b
Applying the point:
3 = (3/5)*2 + b
3 = 6/5 + b
b = 3 - 6/5 = 15/5 - 6/5 = 9/5
Therefore the perpendicular line is given by:
y = (3/5)*x + 9/5
In the coordinate plane shown point not shown) lies on segment AB. Vf the ratioof the length of segment AC to the length of segment CB is 3:1, what is the y-coordinate of point C. Round to the nearest tenth.
We know that the point C lies in the segment AB. Since the ratio between AC and CB is 3:1, we also know it lives in the first 1/4 of the segment.
To calculate the point at 1/4 of the segment, we can find the midpoint, and then the midpoint of the segment made with point A and the midpoint.
The midpoint y-coordinate is given by:
[tex]\frac{y_1+y_2}{2}=\frac{4+10}{2}=7[/tex]Now, calculating the midpoint between the midpoint of the segment AC and A, we have:
[tex]\frac{4+7}{2}=\frac{11}{2}=5.5[/tex]And this is our answer. The y coordinate of the point C is 5.5
I need help with this question
Let:
[tex]\begin{gathered} B=(3,2) \\ B^{\prime}=(-2,-2) \\ \end{gathered}[/tex]After a translation 5 units left
[tex]B\to(3-5,2)\to(-2,2)[/tex]and after a reflection across the x-axis:
[tex]B\to(x,-y)\to(-2,-2)[/tex]I need help with cross products I need to find the value of x in the problem for 4/9=x/15
15 is dividing on the right, then it will multiply on the left
9 is dividing on the left, then it will multiply on the right
[tex]\begin{gathered} 4\cdot15=x\cdot9 \\ 60=x\cdot9 \end{gathered}[/tex]Now, 9 is multiplying on the right, then it will divide on the left
[tex]\frac{60}{9}=x[/tex]Yesterday the snow was 2 feet deep today the depth dropped 1.6 feet because it is so warm what is the percent change in a deck of snow
Given:
Depth dropped from 2 feet to 1.6 feet.
[tex]\begin{gathered} \text{Depth decreased=2-1.6 feet} \\ \text{Depth decreased=}0.4\text{ feet} \end{gathered}[/tex][tex]\begin{gathered} \text{Percentage dropped=}\frac{\text{0.4}}{\text{2}}\times100 \\ \text{Percentage dropped=}20\text{ \%} \end{gathered}[/tex]find the volume of the figure shown here. round your answer to the nearest tenth
The volume of a cylinder is given by the product of the area of its base times its height. If the base of the cylinder has a radius r and the height of the cylinder is h, then its volume V is given by the formula:
[tex]V=\pi r^2h[/tex]Substitute r=5cm, h=8cm and use π=3.14 to find the volume of the cylinder:
[tex]\begin{gathered} V=3.14(5cm)^2(8cm) \\ =3.14\cdot25cm^2\cdot8cm \\ =3.14\cdot200cm^3 \\ =628cm^3 \end{gathered}[/tex]Therefore, the volume of the cylinder is:
[tex]628cm^3[/tex]What is 3/8 of 200?with work
Hello!
To calculate 3/8 of 200, first, we have to divide 200 by 8 (to find 1/8 of it). Then, we will multiply the obtained value by 3 (to find 1/8). Look at the reasoning below:
1/8 of 200 is:[tex]\frac{200}{8}=25[/tex]3/8 of 200 is:[tex]25\cdot3=75[/tex]Answer:3/8 of 200 is 75.
The circle below is centered at (10, 4) and has a radius of 4. What is itsequation?A. (x-10)2 + (y-4)2 = 16O B. (x-4)2 + (x-10)2 = 4C. (x-4)2 + (y-10)2 = 16D. (x-10)2+(y-4)2 = 4
SOLUTION
The general form for the equation of a circle is given as
[tex](x-a)^2+(y-b)^2=r^2_{}[/tex]where the center of the circle is (a,b) and the radius r
Given a center (10,4) and r=4
The equation of the circle becomes
[tex]\begin{gathered} a=10,b=4,r=4 \\ (x-10)^2+(y-4)^2=4^2 \end{gathered}[/tex]Then the equation becomes
[tex](x-10)^2+(y-4)^2=16[/tex]TherefThe right option is A
Justin will choose between two restaurants to purchase pizzas for his party. The first restaurant charges a delivery fee of $9 for the entire purchase and $12 per Pizza. II restaurant charges a delivery fee of $1 for the entire purchase and $13 per pizza.Let x be the number of pizzas purchased.(a) For each restaurant, write an expression for the total change of purchasing x pizzas (with delivery). Total charge from the first restaurant (in dollars) = ??Total charge from the second restuarant (in dollars). = ??(b) Write an equation to show when the total charge (with delivery) would be the same for each restaurant.
(a)
First Restaurants
Let x be number of pizzas purchased
Total charge = 12x + 9
Second Restaurants
Let x be number of pizzas purchased
Total charge = 13x + 1
(b)
Total would be equal when
12x + 9 = 13x + 1
collect like terms
13x - 12x = 9 - 1
x = 8
Total charge for both restaurants will be equal when the total number of pizzas purchased is 8
Can you help me please i will give you a branlist and i did part a i want part b please
Given:
D (2, 6)
E (4, 2)
F (6, 4)
and scale factor of 3/2. To find D'E'F' you have to multiply each coordinate by the scale factor, as follows:
D' (2*3/2, 6*3/2) = D' (3, 9)
E' (4*3/2, 2*3/2) = E' (6, 3)
F' (6*3/2, 4*3/2) = F' (9, 6)
Which of the following identities are true? Select all that apply. 3 answers
Solution
Step 1
Trigonometric inverse theorem
[tex]csc\theta\text{ = }\frac{1}{sin\theta}\text{ TRUE}[/tex]Step 2
[tex]\begin{gathered} Trigonometric\text{ ratio} \\ cot\theta\text{ = }\frac{cos\theta}{sin\theta} \\ Option\text{ B is FALSE} \end{gathered}[/tex]Step 3
[tex]\begin{gathered} 1\text{ + cot}^2\theta \\ =\text{ 1 + }\frac{cos^2\theta}{sin^2\theta} \\ =\text{ }\frac{sin^2\theta+cos^2\theta}{sin^2\theta} \\ \text{= }\frac{1}{sin^2\theta} \\ \text{= csc}^2\theta \\ 1\text{ + cot}^2\theta\text{ = csc}^2\theta\text{ TRUE} \end{gathered}[/tex]Step 4
[tex]\begin{gathered} sin(-\pi) \\ =\text{ sin\lparen0-}\pi) \\ =\text{ sin0cos}\pi\text{ - sin}\pi cos0 \\ =\text{ -sin}\pi \\ sin(-\pi)\text{ = -sin}\pi \\ Option\text{ D is FALSE} \end{gathered}[/tex]Step 5
[tex]\begin{gathered} cos(-\pi) \\ =\text{ cos\lparen0-}\pi) \\ =\text{ cos0cos}\pi\text{ + sin0sin}\pi \\ =\text{ cos}\pi \\ cos(-\pi)\text{ = cos}\pi\text{ TRUE} \end{gathered}[/tex]The box plot below represents some data set. What percentage of the data values aregreater than 78?
To determine the percentage of the data that is greater than 78, the first step is to identify the value on the number line and the corresponding part on the box plot.
Each division on the number line represents 2 units.
The left side of the box is determined by the first quartile (Q1), if you look at the picture above, the first quartile is equal to 66.
The line in the middle of the box represents the second quartile (Q2) or median of the data set, in this case, the second quartile is 72.
The right side of the box is determined by the third quartile (Q3) of the data set, looking at the number line, the value of the third quartile corresponds to 78.
The third quartile is the value of any data set that divides the bottom 75% of the values from the top 25% of the values.
This means that the percentage of values less than 78 is 75%, and the percentage of values greater than 78 is 25%
14 Which number makes this equation D true? 6/8= /4a.2b.3c.4d.6
Answer:
Option B. 3
Because if you divide by 2 both the numerator and denominator, you get 3/4