Answer:
0.6547 or 65.47%
Step-by-step explanation:
One minute equals 1/60 of an hour, the mean number of occurrences in that interval is:
[tex]\lambda =\frac{400}{60}=6.6667[/tex]
The poisson distribution is described by the following equation:
[tex]P(x) =\frac{\lambda^{x}*e^{-\lambda}}{x!}[/tex]
The probability that more than 5 vehicles will arrive is:
[tex]P(x>5)= 1-(P(0)+P(1)+P(2)+P(3)+P(4)+P(5))\\P(x>5) = 1-(\frac{6.667^{0}*e^{-6.667}}{1}+\frac{6.667^{1}*e^{-6.667}}{1}+\frac{6.667^{2}*e^{-6.667}}{2}+\frac{6.667^{3}*e^{-6.667}}{3*2}+\frac{6.667^{4}*e^{-6.667}}{4*3*2}+\frac{6.667^{5}*e^{-6.667}}{5*4*3*2})\\P(x>5)=1-(0.00127+0.00848+0.02827+ 0.06283+0.10473+0.13965)\\P(x>5)=0.6547[/tex]
The probability that more than five vehicles will arrive in a one-minute interval is 0.6547 or 65.47%.
The probability that more than five vehicles will arrive in a one-minute interval is approximately 0.6582.
Step 1
Given that vehicles arrive at an intersection at a rate of 400 vehicles per hour, and this follows a Poisson distribution, we want to find the probability that more than five vehicles will arrive in a one-minute interval.
First, convert the arrival rate to a one-minute interval. Since there are 60 minutes in an hour, the arrival rate per minute is:
[tex]\[ \lambda = \frac{400 \, \text{veh/h}}{60} = \frac{400}{60} \approx 6.67 \, \text{veh/min} \][/tex]
The Poisson distribution formula for the probability of observing k events in an interval is:
[tex]\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \][/tex]
We need the probability that more than five vehicles arrive in one minute:
[tex]\[ P(X > 5) = 1 - P(X \leq 5) \][/tex]
Step 2
First, calculate [tex]\( P(X \leq 5) \)[/tex] by summing the probabilities for k = 0 to k = 5 :
[tex]\[ P(X \leq 5) = \sum_{k=0}^{5} \frac{e^{-6.67} 6.67^k}{k!} \][/tex]
Let's calculate these probabilities:
- For k = 0 :
[tex]\[ P(X = 0) = \frac{e^{-6.67} 6.67^0}{0!} = e^{-6.67} \][/tex]
- For k = 1 :
[tex]\[ P(X = 1) = \frac{e^{-6.67} 6.67^1}{1!} = e^{-6.67} \times 6.67 \][/tex]
- For k = 2 :
[tex]\[ P(X = 2) = \frac{e^{-6.67} 6.67^2}{2!} = e^{-6.67} \times \frac{6.67^2}{2} \][/tex]
- For k = 3:
[tex]\[ P(X = 3) = \frac{e^{-6.67} 6.67^3}{3!} = e^{-6.67} \times \frac{6.67^3}{6} \][/tex]
- For k = 4 :
[tex]\[ P(X = 4) = \frac{e^{-6.67} 6.67^4}{4!} = e^{-6.67} \times \frac{6.67^4}{24} \][/tex]
- For k = 5 :
[tex]\[ P(X = 5) = \frac{e^{-6.67} 6.67^5}{5!} = e^{-6.67} \times \frac{6.67^5}{120} \][/tex]
Sum these probabilities to find [tex]\( P(X \leq 5) \)[/tex].
Step 3
Next, we calculate [tex]\( e^{-6.67} \)[/tex] and the terms:
[tex]\[e^{-6.67} \approx 0.00126\][/tex]
[tex]\[P(X = 0) \approx 0.00126\][/tex]
[tex]\[P(X = 1) \approx 0.00126 \times 6.67 = 0.0084\][/tex]
[tex]\[P(X = 2) \approx 0.00126 \times \frac{6.67^2}{2} = 0.0280\][/tex]
[tex]\[P(X = 3) \approx 0.00126 \times \frac{6.67^3}{6} = 0.0622\][/tex]
[tex]\[P(X = 4) \approx 0.00126 \times \frac{6.67^4}{24} = 0.1037\][/tex]
[tex]\[P(X = 5) \approx 0.00126 \times \frac{6.67^5}{120} = 0.1382\][/tex]
Sum these probabilities:
[tex]\[P(X \leq 5) \approx 0.00126 + 0.0084 + 0.0280 + 0.0622 + 0.1037 + 0.1382 = 0.34176\][/tex]
Finally, the probability that more than five vehicles will arrive in a one-minute interval is:
[tex]\[P(X > 5) = 1 - P(X \leq 5) = 1 - 0.34176 = 0.65824\][/tex]
The probability that more than five vehicles will arrive in a one-minute interval is approximately 0.6582.
If 2 is a factor of n and 3 is a factor of n, then
6 is a factor of n. 2 is not a factor of n or 3 is not a factor of n or 6 is a factor of n.
Answer:
6 is a factor of n
Step-by-step explanation:
2 is a factor of n and 3 is a factor of n means
n = 2×3×k
= 6×k
then n = 6×k
then 6 is a factor of n
Final answer:
If both 2 and 3 are factors of a number n, then 6 must also be a factor of n, because the product of unique prime factors is always a factor of that number.
Explanation:
The product of unique prime factors of a number will be a factor of that number. Since 2 and 3 are prime factors and both are factors of n, their product (which is 6) must also be a factor of n.
For example, consider the number 12. 12 is divisible by 2 and 12 is divisible by 3, and indeed, 12 is divisible by 6 as well. This holds true for any number n that has 2 and 3 as factors. Thus, we can conclude that 6 is a factor of n if both 2 and 3 are factors of n.
Consider a square whose size varies. Let s s represent the side length of the square (in cm) and let P P represent the perimeter of the square (in cm).
Write a formula that expresses P in terms of s.
Answer:
P = 4s
Step-by-step explanation:
The perimeter of a geometric shape is simply the sum of all its sides length. Since the shape in question is a square, which means that all of the four sides have the same length 's', the perimeter can be expressed by:
[tex]P = s+s+s+s\\P=4s[/tex]
For any value of 's', the formula above expresses the perimeter 'P' as a function of 's'
Let x and y be any numbers at all with x ≤ y. Show that the number of integers between x and y is [y] - [x] +1.That is show that the number of integers between x and y is = (the floor of y) - (the ceiling of x) +1
Answer:
the explanation is given below.
Step-by-step explanation:
Here what is applied is assumption of range of values of number from say 1 - 100In total, i stopped at 100 on the dot.from this, the lowest number is 1 and the highest number is 100
hence the range of the numbers = Difference between Highest and Lowestrange = 100 - 1 = 99, the 99 gotten as the range is indicative that a number has been missing.In order to make up the 100, an integer is added to the difference = 99, i.e 99 is added to 1 to make up the 100.Furthermore, if 0 is exclusively out when numbers are counted up 100 with 0 inclusive, in such case, the first and last number are excluded from the counting. as such the integers will be {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20.........., 99} since both 0 and 100 are not included.Here, if we try to get the range = highest - lowest = 99 - 1 = 98, it implies that to make up the 99, an integer is added to the result of the difference = 98+1 = 99As such, the number of integers between two numbers is the difference between the highest and the lowest number plus 1 i.e highest - lowest + 1 = y - x +1 = (the floor of y) - ( the ceiling of x) + 1Fireworks on July4th.A local news outlet reported that 56% of 600 randomly sampled Kansasresidents planned to set off fireworks on July 4th. Determine the margin of error for the 56% point estimateusing a 95% confidence level.1
Answer:
The margin of error is 3.97 percentage points.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 600, p = 0.56[/tex]
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 - 1.96\sqrt{\frac{0.56*0.44}{600}} = 0.5203[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 + 1.96\sqrt{\frac{0.56*0.44}{600}} = 0.5997[/tex]
The margin of error is the upper limit subtracted by the proportion, or the proportion subtracted by the lower limit. They are the same values.
So the margin of error is 0.5997 - 0.56 = 0.56 - 0.5203 = 0.0397 = 3.97 percentage points.
A large university will begin a 13-day period during which students may register for that semester’s courses. Of those 13 days, the number of elapsed days before a randomly selected student registers has a continuous distribution with density function f (t) that is symmetric about t = 6.5 and proportional to 1/(t + 1) between days 0 and 6.5.A student registers at the 60th percentile of this distribution.Calculate the number of elapsed days in the registration period for this student.(A) 4.01
(B) 7.80
(C) 8.99
(D) 10.22
(E) 10.51
Answer:
8.99 days elapsed. Option (C) is correct
Step-by-step explanation:
The distribution has density function k/t+1 for a constant k and t between 0 and 6.5 . Since the distribution is symmetrical in 6.5, the area it forms between 0 and 6.5 should be 1/2, thus
[tex]\frac{1}{2} = \int\limits_0^{6.5} \frac{k}{t+1} \, dt = k *(ln(t+1) \, |_0^{6.5}) = k * (ln(7.5)-ln(1)) = k*ln(7.5)[/tex]
Hence k = 1/(2ln(7.5)), approx 1/4.
We need to find the percentil 0.6, since the integral of the random variable is 1/2 over the first half, we need to find t such that the integral of the random variable between o and 6.5 + t is 0.6. This is equivalent to find t such that the integral between 6.5 and 6.5+t is 0.1. Due to the over 6.5, this t should satisfy that the integral between 6.5-t and 6.5 is also 0.1. Lets compute the integral and find t
[tex]\int\limits^{6.5}_{6.5-t} {\frac{k}{t+1}} \, dx = \frac{1}{2ln(7.5)}*(ln(t+1) \, |_{6.5-t}^{6.5} \, ) = \frac{1}{2ln(7.5)} * (ln(7.5)-ln(7.5-t)) = \\\frac{1}{2} - \frac{ln(7.5-t)}{2ln(7.5)} = 0.1[/tex]
Therefore,
[tex]\frac{ln(7.5-t)}{2ln(7.5)} = 0.4\\\\ln(7.5-t) = 0.8*ln(7.5)\\\\7.5-t = e^{0.8*ln(7.5)}\\\\t = 7.5-e^{0.8*ln(7.5)} = 2.49[/tex]
As a result, the student sould have registered 2.49 days after the day 6.5, thus it should have registeredd at day 8.99. Option (C) is correct.
find the laplace transformation of g(t) = 5te^-5t Us (t) use laplace transforms theorms g
Answer: The required laplace transform of g(t) is [tex]\dfrac{5}{(s+5)^2}.[/tex]
Step-by-step explanation: We are given to find the laplace transform of the following function :
[tex]g(t)=5te^{-5t}.[/tex]
We know the following formulas for laplace transform :
[tex](i)~L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}},\\\\(ii)~L\{cf(t)\}=cL\{f(t)\}.[/tex]
In the given function function, we have
c = 5, n = 1 and a = -5.
Therefore, we get
[tex]L\{g(t)\}\\\\=L\{5te^{-5t}\}\\\\=5L\{te^{-5t}\}\\\\\\=5\times\dfrac{1!}{(s-(-5))^{1+1}}\\\\\\=\dfrac{5}{(s+5)^2}.[/tex]
Thus, the required laplace transform of g is [tex]\dfrac{5}{(s+5)^2}.[/tex]
The Laplace transform of the given function is [tex]\frac{5} { (s + 5)^2}[/tex].
The Laplace transform of a function g(t) is defined as:
[tex]L{(g(t))} = \int\limits^{\infty}_0 e^-^s^tg(t) dt[/tex]
We need to find the Laplace transform of [tex]g(t) = 5te^-^5^t[/tex]. To do this, we use the shifting theorem and known transforms.
First, recall the Laplace transform of [tex]nte^-^a^t[/tex] is:
[tex]L(nte^-^a^t)} = \frac{n! } {(s + a)^n^+^1}[/tex]
For our function [tex]g(t)[/tex] :
a = 5
n = 1
Applying the formula:
[tex]L{(5te^-^5^t)} = \frac{(5 * 1! )}{(s + 5)^2}[/tex]
After simplifying, we get:
[tex]L\leftparanthesis(\ 5te^-^5^t)\rightparanthesis\ = \frac{5 }{ (s + 5)^2}[/tex]
An experiment consists of four outcomes with P(E1) = 0.2, P(E2) = 0.3, and P(E3) = 0.4. The probability of outcome E4 is__________.
Answer: 0.1
Step-by-step explanation:
WE know that the total probability in an experiments = 1.
i.e. Sum of the probabilities of occurring each event is 1.
i.e. If there are n outcomes in any experiment., then the total probability will be:
[tex]P(E_1)+P(E_2)+P(E_3)+...........+P(E_n)=1[/tex]
Given : An experiment consists of four outcomes ,with P(E1) = 0.2, P(E2) = 0.3, and P(E3) = 0.4.
Then , [tex]P(E_1)+P(E_2)+P(E_3)+P(E_4)=1[/tex]
Substitute corresponding values , we get
[tex]0.2+0.3+0.4+P(E_4)=1[/tex]
[tex]0.9+P(E_4)=1[/tex]
[tex]P(E_4)=1-0.9=0.1[/tex]
Hence , the probability of outcome [tex]E_4[/tex] is 0.1.
The probability of event E4 in the given experiment is 0.1, because the sum of probabilities of all outcomes should equal 1.
Explanation:The problem falls under the subject of
Probability Theory
within Mathematics. In probability, the total probability of all possible outcomes is always 1. So, for the given problem where you have four events E1, E2, E3, E4, the total probability P(E1)+P(E2)+P(E3)+P(E4) should equal 1. Given that P(E1) = 0.2, P(E2) = 0.3 and P(E3) = 0.4, we can find P(E4) using the equation
1 - P(E1) - P(E2) - P(E3) = P(E4)
. By substituting the given values into this equation, we find that P(E4) = 1 - 0.2 - 0.3 - 0.4 =
0.1
. Therefore, the probability of outcome E4 is 0.1.
Learn more about Probability here:
https://brainly.com/question/22962752
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Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you to clear the entries below each pivot or of the operations that allow you to swap two rows or to scale a row.
1) 3x₂ - 5x₃ = 89
6x₁ + x₃ = 17
x₁ - x₂ + 8x₃ = -107
2) 4x₁ - x₂ + 3x₃ = 12
2x₁ + 9x₃ = -5
x₁ + 4x₂ + 6x₃ = -32
Answer:
1) The solution of the system is
[tex]\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right[/tex]
2) The solution of the system is
[tex]\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right[/tex]
Step-by-step explanation:
1) To solve the system of equations
[tex]\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right[/tex]
using the row reduction method you must:
Step 1: Write the augmented matrix of the system
[tex]\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]
Step 2: Swap rows 1 and 2
[tex]\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]
Step 3: [tex]\left(R_1=\frac{R_1}{6}\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]
Step 4: [tex]\left(R_3=R_3-R_1\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right][/tex]
Step 5: [tex]\left(R_2=\frac{R_2}{3}\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right][/tex]
Step 6: [tex]\left(R_3=R_3+R_2\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right][/tex]
Step 7: [tex]\left(R_3=\left(\frac{6}{37}\right)R_3\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]
Step 8: [tex]\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]
Step 9: [tex]\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]
Step 10: Rewrite the system using the row reduced matrix:
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right[/tex]
2) To solve the system of equations
[tex]\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right[/tex]
using the row reduction method you must:
Step 1:
[tex]\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]
Step 2: [tex]\left(R_1=\frac{R_1}{4}\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]
Step 3: [tex]\left(R_2=R_2-\left(2\right)R_1\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]
Step 4: [tex]\left(R_3=R_3-R_1\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]
Step 5: [tex]\left(R_2=\left(2\right)R_2\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]
Step 6: [tex]\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)[/tex]
[tex]\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]
Step 7: [tex]\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)[/tex]
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right][/tex]
Step 8: [tex]\left(R_3=\left(- \frac{2}{117}\right)R_3\right)[/tex]
[tex]\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]
Step 9: [tex]\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)[/tex]
[tex]\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]
Step 10: [tex]\left(R_2=R_2-\left(15\right)R_3\right)[/tex]
[tex]\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]
Step 11:
[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right[/tex]
1. Explain or show how you could find 5/ 1/3
by using the value of 5x3
Find 12/ 3/5
Answer:
20
Step-by-step explanation:
You could find 5/⅓
by using 5 × 3
Knowing that:
i. Any number multiplied by 1, gives the number itself.
ii. Dividing any number by itself gives 1.
You would agree with me that
i. (5×3)/(5×3) = 1
ii. Writing 5/⅓ as 5/⅓ × 1 doesn't change the value.
Then I can write 5/⅓ as
5/⅓ × (5×3)/(5×3) = 1
This can become
[5×(5×3)] / [(⅓) × (5×3)]
= 75/(15/3)
= 75/5
= 15
In a similar way,
12/ 3/5
= [12/ (3/5)] × [(5×3)/(5×3)]
= 12×(5×3) / (3/5)×(5×3)
= (12×5×3) / [(3×5×3)/5]
= 180 / (45/5)
= 180 / 9
= 20
Using the laws of logic to prove tautologies.Use the laws of propositional logic to prove that each statement is a tautology.a. ¬r ∨ (¬r → p)b. ¬(p → q) → ¬q
Answer: The proofs are given below.
Step-by-step explanation: We are given to prove that the following statements are tautologies using truth table :
(a) ¬r ∨ (¬r → p) b. ¬(p → q) → ¬q
We know that a statement is a TAUTOLOGY is its value is always TRUE.
(a) The truth table is as follows :
r p ¬r ¬r→p ¬r ∨ (¬r → p)
T T F T T
T F F T T
F T T T T
F F T F T
So, the statement (a) is a tautology.
(b) The truth table is as follows :
p q ¬q p→q ¬(p→q) ¬(p→q)→q
T T F T F T
T F T F T T
F T F T F T
F F T T F T
So, the statement (B) is a tautology.
Hence proved.
Final answer:
To prove that a statement is a tautology using propositional logic, we need to show that the statement is true under all possible truth values of its variables. By applying the laws of implication, disjunction, and contradiction, we can prove that the given statements are tautologies.
Explanation:
To prove that a statement is a tautology using the laws of propositional logic, we need to show that the statement is true under all possible truth value assignments of its variables. Let's consider each statement:
a. ¬r ∨ (¬r → p)
We can use the law of implication, which states that ¬p ∨ q is equivalent to p → q, to rewrite the statement as ¬r ∨ (r → p). By applying the law of disjunction, which states that p ∨ (q ∧ r) is equivalent to (p ∨ q) ∧ (p ∨ r), we can further rewrite the statement as (¬r ∧ r) ∨ (r ∨ p). Using the law of contradiction, which states that p ∧ ¬p is always false, we can simplify the statement to r ∨ p, which is a tautology.
b. ¬(p → q) → ¬q
We can use the law of implication to rewrite the statement as ¬(¬p ∨ q) → ¬q. By applying De Morgan's law, which states that ¬(p ∨ q) is equivalent to ¬p ∧ ¬q, we can simplify the statement to (p ∧ ¬q) → ¬q. Using the law of contradiction, we know that p ∧ ¬p is always false, so the statement simplifies to false → ¬q, which is always true. Therefore, it is a tautology.
Mariel thinks the tens digit goes up by 1 in these numbers. Do you agree? Explain. 864,865,866,867,868,869
Answer:
No, because it is not the tens digit that goes up by 1 in these numbers, it is the unit digit.
Step-by-step explanation:
It is important to know the concepts of units, tenths and cents.
For example
1 = 1 unit
10 = 1*10 + 0 = The tens digit is one the unit digit is 0
21 = 2*10 + 1 = The tens digit is two and the unit digit is 1.
120 = 1*100 + 2*10 + 0 = The cents digit is 1, the tens digit is two and the unit digit is 0.
So
Adding 1 is the same as the unit digit going up by 1.
Adding 10 is the same as the tens digit going up by 1.
Adding 100 is the same as the cents digit going up by 1.
In this problem, we have that:
864,865,866,867,868,869
Each value is the 1 added to the previous value, that is, the unit digit goes up by 1.
Mariel thinks the tens digit goes up by 1 in these numbers. Do you agree?
No, because it is not the tens digit that goes up by 1 in these numbers, it is the unit digit.
Answer:
disagree, its the unit number that goes up by 1
Step-by-step explanation:
evaluate the expression 6÷3+17=
Answer:
=19
Step-by-step explanation:
=6÷3+17
=2+17
=19
HELP PLEASE!
When two functions have an inverse relationship, the line of reflection will be which of the following equations?
Select the correct response(s):
x = 0
y = x
y = 0
y = -x
Thank you!
Answer:
Option C: The line y = x
Step-by-step explanation:
Let us take an ordered pair (x, y) of a function. Then the ordered pair of its inverse function would be (y, x).
That is to say, when we reflect a point (x, y) across the line y = x we get the point (y, x).
Note that since, this function is invertible, it is both 'one - one' and 'onto'.
The line of reflection for two functions with an inverse relationship is the line y = x, as this line shows the symmetry of the original function and its inverse on a graph.
Explanation:When two functions have an inverse relationship, the line of reflection across which their graphs are mirrored is the line y = x. This is because for an inverse relationship, each x value in the first function corresponds to a y value in the second function, and vice versa. Thus, when graphing the original function and its inverse, you will notice that they are symmetric with respect to the line y = x.
A random sample of 11 days were selected from last year's records maintained by the maternity ward in a local hospital, and the number of babies born each day of the days is given below: 3 7710 0 712 530 Find the five number summary (minimum, first quartile, second quartile, third quartile, maximum) of the data.
Answer: ( Min = 0 , [tex]Q_1=1[/tex] , [tex]Q_2=3[/tex] , [tex]Q_3=7[/tex] , Max = 10 )
Step-by-step explanation:
Given : A random sample of 11 days were selected from last year's records maintained by the maternity ward in a local hospital, and the number of babies born each day of the days is given below:
3 7 7 10 0 7 1 2 5 3 0
We first arrange them in increasing order , we get
0 0 1 2 3 3 5 7 7 7 10
Here , N= 11
Now , we can see that
Minimum value = 0
Maximum value = 10
First quartile [tex]Q_1[/tex]= [tex](\dfrac{N+1}{4})^{th}\ term=(\dfrac{12}{4})^{th}\ term = 3^{rd} term =1[/tex]
Second quartile [tex]Q_2[/tex]= Median = Middlemost number = 3
Third quartile [tex]Q_3[/tex] = [tex](\dfrac{3(N+1)}{4})^{th}\ term=(\dfrac{36}{4})^{th}\ term[/tex]
[tex]= 9^{th} term =7[/tex]
∴ The required five number summary : ( Min = 0 , [tex]Q_1=1[/tex] , [tex]Q_2=3[/tex] , [tex]Q_3=7[/tex] , Max = 10 )
Prove that among 502 positive integers, there are always two integers so that either their sum or their difference is divisible by 1000.
Final answer:
Using the pigeonhole principle, we can prove that among 502 positive integers, at least two will have the same remainder when divided by 1000, implying their difference is divisible by 1000.
Explanation:
The question asks to prove that among 502 positive integers, there are always two integers so that either their sum or their difference is divisible by 1000. This statement can be understood through the pigeonhole principle, which in basic terms means if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon.
In this case, consider the remainders when these integers are divided by 1000. Since there are only 1000 possible remainders (from 0 to 999), and we have 502 numbers, at least two of them must have the same remainder when divided by 1000, according to the pigeonhole principle.
Let these two numbers be a and b, where without loss of generality, a ≥ b. If a and b have the same remainder when divided by 1000, then a - b is divisible by 1000. Alternatively, if we had a case where the sum is considered, assuming complementary pairs mod 1000, a similar argument involving the pigeonhole principle can conclude that there must be at least one pair whose sum or difference gives a number divisible by 1000, satisfying the initial claim.
A(n) _________ is a person or object that is a member of the population being studied.
Answer: individual
Step-by-step explanation:
An individual is a person or object that is a member of the population being studied. A population is defined as a group of individuals with a common characteristic living and interbreeding within a given area, in statistics, population is a collection of individuals to be studied. Individuals can also be referred to as the objects/person described by a set of data. For example: when studying the height of students in a school, the students attending that school are individuals.
Use the given information to find the length of a circular arc. Round to two decimal places.the arc of a circle of radius 11 inches subtended by the central angle of pie/4
Answer is in inches(in)
Answer:
The length of the circular arc is 8.64 inches
Step-by-step explanation:
Length of circular arc (L) = central angle/360° × 2πr
central angle = pie/4 = 45°, r (radius) = 11 inches
L = 45°/360° × 2 × 3.142 × 11 = 8.64 inches (to two decimal places)
Answer:
Step-by-step explanation:
The formula for determining the length of an arc is expressed as
Length of arc = θ/360 × 2πr
Where
θ represents the central angle.
r represents the radius of the circle.
π is a constant whose value is 3.14
From the information given,
Radius, r = 11 inches
θ = pi/4
2π = 360 degrees
π = 360/2 = 180
Therefore,
θ = 180/4 = 45 degrees
Therefore,
Length of arc = 45/360 × 2 × 3.14 × 11
Length of arc = 8.64 inches rounded up to 2 decimal places
Solve the following system of equations using Gaussian elimination method. If there are no solutions, type "N" for both xx, yy and zz. If there are infinitely many solutions, type "z" for zz, and expressions in terms of zz for xx and yy.-5x-7y-4z=-66x+2y+3z=-2-1x+2y-7z=0
Answer:
x=-224/229,
y=296/229,
z=118/229
Step-by-step explanation:
-5x-7y-4z=-6
6x+2y+3z=-2
-x+2y-7z=0...........( multiple with (-5) and sum with 1st equation, mult with 6 and sum with 2nd equation)
______________
-x+2y-7z=0
-17y+31z=-6.....(mult with 14)
14y-39z=-2.....(mult with -17) then sum
___________
-x+2y-7z=0
-229z=-118, so here we have z=118/229.
14y-39*(118/229)=-2, from here we have y=296/229
-x+2*(296/229)-7*(118/229)=0, we get that x=-234/229
In the same way you can do this in the matrix form>>
Te probability is 0.5 that an artist makes a craf item with satisfactory quality. Assume the production of each craf item by this artist is independent. What is the probability that at most 3 attempts are required to produce a craf item with satisfactory quality?
Answer:
The probability that at most 3 attempts are required to produce a craft item with satisfactory quality is 0.9375
Step-by-step explanation:
Let E be a random variable denoting the event that an artist makes a craft item with satisfactory level.
Then the random variable E follows a Geometric distribution.
A Geometric distribution is defined as the number of failures (k) before the first success.
The probability function of Geometric distribution is:
[tex]P(X=k)=(1-p)^{k}p[/tex], p = Probability of success and k = 0, 1, 2, 3...
The probability of success is, p = 0.5 and the number of failures is, k = 3.
Compute the probability of at most 3 attempts before the first success is:
[tex]P(X\leq 3) =P(X=3)+P(X = 2)+P(X=1) +P(X = 0)\\=[(1-0.5)^{0}*0.5]+[(1-0.5)^{1}*0.5]+[(1-0.5)^{2}*0.5]+[(1-0.5)^{3}*0.5]\\=0.9375[/tex]
Therefore, the probability that at most 3 attempts are required to produce a craft item with satisfactory quality is 0.9375.
Claim: High School teachers have incomes with a standard deviation that is more than $22,500. A recent study of 126 high school teacher incomes shower a standard deviation of $24,500.
A. Express the original claim in symbolic form.
B. Identify the null and the alternative hypotheses that should be used to arrive at the conclusion that supports the claim.
Answer:
A.
sigma > 22500
B.
Null hypothesis:sigma = 22500
Alternative hypothesis:sigma > 22500
Step-by-step explanation:
A.
The claim states that the standard deviation of high school teachers income is more than 22,500. This can be represented in the symbolic form as sigma > 22500.
B.
The null hypothesis and alternative hypothesis for the given scenario can be written as
Null hypothesis: Standard deviation of income of high school teachers is 22,500.
The standard deviation is represented as sigma.Symbolically it can be written as
Null hypothesis: sigma = 22500
Alternative hypothesis: Standard deviation of income of high school teachers is more than 22,500.
Symbolically it can be written as
Alternative hypothesis: sigma > 22500
Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering your answer below.
The tangent line approximation near x=0 for the function f(x) = \\sqrt{10 + x} is found by first calculating its derivative, then using that derivative to construct the equation of the tangent line at x=0, resulting in the linear approximation y = (1/2)(10)^{-1/2}x + \\sqrt{10}.
Explanation:Finding the tangent line approximation for a function near a point involves using the function's derivative at that point. For the function f(x) = \\sqrt{10 + x}, the derivative at x = 0, denoted as f'(0), will provide the slope of the tangent. To find this, let's differentiate f(x) using the chain rule. The derivative of f(x) with respect to x is (1/2)(10 + x)^{-1/2}. At x = 0, this simplifies to 1/2(\\sqrt{10}), which is the slope of the tangent line at that point. Hence, the tangent line equation is y - f(0) = f'(0)(x - 0), which simplifies to y = (1/2)(10)^{-1/2}x + \\sqrt{10}. This form equation is the linear approximation of f(x) near x = 0.
How many four-letter code words are possible using the letters in IOWA if (a) The letters may not be repeated? (b) The letters may be repeated
Answer:
a. 24ways
b.256ways
Step-by-step explanation:
the letters IOWA contains for letters, since we are to arrange without repeating any letter, we permutate the letters.
For permutation of n object in r ways is expressed as
P(n,r)=n!/(n-r)!
hence for n=4 and r=4, we have P(4,4)=4!/(4-4)!
P(4,4)=4!/(0)!
P(4,4)=4*3*2*1=24ways
b. To arrange the letters such that each letter can be repeated, we can arrange the letter I in four ways, letter O can be arrange in four ways, letter W can be arranged in four ways and letter A can be arranged in four ways ..
Hence we arrive at
4*4*4*4=256ways
Design a sine function with the given properties:
It has a period of 24 hr with a minimum value of 10 at t=4 hr and a maximum value of 16 at t=16 hr.
please show all the steps
To design a sine function with a period of 24 and minima and maxima at given times, we scale the function using a coefficient of 2π/24, shift the function to make the peak occur at x=16, and stretch it by a factor of 3 to make it go from 10 to 16.
Explanation:To design a sine function with a period of 24 and minima and maxima at given times, we first need to understand a few concepts about sine functions. The standard sine function, sine(x), has a period of 2π. Therefore, to stretch it to a period of 24 hours, we would scale the function using a coefficient of 2π/24 or π/12. Thus, our function becomes sine((π/12) x).
Next, we want to shift the function so its maximum occurs at t=16. Normally, the sine function peaks at π/2, so we need to shift the function to the right by an amount that makes the peak occur at x=16. This would be 16 - π/2, which gives us the function sine(π/12 x - 16 + π/2).
Finally, to stretch the function vertically to accommodate the minimum and maximum values of 10 and 16, we note that the amplitude of the sine function is usually 1 (from -1 to 1), so we need to stretch it by a factor of (16-10)/2 = 3 to make it go from 10 to 16. This gives us the function y= 3sin((π/12)x - 16 + π/2)+13.
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A federal report finds that lie detector tests given to truthful persons have probability about 0.2 of suggesting that the person is deceptive.A company asks 12 job applicants about theft from previous employers, using lie detector to assess their truthfulness. Suppose that all 12 answer truthfully. What is the probability that the lie detector says all 12 are truthful? What is the probability that lie detector says at least 1 is deceptive?a. What is the mean number among 12 truthful persons who will be classified as deceptive? What is the standard deviation of this number?b. What is the probability that the number classified as deceptive is less than the mean?c. If the company asks 200 employees to take the lie detector test, what is the probability that at most 10 will be classifies as deceptive?
Answer:
a) [tex]P(X=12)=(12C12)(0.2)^{12} (1-0.2)^{12-12}=4.096x10^{-9}[/tex]
b) [tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313 [/tex]
c) [tex] E(X) = np = 12*0.2= 2.4[/tex]
d) [tex] Sd(X) = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386[/tex]
e) [tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558[/tex]
f) [tex] P(X\leq 10) =1.1x10^{-9}[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=12, p=0.2)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
For this case we want to find this probability:
[tex]P(X=12)=(12C12)(0.2)^{12} (1-0.2)^{12-12}=4.096x10^{-9}[/tex]
Part b
[tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0) [/tex]
[tex]P(X=0)=(12C0)(0.2)^{0} (1-0.2)^{12-0}=0.0687[/tex]
[tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313 [/tex]
Part c
The expected value is given by:
[tex] E(X) = np = 12*0.2= 2.4[/tex]
Part d
The standard deviation is given by:
[tex] Sd(X) = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386[/tex]
Part e
If we want the probability that the number classified as deceptive would be lower than the mean we want:
[tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)[/tex]
[tex]P(X=0)=(12C0)(0.2)^{0} (1-0.2)^{12-0}=0.0687[/tex]
[tex]P(X=1)=(12C1)(0.2)^{1} (1-0.2)^{12-1}=0.2062[/tex]
[tex]P(X=2)=(12C2)(0.2)^{2} (1-0.2)^{12-2}=0.2835[/tex]
[tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558[/tex]
Part f
For this case our random variable would be:
[tex]X \sim Binom(n=200, p=0.2)[/tex]
And we want this probability:
[tex] P(X\leq 10) = P(X=0)+P(X=1)+ .......+P(X=10)[/tex]
And we can use the following excel code to find the answer:
"=BINOM.DIST(10;200;0.2;TRUE)"
And we got: [tex] P(X\leq 10) =1.1x10^{-9}[/tex]
y = c_1e^x + c_2e^-x is a two-parameter family of solutions of the second-order DE y'' - y = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. 11. y(0) = 1, y'(0) = 2 12. y(1) = 0, y'(1) = e 13. y(-1) = 5, y'(-1) = -5 14. y(0) = 0, y'(0) = 0
Answer:
11)y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]
12)y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]
13)y = [tex]5e^{-(x+1)}[/tex]
14)y = 0
Step-by-step explanation:
Given data:
[tex]y=c_{1} e^{x} +c_{2} e^{-x}[/tex]
y''-y=0
The equation is
[tex]m^{r}[/tex]-1 = 0
(m-1)(m+1) = 0
if above equation is zero then either
m - 1 = 0 or m + 1 = 0
m = 1 , m = - 1
11)
y(0) = 1 , y'(0) = 2
[tex]y'=c_{1} e^{x} -c_{2} e^{-x}[/tex]
[tex]c_{1}[/tex] + [tex]c_{2}[/tex] = 1 (y(0) = 1) (1)
[tex]c_{1}[/tex] - [tex]c_{2}[/tex] = 2 (y'(0) = 2) (2)
adding 1 & 2
2[tex]c_{1}[/tex] = 3
[tex]c_{1}[/tex] = 3/2
3/2 + [tex]c_{2}[/tex] = 1
[tex]c_{2}[/tex] = 1 - 3/2
[tex]c_{2}[/tex] = - 1/2
y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]
12)
y(0) = 1 , y'(0) = e
[tex]c_{1}[/tex] + [tex]c_{2}[/tex] = 0 (y(0) = 1) (3)
[tex]c_{1}[/tex] = - [tex]c_{2}[/tex]
[tex]e=c_{1} e -c_{2} e^{-1}[/tex] (y'(0) = 2) (4)
[tex]e=c_{1} e -\frac{c_{2} }{e} }[/tex]
[tex]e =\frac{c_{1} e^{2} -c_{2} }{e} }[/tex]
[tex]e^{2} ={c_{1} e^{2} -c_{2} }[/tex]
replace [tex]c_{2}[/tex] = [tex]c_{1}[/tex] by equation 3
[tex]e^{2} ={c_{1} e^{2} -c_{1} }[/tex]
taking common [tex]c_{1}[/tex]
[tex]e^{2} =c_{1} ({e^{2} -1 })[/tex]
[tex]\frac{e^{2} }{({e^{2} -1 })} =c_{1}[/tex]
[tex]-\frac{e^{2} }{({e^{2} -1 })} =c_{2}[/tex]
∴ y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]
13)
y(-1) = 5 , y'(-1) = -5
[tex]c_{1}[/tex][tex]e^{-1}[/tex] + [tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 (y(-1) = 5 ) (5)
[tex]c_{1}[/tex][tex]e^{-1}[/tex] - [tex]c_{2}[/tex][tex]e^{1}[/tex] = -5 (y'(-1) = -5) (6)
Adding 5&6
2[tex]c_{1}[/tex] [tex]e^{-1}[/tex] = 0
[tex]c_{1}[/tex] = 0
[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - [tex]c_{1}[/tex][tex]e^{-1}[/tex]
[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - 0
[tex]c_{2}[/tex]= 5/e
y = [tex]5e^{-1} e^{-x}[/tex]
y = [tex]5e^{-(x+1)}[/tex]
14)
y(0) = 0 , y'(0) = 0
[tex]c_{1}[/tex] + [tex]c_{2}[/tex] = 0 (y(0) = 0) (7)
[tex]c_{1}[/tex] - [tex]c_{2}[/tex] = 0 (y'(0) = 0) (8)
Adding 7 & 8
2[tex]c_{1}[/tex] = 0
[tex]c_{2}[/tex] =
y = 0
Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 133000 dollars. Assume the population standard deviation is 31000 dollars. Suppose you take a simple random sample of 87 graduates.
1. Find the probability that a single randomly selected salary is at least 134000 dollars.
The probability that a single randomly selected salary is at least $134,000 is approximately 0.5120 or 51.20%.
To find the probability that a single randomly selected salary is at least $134,000, we need to calculate the z-score and use the standard normal distribution.
1. Calculate the z-score:
z = (x - μ) / σ
where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, x = $134,000, μ = $133,000, and σ = $31,000.
z = (134000 - 133000) / 31000
z = 1 / 31000
z ≈ 0.0323
2. Find the probability associated with the z-score:
We can use a z-table or a calculator to find the probability.
From the z-table, we find that the probability corresponding to a z-score of 0.0323 is approximately 0.5120.
Therefore, the probability that a single randomly selected salary is at least $134,000 is 0.5120 or 51.20%.
Determine whether the data set could represent a linear function.
Answer:
yes
Step-by-step explanation:
the rate of change is constant
What is the probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least one digit equal to 5
Answer:
Step-by-step explanation:
The probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least one digit equal to 5 is 6/19,
What is probability?It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.
It is given that:
The randomly chosen number between 1 and 100 is divisible by 3
Applying conditional probability:
Let A is the event: the numbers divisible by 3
Let B is the event: At least one digit equal to 5
P(A|B) = n(A∩B)/n(B)
P(A|B) = 6/19
Thus, the probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least one digit equal to 5 is 6/19.
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consider the following quadratic function f(x)= -2x^2+12x+32=0 select all the statements that are true for the function
Answer:
The answer to your question is
a) 400
b) The function has two real solutions (-2 and 8)
Step-by-step explanation:
Process
1.- Discriminant = b² - 4ac
= 12² -4(-2)(32)
= 144 + 256
= 400
2.- Solutions (using the general formula)
x = [tex]\frac{- b +/- \sqrt{b^{2}- 4ac}}{2a}[/tex]
x = [tex]\frac{- 12 +/- \sqrt{400}}{2(-2)}[/tex]
x = [tex]\frac{-12 +/- 20}{2(-2)}[/tex]
x₁ = [tex]\frac{-12 + 20}{2(-2)} = \frac{-12 + 20}{-4} = \frac{-8}{4} = - 2[/tex]
x₂ = [tex]\frac{- 12 - 20}{- 4} = \frac{-32}{- 4} = 8[/tex]
This function has two real solutions (-2, 8)
Answer: C- The value of the discrimination is 400.
E- The function has 2 real solutions.
Step-by-step explanation: Hope that helped!
This might be hard to explain, but could you try to explain where I should put the points. I’m really confused?
Answer:
im pretty sure you have to find out how far Joel walked and then double joels distance to get Brents
Step-by-step explanation:
Answer:
Step-by-step explanation: