Answer:
9 pieces
Explanation:
From the question, we're told that the length of the stick is 10 1/5 inches, let's convert the mixed fraction into an improper fraction;
[tex]10\frac{1}{5}=\frac{51}{5}[/tex]Also, we're told that the stick was cut into shorter pieces of length 1 2/15 inches each. Converting 1 2/15 into an improper fraction, we'll have;
[tex]1\frac{2}{15}=\frac{17}{15}[/tex]To determine the number of pieces that the stick will be cut into, we'll need to divide 51/5 by 17/15;
[tex]\frac{\frac{51}{5}}{\frac{17}{15}}=\frac{51}{5}\ast\frac{15}{17}=\frac{15}{1}\ast\frac{3}{17}=\frac{153}{17}=9[/tex]Using the z score formula use the information below to find the value of
Explanation
Given that
[tex]\begin{gathered} z=-4.80 \\ x=23.55 \\ \mu=32.67 \end{gathered}[/tex]Using the z-score formula;
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ -4.80=\frac{23.55-32.67}{\sigma} \\ -4.80\sigma=-9.12 \\ \sigma=\frac{-9.12}{-4.80} \\ \sigma=1.9 \end{gathered}[/tex]Answer: 1.9
Not a timed or graded assignment. Need a quick answer showing work. Please DRAW factor tree. Thank you so much.
ANSWER:
[tex]40\sqrt[]{7}m^3[/tex]STEP-BY-STEP EXPLANATION:
Using the following factor tree, we decompose the number 112 to be able to simplify, just like this:
Therefore, it would be:
[tex]\begin{gathered} 10\sqrt[]{2\cdot2\cdot2\cdot2\cdot7\cdot m^6} \\ 10\sqrt[]{2^4\cdot7\cdot m^{3\cdot2}} \\ 10\cdot2^2\cdot m^3\sqrt[]{7} \\ 10\cdot4\cdot m^3\sqrt[]{7}=40\sqrt[]{7}m^3 \end{gathered}[/tex]a quadrilateral has vertices S (0,0) T (4,0) U (5,4) V (1,4). what is the most precise name for the quadrilateral
First, plot the points on a coordinate plane:
Notice that ST is parallel to VU, as well as SV is parallel to TU.
A quadrilateral whose opposite sides are parallel, is called a parallelogram.
Therefore, the most precise name for the quadrilateral is: parallelogram.
Angles P and Q are supplementary and m
Given: P &Q are supplementary. Therefore P+Q = 180 degress
P = 180 - Q
If m
then m +30 = p
m + 30 = 180 - Q
m = 150 - Q
m - 150< -Q
m + 150 > Q
m>q = 150
Zoe and Marsden are working with expressions with rational exponents. Zoe believes V2+ V8 is equivalent to 2.21. Marsden believes v2 + V8 is equivalent to 3.21. Use the properties of exponents to decide who is correct. Write the correct answer in the space provided.
Explanation
the properties of exponents
[tex]\begin{gathered} \sqrt[n]{a}=a^{\frac{1}{n}} \\ \sqrt[n]{a^{n^{}}b^n}\text{ = ab} \\ \sqrt[]{ab}=\sqrt[]{a}\cdot\sqrt[]{b} \end{gathered}[/tex]Step 1
Zoe believes
[tex]\sqrt[]{2}+\sqrt[]{8\text{ }}=\text{ 2.21}[/tex]Marsden Believes
[tex]\sqrt[]{2}+\sqrt[]{8}=3.21[/tex]Step 2
[tex]\sqrt[]{8}=\sqrt[]{4}\cdot\sqrt[]{2}=\text{ 2}\sqrt[]{2}[/tex]then
[tex]\begin{gathered} \sqrt[]{2}+\sqrt[]{8}=\text{ }\sqrt[]{2}+2\sqrt[]{2}=3\sqrt[]{2} \\ \text{also } \\ \sqrt[]{2}=\text{ 1.4142} \\ so, \\ \sqrt[]{2}+\sqrt[]{8}=3\sqrt[]{2}=3\cdot1.4142=4.24 \end{gathered}[/tex]then
[tex]\sqrt[]{2}+\sqrt[]{8}=3\sqrt[]{2}=3\cdot1.4142=4.24[/tex]I hope this helps you
> Use the drop-down menus to explain how Ken' can use the model to find the total weight of the baseballs. Click the arrows to choose an answer from each menu. To represent the weight of one baseball, 0.3 pounds, Ken should shade 3 Choose... To represent the weight of all of the baseballs, he should shade this amount 7 times. The shaded part of the model will represent the expression Choose... The total weight of the baseballs is 2.1 pounds. > My Progress All rights reserved These materials, or any portion thereof may not be reproduced or shared in any manner without expS ENCOME
To represent 0.3 pounds Ken should shade 3 columns
To represent the weight of all of the baseballs He should shade 7 times
The shaded part of this model will represent the expression 7 x 0.3 (7x 0.3 = 2.1)
The Total weight of the baseballs is 2.1 pounds
1) Gathering the data
7 baseballs each one weighs 0.3 pounds
7 x 0.3 = 2.1 ⇔2.1 : 0.3 = 7
Each tiny square corresponds to 0.01 of the block.
2.1 : 0.3 = 7
2) So the total weight (2.1) corresponds to 2 blocks and 0.1 blocks mean 1 column on the third block. Or, 2.1 = 7 x 0.3 Seven times three columns.
Examining the options:
3 ) We can answer each drop-down menu:
• To represent 0.3 pounds Ken should shade ,3 columns
,• To represent the weight of all of the baseballs He should shade ,7 ,times
,• The shaded part of this model will represent the expression ,7 x 0.3 ,(7x 0.3 = 2.1)
,• The Total weight of the baseballs is ,2.1 pounds
XIXIXI
Name:
amount paid (in dollars)
Movie Mania Tickets
ty
72
63
54
45
36
27
10 00
18
9
0
CINEMA
ADMIT ONE
12345
ADMIT ONE
nyum
Movie Ticket Sales
12345
2 4 6
8
# of tickets
72=6=12
X
10
Perfect Picture Tickets
Only $65.50 for 5 tickets!
65÷5=13
# of tickets
Date:
Fantastic Flicks Tickets
3
6
9
12
amount paid (in dollars)
38.25
76.50
114.75
153
Periods:
The Big Screen
Which movie theater is the cheapest?
Pertert Picture
Which movie theater is the most expensive?
Movie mania
What is the constant of proportionality (slope) of each option?
movie minin. = 12
Perfect Pictures (3
·Fantastic Flick5= 12.75
The Big screen= 12.90
Write an equation to represent all four options.
12345
CINEMA
ADMIT ONE
y = 12.90x
NEMA
ADMIT ONE
12345
12.75
1. The cheapest movie theatre is Fantastic Flicks.
2. The most expensive movie theatre is Movie Mania.
3. The slopes for the ticket prices for Movie Mania, Perfect Picture, The Big Screen, and Fantastic Flicks are 13.5, 13.1, 12.9, and 12.75.
4. The equation to represent all the four options is (y - 13.5x)*(y - 13.1x)*(y - 12.9x)*(y - 12.75x) = 0.
We have four movie theaters. The names of the movie theatres are Movie Mania, Perfect Picture, The Big Screen, and Fantastic Flicks. We need to calculate the slopes for each option. Let the slopes for Movie Mania, Perfect Picture, The Big Screen, and Fantastic Flicks be m1, m2, m3, and m4, respectively.
m1 = 27/2 = 13.5
m2 = 65.5/5 = 13.1
m3 = 12.9
m4 = 153/12 = 12.75
The equations for the price of tickets for each theatre are given below.
E1 : y = 13.5x
E2 : y = 13.1x
E3 : y = 12.9x
E4 : y = 12.75x
The theatre with the minimum slope is the cheapest. The theatre with the maximum slope is the most expensive.
The equation to represent all four options is the product of all the other equations.
(y - 13.5x)*(y - 13.1x)*(y - 12.9x)*(y - 12.75x) = 0
To learn more about equations, visit :
https://brainly.com/question/10413253
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Write the inequality statement in x describing the numbers [ 11, ∞)
The inequality [ 11, ∞) represents that value is more than or equal to 11. The interval can be expressed as,
[tex]x\ge11[/tex]In inequality, x is any variable.
So inequality statement in x is,
[tex]x\ge11[/tex]can someone help me find the valu of X &Y?
Given that the triangle ABC and DEF are similar, therefore its corresponding sides must be proportional,
[tex]\begin{gathered} \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF} \\ \frac{4}{6}=\frac{10}{y}=\frac{x}{12} \end{gathered}[/tex]Comparing the first and third terms,
[tex]\begin{gathered} \frac{4}{6}=\frac{x}{12} \\ x=\frac{4}{6}\times12 \\ x=4\times2 \\ x=8 \end{gathered}[/tex]Comparing the first and second terms,
[tex]\begin{gathered} \frac{4}{6}=\frac{10}{y} \\ y=\frac{6}{4}\times10 \\ y=3\times5 \\ y=15 \end{gathered}[/tex]Thus, the values of 'x' and 'y' are 8 and 15, respectively.
Which of the functions is an exponential function? F(x)=-3x^-1F(x)=-3(2)^2F(x)=-3(1)^xF(x)=-3x^2
For this problem we recall the definition of an exponential function:
[tex]\begin{gathered} f(x)\text{ is an exponential function if } \\ f(x)=a\cdot b^{kx} \\ \text{Where a}\ne0,\text{ k}\ne0\text{ and b}\ne1 \end{gathered}[/tex]Answer: F(x)= - 3 (2)^x
Can you please help me out with a question
To determine the green rectangle, each side of the blue rectangle was multiplied by a determined scale factor k.
To determine the measure of x, the first step is to determine the scale factor.
The information that you have to use is the areas of both rectangles.
After dilation, the area of the resulting shape is equal to the area of the original shape multiplied by the square of the scale factor:
[tex]A_{\text{green}}=k^2A_{\text{blue}}[/tex]A.green=50 m²
A.blue= 72m²
[tex]50=72k^2[/tex]-Divide both sides by 72
[tex]\begin{gathered} \frac{50}{72}=\frac{72k^2}{72} \\ \frac{25}{36}=k^2 \end{gathered}[/tex]-Apply the square root to both sides of the equal sign:
[tex]\begin{gathered} \sqrt[]{\frac{25}{36}}=\sqrt[]{k^2} \\ \frac{5}{6}=k \end{gathered}[/tex]Now, to determine the value of x, multiply the length of the corresponding side on the blue rectangle by the scale factor:
[tex]\begin{gathered} x=\frac{5}{6}\cdot12 \\ x=10 \end{gathered}[/tex]The length of the side on the green triangle is 10m
What is - 4 - 2y= - x in standard form ?
The standard form is
[tex]Ax+By=C[/tex]We have
[tex]-4-2y=-x[/tex][tex]x-2y-4=0[/tex][tex]x-2y=4[/tex]The standard form is
[tex]x-2y=4[/tex]ANSWER
Find the perimeter and area of a square with side 9 inches.
The perimeter (P) and area (A) of a square of sides a = 9 in, are given by:
[tex]\begin{gathered} P=4a=4\cdot(9in)=36in, \\ A=a^2=(9in)^2=81in^2. \end{gathered}[/tex]Answer
• Perimeter = 36 in
,• Area =, ,81 in²
What is the value of x in the figure at the right? 60° (2x)°
The angle whose measure is 60° and the angle (2x)° are vertical angles, if two angles are vertical angles, then they are congruent, then we can express:
[tex]2x=60[/tex]From this expression, we can solve for x to get:
[tex]\begin{gathered} \frac{2x}{2}=\frac{60}{2} \\ x=30 \end{gathered}[/tex]Then, x equals 30
Given the function g(x) =x^2 +9x+18, determine the average rate of change of the function over the interval -8_
The given function is:
[tex]undefined[/tex]What is the equation of the circle whose diameter is the segment with endpoints (4,3) and (20,-9).
The standard equation of a circle is expressed according to the equation
[tex](x-a)^2+(y-b)^2=r^2[/tex]where;
(a, b) is the coordinate of the centre of the circle
r is the radius of the circle;
Get the diameter of the circle;
[tex]\begin{gathered} D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ D=\sqrt[]{(20-4)^2+(-9-3)^2} \\ D=\sqrt[]{16^2+(-12)^2} \\ D=\sqrt[]{256+144} \\ D=\sqrt[]{400} \\ D=20\text{units} \end{gathered}[/tex]For the radius of the circle;
[tex]\begin{gathered} r=\frac{D}{2} \\ r=\frac{20}{2} \\ r=10\text{units} \end{gathered}[/tex]Get the centre of the circle. Note that the centre will be the midpoint of the given endpoints as shown;
[tex]\begin{gathered} (a,b)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ (a,b)=(\frac{4+20}{2},\frac{3-9}{2}) \\ (a,b)=(\frac{24}{2},-\frac{6}{2}) \\ (a,b)=(12,-3) \end{gathered}[/tex]Substitute the centre (12, 3) and the radius 10 units into the equation of the circle above to have:
[tex]\begin{gathered} (x-12)^2+(y-(-3))^2=10^2 \\ (x-12)^2+(y+3)=100 \end{gathered}[/tex]This gives the equation of the circle whose diameter is the segment with endpoints (4,3) and (20,-9).
For my practice review, I need help to determine if these are functions or not.
Answer:
1: no
2: no
3: yes
4: no
5: yes
6: yes.
Step-by-step explanation:
Think of a vertical line sweeping across the graph from left to right. If ever this line crosses two points of the graph at the same time, it cannot be a function, since a function can only have max. 1 result per x value.
I need help with 5 and 6. The exponent for part 5 if you can't see it well 2/3
5.
Given the equation to solve for x:
[tex]3(x+1)^{\frac{2}{3}}=12[/tex]The steps for the solution are as follows:
[tex]\begin{gathered} 3(x+1)^{\frac{2}{3}}=12 \\ \frac{3(x+1)^{\frac{2}{3}}}{3}=\frac{12}{3} \\ (x+1)^{\frac{2}{3}}=4 \\ \lbrack(x+1)^{\frac{2}{3}}\rbrack^{\frac{1}{2}}=(4)^{\frac{1}{2}} \\ \lbrack(x+1)^{\frac{1}{3}}\rbrack=\pm2 \\ \lbrack(x+1)^{\frac{1}{3}}\rbrack^3=(\pm2)^3 \\ x+1=\pm8 \end{gathered}[/tex]From the above equation, we have x + 1 = 8 and x + 1 = -8. These imply x = 7 and x = -9.
Check for extraneous solutions:
If x = 7, then the left-hand side of the equation is:
[tex]3(x+1)^{\frac{2}{3}}=3(7+1)^{\frac{2}{3}}=3(4)=12[/tex]Thus, the equation holds true at x = 7.
If x = -9, then the right-hand side of the equation is:
[tex]3(x+1)^{\frac{2}{3}}=3(-9+1)^{\frac{2}{3}}=3(4)=12[/tex]Thus, the equation holds true at x = -9.
There is no extraneous solution. The solutions of the given equation are x = 7 and x = -9.
6.
Given an equation to solve for x:
[tex]\sqrt[]{3x+2}-2\sqrt[]{x}=0[/tex]The steps of the solution are as follows:
[tex]\begin{gathered} \sqrt[]{3x+2}-2\sqrt[]{x}=0 \\ \sqrt[]{3x+2}=2\sqrt[]{x} \\ (\sqrt[]{3x+2})^2=(2\sqrt[]{x})^2 \\ 3x+2=4x \\ 2=4x-3x \\ 2=x \end{gathered}[/tex]Thus, the solution of the equation is x = 2.
what is the answer for this pls answer
Answer: A
Step-by-step explanation: You merge the equations, -2x and 2x cancel out, 4y + 1y = 5y, and 12 + (-7) = 5
You'll be left with 5y = 5
Dividing both sides by 5 to isolate the y results in y = 1
Graph the parabola.y=1/4x^2-1Plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. Then click on the graph-a-function button
This is basic parabola of the form:
[tex]y=ax^2-b[/tex]So, this one is shifted 1 units down.
The vertex is at (0, -1).
To take 2 points to the left of vertex, we find coordinates for x = -2 and x = -4.
To take 2 points to the right of vertex, we find coordinates for x = 2 and x = 4.
Let's find it:
[tex]\begin{gathered} \text{When x = -2,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(-2)^2^{}-1 \\ y=0 \\ When\text{ x = -4,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(-4)^2-1 \\ y=3 \\ \text{When x = 2,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(2)^2-1 \\ y=0 \\ \text{When x= 4,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(4)^2-1 \\ y=3 \end{gathered}[/tex]So, the 4 coordinates are:
(-2,0), (-4,3), (2,0), (4,3)
The graph:
The mean annual salary at the company where Samuel works is $37,000, with standard deviation $4,000. Samuel's salary is $32,500. Based on the mean and standard deviation, is Samuel's salary abnormal compared to other salaries at this company? When choosing your answer, be careful to select the answer with the correct explanation. A. Samuel's salary falls within the standard deviation, so his salary is not abnormal compared to other salaries at this company. B. Samuel's salary falls outside the standard deviation, so his salary is abnormal compared to other salaries at this company. C. Samuel's salary falls within the standard deviation, so his salary is abnormal compared to other salaries at this company. D. Samuel's salary falls outside the standard deviation, so his salary is not abnormal compared to other salaries at this company?
Answer : Samuel salary falls within the standard deviation and his salary is not abnormal
The mean annual salary at the company where samuel works is $37, 000
The standard deviation is given as $4, 000
Samule's annual salary is $32, 500
Using the Z- score formula
[tex]\begin{gathered} z\text{ = }\frac{x\text{ - }\mu}{\sigma} \\ \text{Where x = sample score} \\ \mu\text{ = mean} \\ \sigma\text{ = standard deviation} \end{gathered}[/tex][tex]\begin{gathered} x\text{ = \$32, 500} \\ \mu\text{ = \$37, 000} \\ \sigma=\text{ \$ 4000} \\ z\text{ = }\frac{32,\text{ 500 - 37000}}{4000} \\ z\text{ = }\frac{-4500}{4000} \\ z\text{ = -1.125} \end{gathered}[/tex]Since, the value of Z- score is -1. 125, then, the salary is 1 standard deviation below the mean.
Therefore, Samuel salary falls within the standard deviation and his salary is not abnormal
Which function has the graph shown?O A. y = secx-1)O B. y = - secxO C. y = csexO D. y = csc(x) +1
We will have that the graph of the function shown belongs to:
[tex]y=csc(x)+1[/tex]This can be seeing as follows:
Diagram 3 shows a piece of rectangularcardboard and an open box that is made from the cardboard.The box is made by cutting out four squares of equal size from the cornersof the cardboard then folding up the sides. Finda) the length in cm of sides of the squares to be cut out in order to get a box with largest volume.b) the minimum number of the boxes needed to fill with 5645 cm³ of pudding
SOLUTION:
Step 1:
In this question, we are given the following:
Diagram 3 shows a piece of rectangular cardboard and an open box that is made from the cardboard.
The box is made by cutting out four squares of equal size from the corners
of the cardboard then folding up the sides.
Find
a) the length in cm of sides of the squares to be cut out in order to get a box with the largest volume.
[tex]\begin{gathered} The\text{ volume of the rectangle would be expressed as:} \\ \text{V = ( 30-2x )(16-2x) ( x)} \\ Multiply\in g\text{ out, we have that:} \\ V=480x-92x^2+4x^3 \\ \text{Differentiating V with respect to x, we have that:} \\ \frac{dV}{dx}=480-184x+12x^2=0 \\ \text{Factorizing the quadratic equation, we have that:} \\ \text{x = 12 or x =}\frac{10}{3} \end{gathered}[/tex][tex]\begin{gathered} \text{Differentiating again, we have that:} \\ \frac{d^2V}{dx^2\text{ }}\text{ = -184 + 24 x} \end{gathered}[/tex]To get the maximum, we need to substitute the values of :
[tex]\begin{gathered} x\text{ = 12, we have that:} \\ \frac{d^2V^{}}{dx^2\text{ }}\text{ = -184 + 24( 12) = }-184\text{ +288 = 104} \\ x=\frac{10}{3},\text{ we have that:} \\ \frac{d^2V}{dx^2}\text{ = -184 + 24 (}\frac{10}{3})\text{ = -184 +}\frac{240}{3}\text{ = - 184 + 80 = -104 }<0 \end{gathered}[/tex]At this stage, we can see that:
[tex]\begin{gathered} x\text{ =}\frac{10\text{ }}{3}cm\text{ is the length of the squares to be cut in order to get a box with } \\ \text{largest volume} \end{gathered}[/tex]b) Find the minimum number of the boxes needed to fill with 5645 cm³ of pudding
[tex]\begin{gathered} \text{From the equation,} \\ V=(30-2\text{x )(16-2x)(x)} \\ \text{put x =}\frac{10}{3}\text{ in the equation, we have that:} \\ V\text{ = \lbrack}30\text{ -2(}\frac{10}{3})\rbrack\text{ \lbrack 16-2(}\frac{10}{3}\rbrack\lbrack\frac{10}{3}\rbrack \\ V\text{ = ( 30 -}\frac{20}{3})\text{ ( 16 - }\frac{20}{3})(\frac{10}{3}) \\ V=725.93cm^3 \\ Now\text{, we asked to find the minimum number of boxes ne}eded^{} \\ to^{} \\ \text{fill with 5645cm}^{3\text{ }}\text{ of pudding.} \\ \text{Then, we ne}ed\text{ to do the following:} \end{gathered}[/tex]
Minimum number of boxes =
[tex]\begin{gathered} \frac{5645}{725.93} \\ =\text{ 7.78} \\ \approx\text{ 8} \end{gathered}[/tex]CONCLUSION:
A minimum of 8 boxes will be needed to fill with 5645 cm³ of pudding
A kitche sa tabletop that is a rectangle 24 in long and 18 in wide.Rita is an interior designer and wants to cover the tabletop in small tiles.She knows the area each bag of tiles covers, but only in square centimeters.(a) Find the area of the tabletop in square centimeters. Do notround intermediate computations and round your finalanswer to two decimal places. Use the table of conversionfacts, as needed.cm(b) The designer wants to cover the tabletop with tiles. Shedoesn't have any to begin with and she can't buy partialbags of tiles. Each bag of tiles covers 260 cm². How manywhole bags of tiles does the designer need to buy tocompletely cover the tabletop?bags(c) If each bag of tiles costs $3.76, how much will she need tospend on tile? Write your answer to the nearest cent.ExplanationCheckConversion facts for length2.54 centimeters (cm)= 30.48 centimeters (cm)≈ 0.91 meters (m)1 inch (in)1 foot (ft)1 yard (yd)1 mile (mi)XNote that means "is approximately equal to".For this problem, treat as if it were = .1.61 kilometers (km)5?I need help with this math problem.
Given: a tabletop that is a rectangle 24 in long and 18 in wide.
Find: (a) the area of the tabletop in square centimeters
(b) The designer wants to cover the tabletop with tiles. She doesn't have any to begin with and she can't buy partial bags of tiles. Each bag of tiles covers 260 cm². number of bags of tiles does the designer need to buy to completely cover the tabletop
(c)If each bag of tiles costs $3.76, how much will she need to spend on tiles.
Explanation: (a)
[tex]1\text{ inch= 2.54cm}[/tex]so the length of the tabletop in cm will be
[tex]24\times2.54=60.96cm[/tex]and the breadth of the tabletop in cm will be
[tex]18\times2.54=45.72cm[/tex]the area of the tabletop will be
[tex]\begin{gathered} l\times b \\ =60.96\times45.72 \\ =2787.09cm^2 \end{gathered}[/tex](b) The designer wants to cover the tabletop with tiles and she can't buy partial bags of tiles. Each bag of tiles covers 260 cm² so the numbe rof bags designer needs to buy to cover the tabletop is
[tex]\begin{gathered} \frac{2787.09}{260} \\ =10.71 \end{gathered}[/tex]it means that designer needs to buy 11 bags of tiles to cover the tabletop.
(c) If each bag of tiles costs $3.76.the the total cost will be equal to
[tex]3.76\times11=41.36\text{ \$}[/tex]which expression below has the same value as 9[tex] {9}^{6} [/tex]
Given data:
The given number is 9^6.
The given number can be written as,
[tex]9^6=9\times9\times9\times9\times9\times9[/tex]Thus, second option is correct.
Solve for x. -2x+5≤10
Answer:
x≥-2.5
Explanation:
Given the inequality:
[tex]-2x+5\le10[/tex]Step 1: Subtract 5 from both sides of the inequality
[tex]\begin{gathered} -2x+5-5\le10-5 \\ -2x\le5 \end{gathered}[/tex]Step 2: Divide both sides by -2.
Note: Since we divide by a negative number, the inequality sign is reversed.
[tex]\begin{gathered} -\frac{2x}{-2}\ge\frac{5}{-2} \\ x\ge-\frac{5}{2} \\ x\ge-2.5 \end{gathered}[/tex]The solution to the inequality is x≥-2.5.
A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 10% of the time if the person does not have the virus. (This 10% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive". DRAW A TREE DIAGRAM IN YOUR NOTES AND USE IT TO HELP YOU SOLVE THIS PROBLEM. Find the probability that a person has the virus given that they have tested positive; i.e. find P(AIB). Give your answer as a decimal number and include at least 3 or more non-zero digits. P(AIB)=
In the tree, the first branch will be person has virus or person doesn't have the virus.
P(virus) = 1/300
P(not virus) = 299/300
Now,
Then we branch out from each option. These branches would be positive or negative.
If they have virus:
P(positive) = 0.8
P(negative) = 0.2
If don't have virus:
P(positive) = 0.1
P(negative) = 0.9
Now, solving the question of probability that a person has the virus given that they have tested positive:
We find:
P(A|B)
P(has virus | positive test) = P(positive and has virus) / P(positive test)
P(positive and has virus) = 4/5 * 1/300 = 4/1500
P(positive test) = 1/300 * 4/5 + 1/10 * 299/300
= (4/1500)+(299/3000)
=(8/3000) + (299/3000) = 307/3000
= 0.10233
So,
P(positive and has virus) / P(positive test) = 4/1500 divided by 299/3000 = 0.02675
A warren of bunnies is growing at 14% per month.a. Find the approximate doubling time.b. If the warren is currently at a population of 400 bunnies, use the approximate doubling time to find the size of the warren in 2 years.c. Find the exact doubling time. Round to 2 decimal places.d. If the warren is currently at a population of 400 bunnies, use the exact doubling time to find the size of the warren in 2 years. e. What is the relative error caused by using the approximate doubling time instead of the exact doubling time? Round to 2 decimal places.
If a warren of bunnies is growing at 14% per month then the approximate doubling time is 5 months.
A warren of bunnies is growing at 14% per month
r = 14% = 0.14
Let the initial population of bunnies be N
Then the final population of bunnies = 2N
N(1 + r)ˣ = 2N
here x is time in months
(1 + 0.14)ˣ = ln 2
x ln 0.14 = ln 2
x = 5.29 months
b) doubling time = 5months
number of cycle = 24/5
population = 400(2)²⁴/⁵
= 11143
c) exact doubling time is 5.29 months
Therefore, if a warren of bunnies is growing at 14% per month then the approximate doubling time is 5 months.
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Suppose sin(A) 2/5 Use the trig identity sin(A) + cos(A) = 1 and the trig identity tan(A)= sin(A)/cos(A) to find can(A) in quadrant I. Round to ten thousandth.
Trigonometric identity is tanθ ≅ 0.4364
[tex]$\sin A=\frac{2}{5}$[/tex]
[tex]$\cos ^2 A=1-\sin ^2 A=\frac{21}{25}$[/tex]
[tex]$\cos A=\frac{\sqrt{21}}{5}$[/tex]
[tex]$\tan A=\frac{\sin A}{\cos A}=\frac{\left(\frac{2}{5}\right)}{\left(\frac{\sqrt{21}}{5}\right)}=\frac{2}{\sqrt{21}} \cong 0.4364$[/tex]
Sine, cosine, tangent, cosecant, secant, and cotangent are the functions. All of these trigonometric ratios are defined using the sides of a right triangle, specifically the adjacent, opposite, and hypotenuse sides.
The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. tan ( − θ ) = − tan θ tan ( − θ ) = − tan θ cot ( − θ ) = − cot θ cot ( − θ ) = − cot θ sin ( − θ ) = − sin θ sin ( − θ ) = − sin θ csc ( − θ ) = − csc θ csc ( − θ ) = − csc θ
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Consider the line y=4x-5.Find the equation of the line that is perpendicular to this line and passes through the point (6. 4).Find the equation of the line that is parallel to this line and passes through the point (6, 4).Equation of perpendicular line: Equation of parallel line:
Solution
gradient = 4
Slope for Perpendicular = -1/4
Slope for Parallel = 4
Equation of perpendicular line:
[tex]\begin{gathered} y-4=-\frac{1}{4}(x-6) \\ \\ 4y-16=-x+6 \\ \\ 4y+x=22 \end{gathered}[/tex]Equation of parallel line:
[tex]\begin{gathered} y-4=4(x-6) \\ \\ y-4=4x-24 \\ \\ y=4x-20 \end{gathered}[/tex]