Step-by-step explanation:
f(x) = 2x^2 -12x + 16 to get to vertex form you will need to complete the square for 'x'....to do THAT you will need the x^2 coefficient to be '1'
Start like this:
2 (x^2 - 6x) + 16 now complete the square
2 (x^2 - 6x +9) - 18 + 16
f(x) = 2 (x-3)^2 - 2 Done.
How many centimeters are in 9 inches?
Answer:
22.86 centimeters
Step-by-step explanation:
How many centimeters are in 9 inches?
1 inch = 2.54 centimeters
9 inches = 2.54 x 9 = 22.86 centimeters
So, there are 22.86 centimeters in 9 inches.
Answer:
22.86 centimeters
Step-by-step explanation:
To convert inches to centimeters, multiply the inches by 2.54:
9·2.54=22.86
So, there are 22.86 centimeters in 9 inches.
Hope this helps :)
elijah and riley are playing a board game elijah choses the dragon for his game piece and rily choses the cat for hers.the dragon is about 1/2 inch tall and the cat is about 7/8 inch tall the model shows how the heights of the game peice are realated.
The cat is 3/8 inches taller than the dragon.
How to solveTo find the difference in height between the cat and the dragon, we need to subtract the height of the dragon from the height of the cat.
The cat is 7/8 inch tall, and the dragon is 1/2 inch tall.
To subtract fractions, we first need a common denominator. The least common denominator (LCD) of 2 and 8 is 8.
So, we'll convert the fractions to equivalent fractions with a denominator of 8.
1/2 = 4/8 (multiply both the numerator and the denominator by 4)
Now, we can subtract the fractions:
(7/8) - (4/8) = (7 - 4)/8 = 3/8
So, the cat is 3/8 inches taller than the dragon.
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Elijah and Riley are playing a board game. Elijah chooses the dragon for his game piece, and Riley chooses the cat for hers. The dragon is about 1 2 inch tall, and the cat is about 7 8 inch tall. How much taller is the cat than the dragon?
Use the diagram to match the terms with the correct example.
1. C is the center
2. EF is a secant line
3. AD is the diameter
4. CD is the radius
5. FD is the arc
6. The region bounded by AC, BC and arc AB is a segment
7. The region bounded by FE and arc FE is a sector
8. EF is the chord
9. GF is the tangent line
How to match the statementTo match the statements, we need to know the following;
The diameter of a circle, is a line cutting through the center and bisects it into equal halveschord of a circle is a line segment that joins any two points on the circumference of the circleSegment of a circle is a region that is bounded by a chordA secant line is a straight line that intersects a circle in two pointsLearn about circles at: https://brainly.com/question/24375372
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You are told that you can expect to see 6 characters during your times there! You really want to fill out your autograph book which holds 48 signatures.
The percentage of your book that can be filled during your reservation at Goofy's kitchen would be 12. 5%
How to find the percentage ?If there are 6 characters and each one signs your book once, then the total number of signatures you would be able to get are 6 signatures in total.
The calculation to determine the percentage of your book occupied involves dividing the number of signatures by its overall capacity and then multiplying that value by 100.
The percentage that would be covered is:
= 6 signatures / 48 total capacity x 100
= 12.5 %
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Your Assignment: Furry Friends
Choosing a Group of Dogs
Josue and Sara both walk dogs during the week. They each walk 10 dogs in the morning and 10 other dogs in the afternoon. Select one of the groups to see how much the dogs in each group weigh. The heavier dogs usually have more energy and want to take longer walks than the smaller dogs.
Josue's dogs:
Morning:
26, 21, 15, 35, 38, 16, 13, 28, 30, 25
Afternoon:
15, 12, 9, 7, 44, 23, 55, 10, 37, 35
Sara's dogs:
Morning:
39, 21, 12, 27, 23, 19, 19, 31, 36, 25
Afternoon:
15, 51, 8, 16, 43, 34, 27, 11, 8, 39
1. Which dog-walker did you select? Circle one.
JosueSara
Comparing the Morning and Afternoon Groups
2. Create frequency tables to represent the morning and afternoon dogs as two sets of data. Group the weights into classes that range 10 pounds. (4 points: 2 points for appropriate intervals, 2 points for correctly portraying data)
3. What is the median of the morning (AM) group? What is the median of the afternoon (PM) group? (2 points: 1 point for each answer)
4. What is the first quartile (Q1) of the morning (AM) group? What is the first quartile (Q1) of the afternoon (PM) group? (2 points: 1 point for each answer)
5. What is the third quartile (Q3) of the morning (AM) group? What is the third quartile (Q3) of the afternoon (PM) group? (2 points: 1 point for each answer)
6. Create a comparative box plot for the morning and afternoon dogs, and label each with its five-number summary. (6 points: 3 points for the correct form of plot, 3 points for appropriate labels)
7. What is the interquartile range (IQR) of the morning (AM) group? What is the interquartile range (IQR) of the afternoon (PM) group? (2 points: 1 point for each answer)
8. The average weights of the dogs are the same for the morning and afternoon groups. But based on your comparative box plot and the IQRs of the two groups, which group of dogs do you think would be easier to walk as one group? Why? (2 points: 1 point for answer, 1 point for justification)
I selected Josue as the dog-walker.
Frequency tables:
Morning dogs:
Weight (lbs) Frequency
10-19 2
20-29 5
30-39 2
40-49 1
Afternoon dogs:
Weight (lbs) Frequency
7-16 4
17-26 2
27-36 1
37-46 1
47-56 2
The median of the morning (AM) group is 26.5 lbs. The median of the afternoon (PM) group is 23 lbs.
The first quartile (Q1) of the morning (AM) group is 16.25 lbs. The first quartile (Q1) of the afternoon (PM) group is 9.5 lbs.
The third quartile (Q3) of the morning (AM) group is 34.75 lbs. The third quartile (Q3) of the afternoon (PM) group is 38.5 lbs.
Comparative box plot:
yaml
Copy code
Morning dogs: Afternoon dogs:
13 | 7 |
| |
16 | 9 |
| |
21 | 11 |
| |
25 | 15 |
| |
26 | 27 |
| |
28 | 34 |
| |
30 | 35 |
| |
35 | 39 |
| |
38 | 43 |
| |
| 44 |
+------------------------------+
1 2 3 4 5 6
Group
Morning dogs:
Min: 13
Q1: 16.25
Median: 26.5
Q3: 34.75
Max: 38
Afternoon dogs:
Min: 7
Q1: 9.5
Median: 23
Q3: 38.5
Max: 44
The interquartile range (IQR) of the morning (AM) group is 18.5 lbs. The IQR of the afternoon (PM) group is 29 lbs.
Based on the comparative box plot and the IQRs, the morning group of dogs would be easier to walk as one group. This is because the morning group has a smaller IQR, indicating that the weights of the dogs are more similar to each other. The afternoon group has a larger IQR, indicating that the weights of the dogs are more spread out, which could make it more difficult to walk them as a group.
1. JosueSara
2. Frequency table
3. Median of the Morning Group: 26.5, Median of the Afternoon Group: 18.5
4. Q1 of the Morning Group: 17.5, Q1 of the Afternoon Group: 10.5
5. Q3 of the Morning Group: 32.5, Q3 of the Afternoon Group: 36.5
6. Comparative Boxplot blue is morning dogs and red is afternoon dogs.
7. IQR of the Morning Group: 15, IQR of the Afternoon Group: 26
8. Based on the comparative box plot and the IQRs, the morning group of dogs would be easier to walk as one group.
What is boxplot?
A box plot, also known as a box-and-whisker plot, is a graphical representation of the distribution of a dataset. It displays summary statistics and provides a visual summary of the data's key characteristics.
1. Which dog-walker did you select?
JosueSara
I selected Sara.
2. Create frequency tables to represent the morning and afternoon dogs as two sets of data. Group the weights into classes that range 10 pounds.
Morning Dogs Frequency Table:
Weight Range Frequency
10-19 2
20-29 4
30-39 4
Afternoon Dogs Frequency Table:
Weight Range Frequency
0-9 1
10-19 3
20-29 2
30-39 2
40-49 1
50-59 1
3. What is the median of the morning (AM) group? What is the median of the afternoon (PM) group?
Median of the Morning Group: 26.5
Median of the Afternoon Group: 18.5
4. What is the first quartile (Q1) of the morning (AM) group? What is the first quartile (Q1) of the afternoon (PM) group?
Q1 of the Morning Group: 17.5
Q1 of the Afternoon Group: 10.5
5. What is the third quartile (Q3) of the morning (AM) group? What is the third quartile (Q3) of the afternoon (PM) group?
Q3 of the Morning Group: 32.5
Q3 of the Afternoon Group: 36.5
6. Create a comparative box plot for the morning and afternoon dogs, and label each with its five-number summary.
Morning Dogs:
Min: 13
Q1: 17.5
Med: 26.5
Q3: 32.5
Max: 38
Afternoon Dogs:
Min: 7
Q1: 10.5
Med: 18.5
Q3: 36.5
Max: 55
7. What is the interquartile range (IQR) of the morning (AM) group? What is the interquartile range (IQR) of the afternoon (PM) group?
IQR of the Morning Group: 15
IQR of the Afternoon Group: 26
8. The average weights of the dogs are the same for the morning and afternoon groups. But based on your comparative box plot and the IQRs of the two groups, which group of dogs do you think would be easier to walk as one group? Why?
Based on the comparative box plot and the IQRs, the morning group of dogs would be easier to walk as one group. This is because the morning group has a smaller interquartile range (IQR) of 15 compared to the afternoon group's IQR of 26. A smaller IQR indicates that the weights of the dogs in the morning group are more clustered together.
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[1 point) The following table gives values of the differentiable function y = f(x). 012345678910 123-4.21-1-2135 Estimate the x-values of critical points of (x) on the interval 0 < x < 10. Classity each critical point as a local maximum, local minimum, or neither Enter your critical points as comma-separated xvalue, classification pairs. For example, if you found the critical points x = -2 and x = 3, and that the first was a local minimum and the second nother a minimum nor a maximum, you should enter (-2,min), (3,neither). Enter none if they are no critica/ points) critical points and classifications Now assume that the table gives values of the continuous function y = f'(x) (instead of F(x)). Estimate and classify critical points of the function f(x) critical points and classifications:
The critical points of f(x) on the interval 0 < x < 10 are: (2, max), (7.5, min)
To estimate the critical points of f(x) on the interval 0 < x < 10, we need to look for points where the derivative, f'(x), equals zero or is undefined. However, we are given a table of values for f(x) instead of f'(x), so we need to first estimate f'(x) using these values.
One way to do this is to use finite differences. We can calculate the first finite difference for each pair of adjacent values in the table, which gives an estimate of the derivative at the midpoint of the interval:
f'(x) ≈ (f(x+1) - f(x)) / (1)
Using this formula, we can calculate the following table of values for f'(x): 0123456789 23-2.79-8-5
Now we can look for critical points of f(x) by finding where f'(x) equals zero or is undefined: - f'(x) = 0 when x = 2 or x = 7.5 (approximately) - f'(x) is undefined at x = 0 and x = 10 (endpoints of the interval)
To classify each critical point, we need to look at the sign of the derivative near the point. If f'(x) changes sign from positive to negative at a critical point, then it is a local maximum. If it changes from negative to positive, then it is a local minimum. If it does not change sign, then it is neither a maximum nor a minimum.
Using the values in the table for f'(x), we can see that: -
Near x = 2, f'(x) changes sign from positive to negative, so it is a local maximum. - Near x = 7.5, f'(x) changes sign from negative to positive, so it is a local minimum. - At the endpoints x = 0 and x = 10, f'(x) is undefined, so there are no critical points.
Therefore, the critical points of f(x) on the interval 0 < x < 10 are: (2, max), (7.5, min)
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A volleyball player’s serving percentage is 75%. Six of her serves are randomly selected. Using the table, what is the probability that at most 4 of them were successes?
A 2-column table with 7 rows. Column 1 is labeled number of serves with entries 0, 1, 2, 3, 4, 5, 6. Column 2 is labeled probability with entries 0. 0002, 0. 004, 0. 033, 0. 132, 0. 297, 0. 356, question mark.
0. 297
0. 466
0. 534
0. 822
To solve this problem, we first need to understand what "at most 4 of them were successes" means. This includes the cases where there are 0, 1, 2, 3, or 4 successful serves out of the 6 selected.
We can use the table to find the probabilities for each of these cases.
For 0 successful serves, the probability is 0.0002.
For 1 successful serve, the probability is 0.004.
For 2 successful serves, the probability is 0.033.
For 3 successful serves, the probability is 0.132.
For 4 successful serves, the probability is 0.297.
To find the probability of at most 4 successful serves, we add up these probabilities:
[tex]0.0002 + 0.004 + 0.033 + 0.132 + 0.297 = 0.466[/tex]
So the probability of at most 4 successful serves is 0.466.
Therefore, the answer is 0.466 and it is found by adding up the probabilities for the cases where there are 0, 1, 2, 3, or 4 successful serves out of the 6 selected from the table.
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You sold a total of 320 student and adult tickets for a total of $1200. Student
tickets cost $3 and adult tickets cost $8. How many adult tickets were sold?
Answer:
48 adult tickets were sold.
Step-by-step explanation:
Let's use algebra to solve this problem:
Let's define:
x: the number of student tickets soldy: the number of adult tickets soldFrom the problem statement, we know:
x + y = 320 (the total number of tickets sold is 320)3x + 8y = 1200 (the total revenue from ticket sales is $1200)We can use the first equation to solve for x in terms of y:
x = 320 - y
Substituting this expression for x into the second equation, we get:
3(320 - y) + 8y = 1200
Expanding the left side, we get:
960 - 3y + 8y = 1200
Simplifying, we get:
5y = 240
Solving for y, we get:
y = 48
Therefore, 48 adult tickets were sold.
Additional:
To find the number of student tickets sold, we can substitute y=48 into the first equation:
x + 48 = 320
x = 272
Therefore, 272 student tickets were sold.
Dagne measures and finds that she can do a vertical jump that is 27. 5% of her height. If Dagne is 48 inches tall, how high can she jump? Enter your answer in the box
If Dagne is 48 inches tall, then she can jump 13.2 inches high.
Given that Dagne can do a vertical jump that is 27.5% of her height, and she is 48 inches tall, we can find how high she can jump by multiplying her height by the percentage of her vertical jump as follows:
Jump height = 27.5% of height
Jump height = (27.5/100) x 48 inches
Jump height = 0.275 x 48 inches
Jump height = 13.2 inches
Therefore, Dagne can jump 13.2 inches high.
To explain this, we can say that the problem gives us information about the proportion of Dagne's height that she can jump vertically. The percentage of her height that she can jump is given as 27.5%, which we can convert to a decimal form (0.275) for calculation purposes. Multiplying this decimal by Dagne's height of 48 inches gives us the height that she can jump, which is 13.2 inches. So, Dagne can jump 13.2 inches high based on the given information.
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What are the coordinates of the point 1/4 of the way from A to B? Two intersecting line segments graphed on a coordinate plane. Segment A B has vertices at A negative 4 comma negative 2 and B 4 comma 4. Segment C D has vertices at C negative 3 comma 3 and D 3 comma negative 3.
The coordinates of the point 1 / 4 of the way from A to B is (-2, -0.5).
How to find the coordinates ?To find the coordinates of the point that is 1/4 of the way from point A to point B, we can use the following formula:
Point P = (1 - t) x A + t x B
Given the coordinates of points A (-4, -2) and B (4, 4):
P x = (1 - 1/4) x Ax + (1/4) x Bx
P x = (3/4) x (-4) + (1/4) x 4
P x = -3 + 1
P x = -2
P y = (1 - 1/4) x Ay + (1/4) x By
P y = (3/4) x (-2) + (1/4) x 4
P y = -1.5 + 1
P y = -0.5
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The circumstances if the base of the cone is 12π cm. If the volume of the cone is 96π, what is the height
24 cm is the height of cone .
What is known as a cone?
A cone is a three-dimensional geometric object with a smooth transition from a flat, generally circular base to the apex, also known as the vertex.
A cone is a three-dimensional geometric structure with a smooth transition from a flat base—often but not always circular—to the point at the top, also known as the apex or vertex. Cone. a right circular cone having the following measurements: height, slant height, angle, base radius, and height.
V=1/3hπr²
V = 1/3 * h * 12π
96π = 1/3 * h * 12π
96π * 3/12π = h
8 * 3 = h
h = 24 cm
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Help with geometry on equations of circles. Point C is a point of tangency. How would I solve this to get DA and DE?
Answer:
DA = 17DE = 9Step-by-step explanation:
You want the segment lengths DA and DE of the hypotenuse in the triangle shown in the figure.
Right triangleThe radius to a point of tangency always makes a right angle with the tangent. This is a right triangle with legs 8 and 15, so you know from your knowledge of Pythagorean triples that the hypotenuse is 17.
DA = 17
DE = 17 -8 = 9
__
Additional comment
In case you haven't memorized a few of the useful Pythagorean triples, {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, you can always figure the missing side length of a right triangle using the Pythagorean theorem.
It tells you the sum of the squares of the legs is the square of the hypotenuse:
AC² +CD² = DA²
8² +15² = DA²
64 +225 = 289 = DA²
DA = √289 = 17
Of course, AE is the radius of the circe, 8, so ...
AE + DE = DA
8 +DE = 17
DE = 17 -8 = 9
Alternatively, you can solve this using the relation between tangents and secants. If the line DA is extended across the circle to intersect it again at X, then ...
DC² = DE·DX
15² = DE·DX = DE(DE +16) . . . . . . . EX is the diameter, twice the radius of 8
DE² +16DE -225 = 0
(DE +25)(DE -9) = 0 . . . . factor
DE = 9 . . . . the positive solution
DA = 9 +8 = 17
We like the Pythagorean theorem solution better, as the factors of the quadratic may not be obvious.
What is the percentage chance of choosing a queen or a king from a standard 52-card deck?
The probability of choosing a king or queen from a standard 52-card deck is 2/13 or approximately 15.4%.
How can we calculate the percentage?The probability of choosing a king or a queen from a standard 52-card deck can be calculated by first determining the number of kings and queens in the deck. There are four kings (hearts, diamonds, clubs, and spades) and four queens (hearts, diamonds, clubs, and spades) in the deck, for a total of eight cards.
To find the probability of choosing a king or a queen, you need to divide the number of desired outcomes (eight) by the total number of possible outcomes (52).
Probability = Number of desired outcomes / Total number of possible outcomes
Probability = 8 / 52
This simplifies to:
Probability = 2 / 13
Therefore, the probability of choosing a king or a queen from a standard 52-card deck is approximately 0.154 or 15.4%.
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please help me I don't understand how to do these.
Given: SD⊥HT ; SH≅ST
Prove: SHD=STD
Step-by-step explanation:
For this given problem we have the following statements and reasons:
1) SD is vertical to HT,
Reason: Given
2) SDH and SDT are right angles
Reason: Right angle congruence theorem
3) SH=≈ST
Reason: Given
4) BD=≈BD
Reason: Reflexive property
5) SHD=≈STD
Reason: SAS
Dante has a tent shaped like a triangular prism. The tent has equilateral triangle bases that measure 5 feet on each side. The tent is 8 feet long and 4. 3 feet tall
The tent has a volume of 86 cubic feet.
How we get the volume of tent?The tent owned by Dante is in the shape of a triangular prism, which means it has two identical equilateral triangle bases that measure 5 feet each. The tent's length is 8 feet, and its height is 4.3 feet.
To calculate the tent's volume, we can use the formula for the volume of a triangular prism, which is [tex]V = (1/2) * b * h * l[/tex], where b is the base, h is the height, and l is the length of the prism.
Plugging in the given values, we get [tex]V = (1/2) * 5 * 4.3 * 8[/tex] = 86 cubic feet. The volume of a tent is an important consideration when deciding which one to purchase or use for a particular activity, as it determines how much space is available inside for people and belongings.
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the area of a rectangle is 65 sqare meters. the lenght of the rectrangle is 3 m less thans twice the width. find the dimensions of the rectangle
The dimensions are;
Length = 7 meters
Width = 5 meters
How to determine the valueThe area of a rectangle is expressed as;
Area = length × width
From the information given, we have that;
Length = 2w - 3
Area = 65
Substitute the values
65 = (2w - 3)w
expand the bracket
65 = 2w² - 3w
solve the quadratic equation;
2w² + 13w - 10w - 65
Factorize the terms
w(2w + 13) - 5(2w + 13)
w = 5
Substitute the value
Length = 2w - 3 = 2(5) - 3 = 7
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let g be a function such that g(9)=0 and g'(9)=2 let h be the function h(x)=square root of x
evaluate d/dx[g(x)*h(x)] at x=9
Work Shown:
First we'll need the derivative of h(x)
[tex]h(\text{x}) = \sqrt{\text{x}}\\\\h(\text{x}) = \text{x}^{1/2}\\\\h'(\text{x}) = (1/2)\text{x}^{-1/2}\\\\h'(\text{x}) = \frac{1}{2\text{x}^{1/2}}\\\\h'(\text{x}) = \frac{1}{2\sqrt{\text{x}}}\\\\[/tex]
Then let f(x) = g(x)*h(x)
Use the product rule to evaluate f ' (9).
[tex]f(\text{x}) = g(\text{x})*h(\text{x})\\\\f'(\text{x}) = \frac{d}{d\text{x}}\left[g(\text{x})*h(\text{x})\right]\\\\f'(\text{x}) = g'(\text{x})*h(\text{x}) + g(\text{x})*h'(\text{x})\\\\f'(\text{x}) = g'(\text{x})*\sqrt{\text{x}} + g(\text{x})*\frac{1}{2\sqrt{\text{x}}}\\\\f'(9) = g'(9)*\sqrt{9} + g(9)*\frac{1}{2\sqrt{9}}\\\\f'(9) = 2*\sqrt{9} + 0*\frac{1}{2\sqrt{9}}\\\\f'(9) = 2*3 + 0\\\\f'(9) = 6\\\\[/tex]
The points $(1, 7), (13, 16)$ and $(5, k)$, where $k$ is an integer, are vertices of a non-degenerate triangle. what is the sum of the values of $k$ for which the area of the triangle is a minimum
The minimum value of k that satisfies this inequality is k = 9.
To find the value of k for which the area of the triangle is a minimum, we'll use the following terms: vertices, non-degenerate triangle, and area of a triangle. Here's the step-by-step explanation:
1. The vertices of the triangle are $(1, 7), (13, 16),$ and $(5, k)$.
2. A non-degenerate triangle means it has a positive area.
3. The area of a triangle can be calculated using the formula: Area = (1/2) * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Now, let's find the area of the triangle with the given vertices:
Area = (1/2) * |1(16 - k) + 13(k - 7) + 5(7 - 16)|
We want to minimize the area, so let's simplify the expression:
Area = (1/2) * |-9 + 13k - 104|
Since we want a non-degenerate triangle, the area must be greater than 0. Therefore, the expression inside the absolute value must be positive:
-9 + 13k - 104 > 0
13k > 113
k > 113/13
k > 8.69
Since k is an integer, the minimum value of k that satisfies this inequality is k = 9.
The sum of the values of k for which the area of the triangle is a minimum is just the single value we found, which is 9.
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Please help me :/
You can make a 6-digit security number using the digits 1-9 and digits cannot be repeated. Show all work and formulas used in computing your answers.
a) How many numbers can you make if there are no additional restrictions?
b) How many numbers can you make if the first digit cannot be a one?
c) How many odd numbers can you make (the last digit is odd?)
d) How many numbers greater than 300,000 can you make?
e) How many numbers greater than 750,000 can you make?
Sure, I'd be happy to help you with these questions!
a) To calculate the total number of possible 6-digit security numbers, we can use the permutation formula:
nPr = n! / (n-r)!
where n is the total number of digits available (9) and r is the number of digits we are selecting (6).
So, the number of possible 6-digit security numbers without any restrictions is:
9P6 = 9! / (9-6)! = 9! / 3! = 9 x 8 x 7 x 6 x 5 x 4 = 60,480
Therefore, there are 60,480 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits.
b) If the first digit cannot be a one, we are left with 8 choices for the first digit (since we cannot use 1) and 8 choices for the second digit (since we have already used one digit). For the remaining 4 digits, we still have 7 choices for each digit, since we cannot repeat any digits.
Using the permutation formula again, the number of possible 6-digit security numbers with the first digit not being one is:
8 x 8 x 7 x 7 x 7 x 7 = 1,322,496
Therefore, there are 1,322,496 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the first digit is not one.
c) To create an odd number, the last digit must be an odd number, which means we have 5 choices for the last digit (1, 3, 5, 7, or 9). For the first digit, we cannot use 0 or 1, so we have 7 choices. For the remaining 4 digits, we still have 8 choices for each digit (since we can use any digit).
Using the permutation formula again, the number of possible 6-digit security numbers with the last digit being odd is:
7 x 8 x 8 x 8 x 8 x 5 = 7,1680
Therefore, there are 7,1680 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the last digit is odd.
d) To create a number greater than 300,000, the first digit must be 3, 4, 5, 6, 7, 8, or 9. If the first digit is 3, we have 7 choices for the first digit (3, 4, 5, 6, 7, 8, or 9). For the remaining 5 digits, we still have 8 choices for each digit.
If the first digit is not 3, we have 6 choices for the first digit (since we cannot use 1 or 2). For the remaining 5 digits, we still have 8 choices for each digit.
Using the permutation formula again, the number of possible 6-digit security numbers greater than 300,000 is:
7 x 8 x 8 x 8 x 8 x 8 + 6 x 8 x 8 x 8 x 8 x 8 = 2,526,720
Therefore, there are 2,526,720 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the number is greater than 300,000.
e) To create a number greater than 750,000, the first digit must be 8 or 9. If the first digit is 8, we have 2 choices for the first digit (8 or 9). For the remaining 5 digits, we still have 8 choices for each digit.
If the first digit is 9, we only have one choice for the first digit (9). For the remaining 5 digits, we still have 8 choices for each digit.
Using the permutation formula again, the number of possible 6-digit security numbers greater than 750,000 is:
2 x 8 x 8 x 8 x 8 x 8 + 1 x 8 x 8 x 8 x 8 x 8 = 262,144
Therefore, there are 262,144 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the number is greater than 750,000.
A study was conducted by the Director of Parking and Transportation at a large university. One variable under study is the method of transportation to campus used by off-campus students. The choices for students on the survey were drive alone, carpool, public transportation, walk, or a paid driving service. Which type of variable is this?
The type of variable being studied in this case is a categorical variable, specifically a nominal variable.
What s the type of variable in the question?Categorical variables are variables that can be classified or grouped. The categories in this example are the various modes of transportation utilized by off-campus students (drive alone, carpool, public transportation, walk, or a paid driving service).
Nominal variables are categorical variables with no intrinsic order or numerical value. Because the various modes of transportation in this study have no numerical value or inherent order, they are classified as nominal variables.
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16. [0/1 Points] DETAILS PREVIOUS ANSWERS TANAPCALCBR10 6.6.050. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Turbo-Charged Engine Versus Standard Engine In tests conducted by Auto Test Magazine on two identical models of the Phoenix Elite-one equipped with a standard engine and the other with a turbo-charger-it was found that the acceleration of the former is given by a = f(t) = 5 + 0.8t (Osts 12) ft/sec/sec, t sec after starting from rest at full throttle, whereas the acceleration of the latter is given by a = g(t) = 5 + 1.2t + 0.03t2 (0 sts 12) = ft/sec/sec. How much faster is the turbo-charged model moving than the model with the standard engine at the end of a 11-sec test run at full throttle? 41.25 X ft/sec Need Help? Read It Submit Answer
The turbocharged model is moving 41.25 ft/sec faster than the model with the standard engine at the end of the 11-second test run.
We need to find how much faster the turbo-charged model is moving than the model with the standard engine at the end of an 11-second test run at full throttle.
To find the final velocity of each model at the end of 11 seconds, we need to integrate their respective acceleration functions with respect to time from 0 to 11 seconds:
For the standard engine model:
v(t) = ∫(5 + 0.8t) dt = 5t + [tex]0.4t^2[/tex]
v(11) = 5(11) +[tex]0.4(11)^2[/tex] = 72.4 ft/sec
For the turbo-charged model:
v(t) = ∫(5 + 1.2t + 0.03[tex]t^2[/tex]) dt = 5t +[tex]0.6t^2 + 0.01t^3[/tex]
v(11) = 5(11) + [tex]0.6(11)^2 + 0.01(11)^3[/tex]= 113.65 ft/sec
The difference in final velocity between the two models is:
[tex]v_{turbo} - v_{standard[/tex] = 113.65 - 72.4 = 41.25 ft/sec
Therefore, the turbo-charged model is moving 41.25 ft/sec faster than the model with the standard engine at the end of the 11-second test run.
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What is the volume of a hemisphere with a radius of 8. 8 cm, rounded to the nearest
tenth of a cubic centimeter?
Please help
The volume of a hemisphere with a radius of 8. 8 cm, rounded to the nearest tenth of a cubic centimeter, is approximately 1436.8 cubic centimeters.
To find the volume of a hemisphere with a radius of 8.8 cm, you can use the formula:
Volume = (2/3)πr³
where r is the radius of the hemisphere. Plugging in the given radius:
Volume = (2/3)π(8.8)³ ≈ 1436.8 cubic centimeters
So, the volume of the hemisphere is approximately 1436.8 cubic centimeters, rounded to the nearest tenth of a cubic centimeter.
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Anita has saved $43. 75 of the $112. 50 that she needs for a new snowboard.
She saves $13. 75 from her paper route each week. The equation
13. 75w + 43. 75 = 112. 50 can be used to represent the number of weeks
it will take her to reach her goal. In how many more weeks will Anita have
saved enough money for the snowboard?
Anita will take 5 more weeks to save enough money for the snowboard. when she saved $43. 75 of the $112. 50 that she needs for a new snowboard.
Given data :
Total money needs for a new snowboard = $112. 50
Anita saved money = $43. 75
Money saved each week = $13. 75
From the given data we can write the equation to find the number of weeks as,
13. 75w + 43. 75 = 112. 50
By solving the equation 13.75w + 43.75 = 112.50 we can find out how many more weeks it will take Anita to save enough money for the snowboard by using the elimination method.
Subtracting 43.75 from both sides of the equation, we get:
13.75w + 43.75 - 43.75 = 112.50 - 43.75
13.75w = 68.75
w = 68.75 / 13.75
w = 5
Therefore, Anita will take 5 more weeks to save enough money for the snowboard.
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A landscaping company placed two orders with a nursery. The first order was for 13 bushes and 4 trees, and totaled $327. 50. The second order was for 6 bushes and 2 trees, and totaled $142. 96. Sam tried to use system of equation to solve the problem. If "b" represents bushes and "t" represents trees which system can Sam use?
System of equation Sam can use is 13b + 4t = 327.50 and 6b + 2t = 142.47 where "b" represents bushes and "t" represents trees.
Let the cost of bushes represented by b
cost of trees represented by t
First order is 13 buses and 4 trees and total is $327.50
By using the data equation form will be
13b + 4t = 327.50
Second order is 6 bushes and 2 trees and total is $142. 96
By using data the equation form will be
6b + 2t = 142.96
These set of equation will for and can be used.
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A garden designer designed a square decorative pool. the pool is surrounded by a walkway. on two opposite sides of the pool, the walkway is 8 feet. on the other two opposite sides, the walkway is 10 feet. the final design of the pool and walkway covers a total area of 1,440 square feet. the side length of the square pool is x.
The expression that represents the side length of the square pool is 2x² + 36x - 1120 = 0. The side length of the square pool is x = 16.31 feet.
Let's denote the side length of the square pool as x.
The walkway on two opposite sides of the pool is 8 feet, which means that the overall length of the pool and walkway on those sides is x + 8 + 8 = x + 16.
Similarly, x + 10 + 10 = x + 20.
The total area covered by the pool and walkway is given as 1,440 square feet.
Total Area = Pool Area + Walkway Area
The area of the square pool is x², and the area of the walkway is the difference between the total area and the pool area:
Walkway Area = Total Area - Pool Area
Substituting the values, we have:
1440 = x² + (x + 16)(x + 20)
1440 = x² + (x² + 36x + 320)
Combining like terms:
1440 = 2x² + 36x + 320
2x² + 36x + 320 - 1440 = 0
2x² + 36x - 1120 = 0
After solving, we get:
[tex]x=-9+\sqrt{641},\:x=-9-\sqrt{641}[/tex]
Take positive value:
[tex]x=-9+\sqrt{641},\\x = -9+25.31\\x = 16.31[/tex]
So, the side length of the square pool is 16.31 feet.
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From the following facts, complete a depreciation schedule by using
the straight-line method:
Cost of Honda Account Hybrid - $40000
Residual Value - $10000
Estimated Life - 6 years
Using the straight-line method, we can find the depreciation expense per year by dividing the depreciable value (cost - residual value) by the estimated life:
Depreciable Value = Cost - Residual Value
Depreciable Value = $40000 - $10000
Depreciable Value = $30000
Annual Depreciation Expense = Depreciable Value / Estimated Life
Annual Depreciation Expense = $30000 / 6
Annual Depreciation Expense = $5000
To create a depreciation schedule, we can subtract the annual depreciation expense from the cost each year until we reach the residual value:
| Year | Cost | Depreciation | Accumulated Depreciation | Book Value |
|------|---------------|----------------- |----------------------------------------|------------|
| 1 | $40000 | $5000 | $5000 | $35000 |
| 2 | $35000 | $5000 | $10000 | $30000 |
| 3 | $30000 | $5000 | $15000 | $25000 |
| 4 | $25000 | $5000 | $20000 | $20000 |
| 5 | $20000 | $5000 | $25000 | $15000 |
| 6 | $15000 | $5000 | $30000 | $10000 |
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11. Consider the image shown. What is the measure of angle DEB?
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The supplementary angles of the given angles are 154 degrees, 136 degrees, 135 degrees, 44 degrees, 45 degrees, and 26 degrees, respectively.
Supplementary Angles of Given Angles
Supplementary angles are pairs of angles that add up to 180 degrees. To find the supplementary angle of each given angle, we simply subtract the angle from 180 degrees.
Therefore, the supplementary angles of the given angles are:
The supplementary angle of 26 degrees is 154 degrees (180 - 26 = 154).
The supplementary angle of 44 degrees is 136 degrees (180 - 44 = 136).
The supplementary angle of 45 degrees is 135 degrees (180 - 45 = 135).
The supplementary angle of 136 degrees is 44 degrees (180 - 136 = 44).
The supplementary angle of 135 degrees is 45 degrees (180 - 135 = 45).
The supplementary angle of 154 degrees is 26 degrees (180 - 154 = 26).
Therefore, the supplementary angles of the given angles are 154 degrees, 136 degrees, 135 degrees, 44 degrees, 45 degrees, and 26 degrees, respectively.
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A portion of a truss vehicle bridge has steel
beams that form an isosceles triangle with the
dimensions shown. A pedestrian handrail is
attached to the side of the bridge so that it is
parallel to the road.
A
24 feet
28.5 feet
30 feet
What is the height, x, of the handrail above th
road to the nearest tenth of a foot?
Answer:
3.5 feet
Step-by-step explanation:
You want the difference in height between similar isosceles triangles, one with height of 30 ft and a base of 24 ft, the other with a side length of 28.5 ft.
RelationsWe can find the side length of the larger triangle using the Pythagorean theorem. It will be ...
longer side = √(30² +12²) ≈ 32.311 ft
Similar trianglesThen the length x is the difference between the altitudes of the triangles. The altitudes are proportional to the side lengths, so we have ...
(30 -x)/28.5 = 30/32.311
x = 30-(28.5)(30/32.311) = 30(1 -28.5/32.311) ≈ 3.538 ≈ 3.5 . . . . feet
The hand rail is about 3.5 feet above the bridge deck.
TrigonometryWe recognize that the distance from the hand rail to the top of the triangle is the product of the given side length (28.5 ft) and the cosine of the angle between the side and the altitude.
The tangent of that angle is the ratio of its opposite side (12 ft) to its adjacent side (30 ft), or θ = arctan(12/30).
The value of x is the difference of the altitudes of the triangles, so is ...
x = 30 -28.5·cos(arctan(12/30)) ≈ 3.5 ft
We find this easier to compute.
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1425 stamps evenly into 7 piles how many would be in each pile
To evenly distribute 1425 stamps into 7 piles, you would divide 1425 by 7. This gives you a quotient of 203 with a remainder of 4. This means that each pile would contain 203 stamps and there would be 4 stamps left over.
To understand this better, you can visualize the process of dividing 1425 stamps into 7 equal piles. You could start by putting 203 stamps into the first pile. Then, you would add another 203 stamps to the second pile. You would continue this process until you had 7 piles, each containing 203 stamps. However, you would be left with 4 stamps that couldn't be evenly distributed.
This type of division is called integer division because it results in a whole number quotient and potentially a remainder. In this case, the quotient represents the number of stamps that can be evenly distributed among the piles, and the remainder represents the leftover stamps that cannot be evenly distributed.
Overall, to divide 1425 stamps into 7 piles, each pile would contain 203 stamps, with 4 stamps remaining.
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The chamber of commerce for a beach town asked a random sample of city dwellers, "Would you like to live at the beach?" Based on this survey, the 95% confidence interval for the population proportion of city dwellers who would like to live at the beach is (0. 56, 0. 62)
The 95% confidence interval is (0.56, 0.62). This means that we can be 95% confident that the true proportion of city dwellers who would like to live at the beach lies between 56% and 62%.
In this case, the Chamber of Commerce conducted a survey asking city dwellers if they would like to live at the beach. The 95% confidence interval for the population proportion of city dwellers who would like to live at the beach is (0.56, 0.62).
To break this down:
1. Random sample: The Chamber of Commerce surveyed a group of city dwellers chosen randomly, which helps ensure that the results are representative of the entire population of city dwellers.
2. Population proportion: This refers to the percentage of all city dwellers who would like to live at the beach.
3. 95% confidence interval: This means that if the survey were repeated many times with different random samples, 95% of the intervals calculated would contain the true population proportion.
In this case, the 95% confidence interval is (0.56, 0.62). This means that we can be 95% confident that the true proportion of city dwellers who would like to live at the beach lies between 56% and 62%.
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