Answer: The odds against an event occurring are the ratio of the probability that the event will not occur to the probability that the event will occur. To find the odds against an event, you can use the formula:
Odds against = (1 - probability of event occurring) / probability of event occurring
In this case, the event is that exactly 5 cars will pass over the bridge in a 4-minute period. The probability that this event will occur is 0.16. Therefore, the odds against this event occurring are:
Odds against = (1 - 0.16) / 0.16
= 0.84 / 0.16
= 5.25
So, the odds against exactly 5 cars passing over the bridge in a 4-minute period are 5.25 to 1.
It means that there are 5.25 times more likely that it will not happen compare to happen.
Step-by-step explanation:
How do you solve 5t 35?
The answer to this question will t = 8. The first step to solving this equation is to combine the like terms by adding the two numbers together.
In this case, 5t + 35 = 40t. The next step is to divide both sides of the equation by the coefficient of the variable, which in this case is 5. When you do this, you get 40t/5 = t. Therefore, the answer to 5t + 35 is t = 8.
A coefficient is an integer that is either multiplied by the variable it is associated with or written alongside the variable. A coefficient is, in other words, the numerical factor of a term that contains both constants and variables. For instance, the coefficient in the expression 2x is 2.
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4x + 2y = 8 and 16x – y = 14 what does x and y equal?
Answer:
here is ur answer
Step-by-step explanation:
4x +2y = 4
4x = 4 -2y
x = 4/4 -2y
x= -2 y
multiple choice for final pls help me
Answer:
I hope this helps!
Step-by-step explanation:
ACB
What is the solution of 2x^2 5x 3 0?
The Given quadratic equation has two solutions: 1 and 3/2. A quadratic problem was solved by utilizing the factoring method.
A quadratic equation is defined.
At least one squared term must be present because a quadratic is a second-degree polynomial equation. It is also known as quadratic equations. ax^2 + b*x + c = 0 is the quadratic equation's general form.
where x is the unknown variable and a, b, and c are constant terms.
In this case, I'm solving a quadratic problem by factoring.
The following quadratic equation is
2x^2 - 5x + 3 = 0.
2x^2 - (3+2)x + 3 = 0
2x^2 -3x -2x + 3 = 0.
x(2x-3) -1(2x -3) = 0.
(2x-3)(x-1) = 0.
2x - 3 =0 or x -1 =0
x = 3/2 or x =1
So the solutions of a quadratic equation are 3/2 and 1.
Given Question is incomplete complete question here:
What are the solutions of the quadratic equation 2x^2-5x+3 = 0?
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a male animal can weigh up to 8,000 pounds. How many tons is 8,000 pounds
Answer:
4
Step-by-step explanation:
Evie read an article that said 6\%6%6, percent of teenagers were vegetarians, but she thinks it's higher for students at her large school. To test her theory, Evie took a random sample of 252525 students at her school, and 20\%20%20, percent of them were vegetarians. To see how likely a sample like this was to happen by random chance alone, Evie performed a simulation. She simulated 404040 samples of n=25n=25n, equals, 25 students from a large population where 6\%6%6, percent of the students were vegetarian. She recorded the proportion of vegetarians in each sample. Here are the sample proportions from her 404040 samples: A dot plot for simulated sample proportions has a scale from 0.00 to 0.25 in increments of 0.01. The distribution is right skewed with dots plotted as follows. 0.00, 8. 0.004, 17. 0.08, 5. 0.12, 5. 0.16, 2. 0.20, 2. 0.24, 1. Evie wants to test H_0: p=6\%H 0 :p=6%H, start subscript, 0, end subscript, colon, p, equals, 6, percent vs. H_\text{a}: p>6\%H a :p>6%H, start subscript, start text, a, end text, end subscript, colon, p, is greater than, 6, percent where ppp is the true proportion of students who are vegetarian at her school. Based on these simulated results, what is the approximate ppp-value of the test? Note: The sample result was \hat p=20\% p ^ =20%p, with, hat, on top, equals, 20, percent. Choose 1 answer: Choose 1 answer: (Choice A) A p\text{-value}\approx0.01p-value≈0.01p, start text, negative, v, a, l, u, e, end text, approximately equals, 0, point, 01 (Choice B) B p\text{-value}\approx0.025p-value≈0.025p, start text, negative, v, a, l, u, e, end text, approximately equals, 0, point, 025 (Choice C) C p\text{-value}\approx0.03p-value≈0.03p, start text, negative, v, a, l, u, e, end text, approximately equals, 0, point, 03 (Choice D) D p\text{-value}\approx0.075p-value≈0.075
Answer:
p-value~ 0.075
Step-by-step explanation:
( Choice D on Khan Academy)
The closest answer choice is (Choice A) A p-value ≈ 0.01, is the best approximation among the answer choices given.
What is p-value?In a statistical hypothesis test, the p-value is a probability value that measures the strength of evidence against the null hypothesis.
It is the likelihood of observing a test statistic that is as extreme as or more extreme than the observed test statistic, assuming the null hypothesis is true.
To find the approximate p-value of the test, we need to calculate the proportion of simulated sample proportions that are greater than or equal to the sample proportion of 20%.
From the dot plot, we can see that only a few simulated sample proportions are greater than or equal to 20%, which suggests that the p-value is small.
Counting the dots from the dot plot, we can see that there are a total of 8 + 17 + 5 + 5 + 2 + 2 + 1 = 40 simulated sample proportions that are greater than or equal to 20%.
Therefore, the proportion of simulated sample proportions that are greater than or equal to 20% is 40/40 = 1.
Since the p-value is the probability of observing a sample proportion as extreme or more extreme than the observed sample proportion, given that the null hypothesis is true, we can approximate the p-value as the proportion of simulated sample proportions that are greater than or equal to the observed sample proportion.
Therefore, the approximate p-value is 1, which is very small and suggests strong evidence against the null hypothesis.
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The perimeter of triangle ABC is 26 units. The altitudes of triangle ABC are in the ratio 2: 3: 4. Compute the area of triangle ABC.
please help me w step by step explanation and work tysm !
Answer:
59
Step-by-step explanation:
I need a verbal description for the equation using the graph texts. Thank you!
Answer:
The y intercept is at (0,12) and the line has a positive slope of rising up 1.5 squares and moving horizontly 1 square
Step-by-step explanation:
What is the measure of angle P?
Enter your answer as a decimal in the box. Round only your final answer to the nearest hundredth.
m∠P= °
The measurement of angle P in the diagram given in the question is 61.93 °
Trigonometric ratiosThe trigonometric ratios which has been proven as regarding right angle triangles are given below:
Sine θ = Opposite / Hypothenus
Cos θ = Adjacent / Hypothenus
Tan θ = Opposite / Adjacent
With above information in mind, we can easily determine the value of angle P in the question. This is illustrated below:
How to determine the value of angle PFrom the question given above, the following data were obtained:
Hypothenus = 17 cmOpposite = 15 cmAdjacent = 8 cmAngle P =?Sine θ = Opposite / Hypothenus
Sine P = 15 / 17
Sine P = 0.8824
Take the inverse of sine
P = Sine⁻¹ 0.8824
P = 61.93 °
Thus, the value of angle P in the diagram is 61.93 °
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20
Mrs. Trailer has 1283.4 milliliters to divide
equally into 31 beakers. How many liters
will be in each beaker?
Answer:
about 5 milliliters
Step-by-step explanation:
Find a basis for the eigenspace corresponding to each listed eigenvalue of A below.
A = 4 0 -1 14 5 -10 2 0 1 λ=5,2,3
A basis for the eigenspace corresponding to λ = 5 is { }. (Use a comma to separate answers as needed.) A basis for the eigenspace corresponding to λ = 2 is { }. (Use a comma to separate answers as needed.) A basis for the eigenspace corresponding to λ = 3 is . { }. (Use a comma to separate answers as needed.)
The basis for the eigenspace corresponding to lambda=5,1,4 are None,[tex]\left[\begin{array}{c}-1 \\\frac{1}{2} \\0\end{array}\right][/tex] and [tex]$\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$[/tex]
[tex]$$A=\left[\begin{array}{ccc}5 & -12 & 10 \\0 & 7 & -3 \\0 & 6 & -2\end{array}\right]$$[/tex]
Eigenspace corresponding to lambda=5,1,4
The eigenspace E_lambda corresponding to the eigenvalue lambda is the null space of the matrix a [tex]\mathrm{A}-(\lambda) \mathrm{I}"[/tex]
for lambda=5
[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-5 \mathrm{I})$$[/tex]
Reducing the matrix A-5I by elementary row operations
[tex]$$\begin{aligned}A-5 I & =\left[\begin{array}{ccc}5-5 & -12 & 10 \\0 & 7-5 & -3 \\0 & 6 & -2-5\end{array}\right] \\& =\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 2 & -3 \\0 & 6 & -7\end{array}\right] \\& \sim\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_2 \rightarrow \frac{R_2}{2} \\& \sim\left[\begin{array}{ccc}1 & 0 & -8 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_1 \rightarrow R_1+2 R_2\end{aligned}$$[/tex]
[tex]\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 2\end{array}\right] R_3 \rightarrow R_3-6 R_2$$\\\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] R_3 \rightarrow \frac{\mathrm{R}_3}{2}$$\\\sim\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+8 \mathrm{R}_3$[/tex]
[tex]$\sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] R_2 \rightarrow R_2+\frac{2 R_3}{2}$[/tex]
The solutions x of A-5I=0 satisfy x_1=x_2=x_3=0 that is, the null space solves the matrix
[tex]$$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]
Hence The null space is [tex]\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] E_5[/tex] has no basis
[tex]$$\begin{aligned}& \text { case: } 2 \\& \text { for } \lambda=1 \\& \mathrm{E}_5=\mathrm{N}(\mathrm{A}-(1) \mathrm{I})\end{aligned}$$[/tex]
we reduce the matrix A-I by elementary row operations as follows.
[tex]$$\begin{aligned}A-1 & =\left[\begin{array}{ccc}5-1 & -12 & 10 \\0 & 7-1 & -3 \\0 & 6 & -2-1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 6 & -3 \\0 & 6 & -3\end{array}\right] R_1 \rightarrow \frac{R_1}{4} \\& \sim\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 1 & -\frac{1}{2} \\0 & 6 & -3\end{array}\right] R_2 \rightarrow \frac{R_2}{6}\end{aligned}[/tex]
[tex]$$$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 6 & -3\end{array}\right] R_1 \rightarrow R_1+3 R_2$\\$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right] R_3 \rightarrow R_3-6 R_2$[/tex]
Thus, the solutions x of (A-I) X=0 satisfy
[tex]$\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$[/tex]
x_3=t
[tex]$\Rightarrow \mathrm{x}_1=-\mathrm{t}, \mathrm{x}_2=\frac{\mathrm{t}}{2}$[/tex]
[tex]$\vec{x}=\left[\begin{array}{c}-t \\ \frac{t}{2} \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right] t$[/tex]
The Basis for the nullspace A-I will be: [tex]$\left.\left(\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right]\right)$[/tex]
case:3
lambda=4
[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-(4) \mathrm{I})$$[/tex]
we reduce the matrix A-4I by elementary row operations as follows.
[tex]$\begin{aligned} A-4 \mid & =\left[\begin{array}{ccc}5-4 & -12 & 10 \\ 0 & 7-4 & -3 \\ 0 & 6 & -2-4\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 3 & -3 \\ 0 & 6 & -6\end{array}\right] \\ & \sim\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] R_2 \rightarrow \frac{R_2}{3}\end{aligned}$[/tex]
[tex]$\begin{aligned} & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+12 \mathrm{R}_2 \\ & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2\end{aligned}$[/tex]
Thus, the solutions x of (A-4IX)=0 satisfy
[tex]$$\left[\begin{array}{ccc}1 & 0 & -2 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]
x_3=t
[tex]$\Rightarrow \mathrm{x}_1=2 \mathrm{t}, \mathrm{x}_2=\mathrm{t}$[/tex]
[tex]$$\vec{x}=\left[\begin{array}{c}2 t \\t \\t\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1\end{array}\right] t$$[/tex]
The Basis for the nullspace A-4 I will be [tex]\left(\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right)[/tex]
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Composite Figures Question 9.
A. 57 ft
B. 288 ft
C. 208 ft
D. 264 ft
Answer:
D
Step-by-step explanation:
find area of first rectangle
18 x 6 = 108
find area of second rectangle
20 x 5 = 100
find area of two triangles
7 x 8 =56
56/2 = 28
each triangle has an area of 28
28+28+100+108 = 264ft
A jetski races down the river at a constant speed for t seconds, covering m metres. How long, in
minutes, will it take for the jetski to cover n metres at the same speed?
It will take a time of n/60m minutes for the jetski to cover n metres at the same speed.
Define the term linear speed?The measurement of a moving object's actual distance travelled is called linear speed. Linear speed is the rate of motion of an object along a straight line.Thus, the angular speed times the r-dimensional distance equals the linear speed of such a point located on the object. The measuring unit is metres per second and metres per second. = the rate of change in radians per second.Speed = distance/ time
Speed = m/s
For the distance of n meters.
distance = speed * time
n = m/s*time
n = m /s* time
Time = n(m/sec)
Time = n/m * sec
Time = n/60m min
Thus, it will take n/60m minutes for the jetski to cover n metres at the same speed.
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the difference of a number x and 7.
Answer: difference means subtract
x-7
HOPE THIS HELPS
The pitchers for the home team had 12 strikeouts for 32 batters, while the pitchers for the visiting team had 15 strikeouts for 35 batters. Which pitching team had a greater fraction of strikeouts.(Hint: Just make the 12 for 32 and 15 for 35 into fractions and see which is greater or less than.)
Answer:
2958843/5 5/1000775747373838388484737372737375823Estimate the difference by rounding each number to the nearest whole number and then subtracting.
10.04 − 6.3
Answer:
Estimate of difference is 4.
Step-by-step explanation:
10.04 = 10 to nearest whole number
6.3 = 6 to nearest whole number.
10 - 6 = 4.
What Is 8 1/8 - 4 3/8
Answer:
3 3/4 or 3 6/8
Step-by-step explanation:
hope this helps <3
Given the matrices, calculate AB
Answer:
C11=-2
C22=8
hope this will help you more
good luck
WILL MARK BRRRRRRRRRRRRRRRRRRAIIIIIINNNNNNNNNNNNNLLLLLLLLLLIIIIIIIIIIIEEEEEEEEESSSSSSSSSSTTTTTTTTTTT
The volume of the cube, given the lengths of the small cubes used to make the larger cube, is 8 cm ³
The surface area of the cube is 24 cm ²
The total length of the edges of the cube is 24 cm
How to find the volume of a cube ?The volume of a cube can be found by the formula :
= Side x Side x Side
This is because all the sides of a cube at the same and so the volume would measure them three - dimensionally.
The side length of this cube is the length of two of the smaller cubes as these make up a side of the cube :
= 2 x 1
= 2 cm
The volume is therefore :
= 2 x 2 x 2
= 8 cm ³
How to find the area of a cube ?The surface area of a cube is:
= 6 x side ²
= 6 x 2 ²
= 24 cm ²
How to find the length of a cube ?The total length of the edges would be:
= 12 edges on a cube x length of each edge
= 12 x 2
= 24 cm
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Calculate the bearing of A from B.
N
B
73°
N
A
The direction of A with respect to B. N B 73° N A is 287
Explain about the direction?A direction is the way that something is moving, directing, or moving toward something or someone. The directions could be Up, Down, Left, and Right, or North, South, East, and West.
The ability to specify where one thing is in relation to another is known as position in mathematics. Defining direction entails explaining the direction in which something moves, such as forward, backward, or in a full or half turn.
It is possible to define direction using relative terms like up, down, in, out, left, right, forward, backward, or sideways. A direction can also be denoted by one of the four cardinal directions: north, south, east, or west.
Angles at a point add up to 360 degrees.
⇒ Bearing from A to B is = 360° - 73° = 287°.
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!!!!!!NEEED HELP ASAP DUE SOON!!!!!!!!
someone help me plsss
x2 - 3x - 5 = 0 has a unique solution of type -11, which is what the equation entails.
what is discriminant ?The "discriminant" is the value that aids in identifying the many types of solutions when using the quadratic formula to solve a problem. It is a unique function of a co-efficient in a mathematical equation (a quadratic equation), and the value it yields provides details about the roots of the equation. Calculating the discriminant involves square rooting the "b" term and subtracting four times the "a" term from the "c" term. Alongside "D," there is a (delta) that represents discrimination. A quadratic equation with two distinct real number solutions is said to have a positive discriminant. A quadratic equation with a discriminant of zero has repeated real number solutions. Both of the solutions are not real numbers, as indicated by a negative discriminant.
given
x2 - 3x - 5 = 0
D=b2−4ac
D= −32−4×1×5
D=92−20
D= −11
x2 - 3x - 5 = 0 has a unique solution of type -11, which is what the equation entails.
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On a bookshelf, there are 8 fiction and 7 nonfiction books. Paul randomly selects one, puts it back, and then randomly selects another. What is the probability that both selections were fiction books?
Answer:
28%
Step-by-step explanation:
The distance between two points A and B on latitude 60°S along their parallel of latitude is 2816km. (I) calculate the angular difference in their longitudes (ii) if A is due west of B and on longitude 20°W, find the longitude of B
Answer:
i) The angular difference in their longitude is approximately 50.65°
ii) The line of longitude where B is located is approximately 30.65°E
Step-by-step explanation:
The given parameters are;
The angle of latitude where the points A and B are located = 60°S
The distance between points A and B = 2,816 km
(i) The angular difference in their longitude, θ, is given as follows;
θ = 2,816/(2 × π × 6,371 × cos(60°)) × 360 ≈ 50.65°
The angular difference in their longitude, θ ≈ 50.65°
(ii) The location of the point A relative to B = Due west
The line of longitude on which A is located = 20°W
∴ The line of longitude where B is located, P = θ - 20°
∴ P = 50.65° - 20°W = 30.65°E
The line of longitude where B is located ≈ 30.65°E.
find the area of the deck around this pool
15 m
25 m
50 m
90 m
Answer:
A = 1800m^2
Step-by-step explanation:
(90)(25) - (30)(15)
2250 - 450 = 1800
If m^2 = 3 then what is the value of 5m^6
a. 15
b. 30
c. 45
d. 135
Answer:
135
Step-by-step explanation:
m^2 = 3
Cube m^2
(m^2)^3 = 3^3
m^(2*3) = 27
m^6 = 27
Now multiply by 5
5 * m^6 = 5*27
5 m^6 =135
The point P has coordinates (8, 9). In which quadrant does point P lie?
first quadrant
fourth quadrant
second quadrant
third quadrant
Answer:
The coordinates (8,9) lie in quadrant 1
Step-by-step explanation:
The coordinates on quadrant ___ go like this:
Quadrant l : (+,+)
Quadrant ll : (-,+)
Quadrant lll : (-,-)
Quadrant llll : (+,-)
Is the following relation a function?
Answer:
yes
Step-by-step explanation:
A 12 pack of soda cost $4.80. What is the unit cost?
Answer:The cost of each can is $ 0.31
.
Step-by-step explanation: You need to take the amount of money each pack costs and divide it by the number of cans to find the unit rate. Hope this helps!!
brianlist please ?
To find the unit cost, simply divide the amount of money by the amount.
4.80÷12= 0.4 dollars
So each soda is only 0.4 dollars.
How can you check this?
0.4 dollars times 12 should equal the total cost of soda.
0.4x12=4.80, so it's correct.
Find the equation of the normal to the circle, whose equation is given below, at the point (1,2).( x − 3 ) 2 + ( y + 2 ) 2 = 20
Answer:
The equation of the normal is y = 4 - 2·x
Step-by-step explanation:
The given equation of the circle, is presented as follows;
(x - 3)² + (y + 2)² = 20
The point of the normal of the circle = (1, 2)
The equation of the normal to a circle, x² + y² + D·x + E·y + F = 0 at a point P(x₁, y₁) is given as follows;
[tex]\dfrac{y - y_1}{x - x_1} = \dfrac{2 \cdot y_1 + D}{2 \cdot x_1 + E}[/tex]
Expanding the given equation of the circle, gives;
(x - 3)² + (y + 2)² = x² + y² - 6·x + 4·y + 13 = 20
∴ x² + y² - 6·x + 4·y + 13 - 20 = x² + y² - 6·x + 4·y - 7 = 0
x² + y² - 6·x + 4·y - 7 = 0
∴ x₁ = 1, y₁ = 2, D = -6, E = 4, and F = 13
Which gives;
[tex]\dfrac{y - 2}{x - 1} = \dfrac{2 \times 2 + 4}{2 \times 1 + (-6)} = \dfrac{8}{-4} = -2[/tex]
∴ y - 2 = -2 × (x - 1) = 2 - 2·x
y = 2 - 2·x + 2 = 4 - 2·x
The equation of the normal to the circle with equation (x - 3)² + (y + 2)² = 20, at the point (1, 2) is y = 4 - 2·x