Answer:
See below
Step-by-step explanation:
a) By the base case, a∈S. By the recursive step, bsb∈S if s∈S. Then, for s=a, bab∈S.
b) By the base case, ∅∈S. By the recursive step, asa∈S if s∈S. Then, for s=∅, a∅a=aa∈S. By the recursive step, bsb∈S if s∈S. Then, for s=aa, baab∈S.
c) Structural induction: We want to prove that s is palindrome for all s∈S.
Inductive basis: If s=a,b or ∅, then s is palindrome because s has either 0 or 1 characters.
Inductive hypothesis: Suppose that r∈S is a palindrome.
Inductive step: We will prove that every element constructed from r using the recursion is also a palindrome. Because of the restriction, all elements of S are constructed in this way, except for the base case. Thus, combining this with the inductive step, we will prove that every element of S is a palindrome.
Let t be an element constructed from r by recursion. Then t=ara or t=brb. If t=ara, then t is a palindrome, because reversing the word (denote the reverse word by capitals) gives T=aRa, with R being the reverse word of r. But r is a palindrome, hence r=R and T=aRa=ara=t. Again, if t=brb, T=bRb=brb=t.
We have completed the inductive step, hence by structural induction, every element of S is palindrome.
d) By recursion and the restiction, the only elements of S of length 3 are aaa,bbb,aba,bab. abb is none of those, hence abb∉S. (note that abb is not a palindrome, so by part c), abb∉S).
It is known that the A matrix of a single-input-single-output state space system has 4 eigenvalues at -1, 1, 2, 3, and the D matrix is 0. Furthermore, when the input is u(t) = 1 and the initial states are all zero, the steady-state output of the system is 5. Find the controllable canonical form of the system
Answer:
The controllable canonical form of the system =
Y = (5 0 0 0) (x1)
(x2)
(x3)
(x4)
Step-by-step explanation:
The detailed explanation is as shown in the attached file.
A house hunter on Long Island estimates that 20% of the available houses in her price range are in acceptable condition. Furthermore, she has time to look at only one house each week. What is the probability that she will find an acceptable house in the first two weeks that she looks (round off to second decimal place)?
Answer:
36% probability that she will find an acceptable house in the first two weeks that she looks.
Step-by-step explanation:
For each house, there is a 20% probability of it being acceptable. And a 100-20 = 80% of not being acceptable
What is the probability that she will find an acceptable house in the first two weeks that she looks (round off to second decimal place)?
She only looks one house a week.
So this is the same as the probability of taking two or less weeks to find a house.
The are two outcomes
Finding an acceptable house in the first week, with 20% probability
Not finding an acceptable house in the first week, with 80% probability, and then finding an acceptable house in the second week, with 20% probability.
Probability:
[tex]P = 0.2 + 0.8*0.2 = 0.36[/tex]
36% probability that she will find an acceptable house in the first two weeks that she looks.
The correct answer is 0.36 or 36%.
To solve this problem, we need to calculate the probability that the house hunter does not find an acceptable house in the first two weeks and then subtract this probability from 1 to find the probability that she does find an acceptable house within that time frame.
Let's denote the probability of finding an acceptable house in a given week as[tex]\( p \)[/tex] and the probability of not finding an acceptable house in a given week as [tex]\( q \)[/tex]. Since 20% of the houses are in acceptable condition, [tex]\( p = 0.20 \) and \( q = 1 - p = 1 - 0.20 = 0.80 \).[/tex]
The probability that the house hunter does not find an acceptable house in the first week is[tex]\( q \).[/tex] The probability that she does not find an acceptable house in the first two weeks is the product of the probabilities that she does not find one in each of the two weeks separately, which is [tex]\( q \times q \) or \( q^2 \).[/tex]
Now, we calculate [tex]\( q^2 \):[/tex]
[tex]\[ q^2 = (0.80)^2 = 0.64 \][/tex]
This is the probability that she will not find an acceptable house in the first two weeks. To find the probability that she will find an acceptable house in the first two weeks, we subtract this value from 1:
[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 1 - q^2 \][/tex]
[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 1 - 0.64 \][/tex]
[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 0.36 \][/tex]
In a high school graduating class of 202 students, 95 are on the honor roll. Of these, 71 are going on to college, and of the other 107 students, 53 are going on to college. A student is selected at random from the class. What are the probabilities that the person chosen is (a) going to college, (b) not going to college, and (c) on the honor roll, but not going to college?
Answer:
Step-by-step explanation:
Given that in a high school graduating class of 202 students, 95 are on the honor roll. Of these, 71 are going on to college, and of the other 107 students, 53 are going on to college.
Total students = 202
Honor roll = 95
College going out of honor = 71
Non college going out of honor = 24
Not in honor roll = 107
Not going to college from not in honour = 53
The probabilities that the person chosen is
(a) going to college, = [tex]\frac{71+53}{202} \\=0.614[/tex]
(b) not going to college, =[tex]\frac{202-124}{202} \\=0.485[/tex]
(c) on the honor roll, but not going to college
=[tex]\frac{24}{202} \\=0.119[/tex]
Final answer:
The probabilities for a student selected at random from the class are: going to college, not going to college, and being on the honor roll but not going to college, calculated based on provided numbers for a graduating class of 202 students.
Explanation:
To solve this problem, we first need to understand the given information about the high school graduating class consisting of 202 students, with 95 on the honor roll and 107 not on the honor roll. Out of those on the honor roll, 71 are going to college, and of those not on the honor roll, 53 are going to college. Let's calculate the probabilities for each scenario.
Probability of Going to College
Total students going to college = Students on the honor roll going to college + Students not on the honor roll going to college = 71 + 53 = 124
Probability = (Total students going to college) / (Total students) = 124 / 202
Probability of Not Going to College
Total students not going to college = Total students - Total students going to college = 202 - 124 = 78
Probability = 78 / 202
Probability of Being on the Honor Roll but Not Going to College
Total students on the honor roll but not going to college = Total students on the honor roll - Students on the honor roll going to college = 95 - 71 = 24
Probability = 24 / 202
Five urns are numbered 3,4,5,6 and 7, respectively. Inside each urn, there are n² dollars where n is the number on the urn.
The following experiment is performed:
An urn is selected at random. If its number is a prime number the experimenter receives the amount in the urn and the experiment is over. If its number is not a prime number, a second urn is selected from the remaining four and the experimenter receives the total amount in the two urns selected.
What is the probability that the experimenter ends up with exactly twenty- five dollars?
Answer:
0.25 or 25%
Step-by-step explanation:
3, 5 and 7 are prime numbers.
There are two possible outcomes for which the experimenter ends up with exactly twenty-five dollars:
A) Choosing urn 5 (5 x 5 = 25).
[tex]P(A) = \frac{1}{5}[/tex]
B) Choosing urn 4 and then urn 3 ([4 x 4] + [3 x 3] = 25).
[tex]P(B)= \frac{1}{5} *\frac{1}{4}=\frac{1}{20}[/tex]
The probability that the experimenter ends up with exactly $25 is:
[tex]P(x=\$25)=P(A)+P(B)= \frac{1}{5}+\frac{1}{20}\\P(x=\$25)=0.25=25\%[/tex]
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines simultaneously tangent to both graphs.(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation: (Two decimal places of accuracy.)y = ___ x + ___(b) The other line simultaneously tangent to both graphs has equation:(Two decimal places of accuracy.)y = ___ x + ___
Answer:
a) y = 7.74*x + 7.5
b) y = 1.148*x + 6.036
Step-by-step explanation:
Given:
f(x) = 6 - 10*x^2
g(x) = 8 - (x-2)^2
Find:
(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation
(b) The other line simultaneously tangent to both graphs has equation,
Solution:
- Find the derivatives of the two functions given:
f'(x) = -20*x
g'(x) = -2*(x-2)
- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,
g'(x_o) = -2*(x_o -2)
- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):
m = (g(x_o) - f(x)) / (x_o - x)
m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)
m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)
m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)
m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)
- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:
m = f'(x) = g'(x_o)
- We will develop the first expression:
m = f'(x)
( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x
Eq 1. (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2
And,
m = g'(x_o)
( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x
-2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)
Eq 2 -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)
- Now subtract the two equations (Eq 1 - Eq 2):
-20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0
-22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0
- Form factors: 20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0
20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0
(x - x_o)(20*x - 2*x_o + 4) = 0
x = x_o , x_o = 10x + 2
- For x_o = 10x + 2 ,
(g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x
(8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x
(-90*x^2 + 2) = -180*x^2 - 40*x
90*x^2 + 40*x + 2 = 0
- Solve the quadratic equation above:
x = -0.0574, -0.387
- Largest slope is at x = -0.387 where equation of line is:
y - 4.502 = -20*(-0.387)*(x + 0.387)
y = 7.74*x + 7.5
- Other tangent line:
y - 5.97 = 1.148*(x + 0.0574)
y = 1.148*x + 6.036
Suppose we have a population of N deer in a study area. Initially n deer from this population are captured, marked so that they can be identified as having been captured, and returned to the population. After the deer are allowed to mix together, m deer are captured from the population and the number k of these deer having marks from the first capture is observed. Assuming that the first and second captures can be considered random selections from the population and that no deer have either entered or left the study area during the sampling period, what is the probability of observing k marked deer in the second sample of m deer
Answer:
[tex]P(k) = \frac{(n C k) [(N-n) C (m-k)]}{(NCm)}[/tex]
Step-by-step explanation:
Step 1: Number of possible combination of selecting ‘m’ deer in second sample
Total number of deer are N and therefore the combinations can be calculated as (N С m).
Step 2: Number of possible combination of marked deer ‘k’ in second sample
Total number of marked deer in total population is ‘n’. Therefore, the possible number of selecting marked deer is (n C k).
Step 3: Number of possible combination unmarked deer in second sample
Since we have already calculated the total combinations of selecting marked deer in the second sample. Hence, we have to calculate the total unmarked deer in total population which is N-n and number of unmarked deer in the second sample which is m-k.
Therefore, total possible combination of unmarked deer in second sample is [(N-n) C (m-k)].
Step 4: Probability of selecting unmarked deer in the second sample is
Let the probability of selecting unmarked deer in the second sample be P(k)
Therefore,
[tex]P(k) = \frac{(n C k) [(N-n) C (m-k)]}{(NCm)}[/tex]
Which of the following is equivalent to finding the zeros of a function
A. X-intercepts
B. Y-intercepts
C. Slope
D. Origin
Answer:
A. X-intercepts
Step-by-step explanation:
The zeros of a function are the values of x for which y is 0.
For example:
y = 2x - 4
Has zeros
2x - 4 = 0
2x = 4
x = 2
Which means that when x = 2, y = 0. Looking at the graphic of this function, it crosses the x line when x = 2. So the zeros are also called x intercepts.
So the correct answer is
A. X-intercepts
Answer:
X-intercepts
Step-by-step explanation:
TRUST ME
The measures of the angles of ABC are given by the expressions in the table A( 6X - 1 ) ° B 20° C ( x + 14 ) ° what are the measures of angles A and C ? Enter your answers in the boxes mA = mC =
Answer:
Step-by-step explanation:
The sum of the angles in a triangle is 180 degrees. This means that in triangle ABC,
Angle A + angle B + angle C = 180
Therefore,
6x - 1 + 20 + x + 14 = 180
6x + x + 20 + 14 - 1 = 180
7x + 33 = 180
Subtracting 33 from the left hand side and the right hand side of the equation, it becomes
7x + 33 - 33 = 180 - 33
7x = 147
Dividing the left hand side and the right hand side of the equation by 7, it becomes
7x/7 = 147/7
x = 21
Therefore
Angle A = 6x - 1 = 6 × 21 - 1
Angle A = 125 degrees
Angle C = x + 14 = 21 + 14
Angle C = 35 degrees.
Find the 95% confidence intervalfor the variance and standard deviation for the time ittakes a customer to place a telephone order with a largecatalogue company if a sample of 23 telephone ordershas a standard deviation of 3.8 minutes. Assume thevariable is normally distributed. Do you think that thetimes are relatively consistent?
Answer:
[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]
[tex] 2.939 \leq \sigma \leq 5.379[/tex]
Step-by-step explanation:
Data given and notation
s represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=23 the sample size
Confidence=95% or 0.95
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
The Chi Square distribution is the distribution of the sum of squared standard normal deviates .
Calculating the confidence interval
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
The sample standard deviation for this case was s = 3.8
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=23-1 =22[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabele to find the critical values.
The excel commands would be: "=CHISQ.INV(0.025,22)" "=CHISQ.INV(0.975,22)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=36.78[/tex]
[tex]\chi^2_{1- \alpha/2}=10.98[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(22)(3.8)^2}{36.78} \leq \sigma^2 \leq \frac{(22)(3.8)^2}{10.98}[/tex]
[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 2.939 \leq \sigma \leq 5.379[/tex]
you have 200 grams of a radioactive kind of iodine how much will ne left after 120 days if its half life is 60 days?
By definition, the half life is a quantity is the time it takes to lose half of that quantity.
So, if the half life is 60 days, it means that after 60 days you have lost half the initial amount, i.e. you're left with 100 grams.
After another 60 days, you've lost, again, half of that amount, so you're left with 50 grams.
In other words, every time a half life period passes, you're left with half the quantity you had at the beginning of that period.
So, after two half-life periods (i.e. 120 days), you'll have half of the half of the initial quantity, i.e. one quarter.
According to data from the state blood program, 40 percent of all individuals have group A blood. If six individuals give blood, find the probability that exactly three of the individuals have group A blood.
The probability that exactly three out of six individuals have group A blood, with a 40 percent chance for each individual, is found using the binomial probability formula. The calculation yields approximately 27.648% as the probability for exactly three individuals having group A blood.
Explanation:The question asks for the probability that exactly three out of six individuals have group A blood, given that 40 percent of all individuals have group A blood. This is a binomial probability problem because there are two outcomes (having group A blood or not) and a fixed number of trials (six individuals).
Let's denote X as the random variable representing the number of individuals with group A blood. The probability of success on any given trial is p = 0.40 (having group A blood). The probability of failure is q = 1 - p = 0.60 (not having group A blood).
The binomial probability formula is:
[tex]P(X = k) = C(n, k) * p^k * q^(n-k)[/tex]
where:
For our problem:
n = 6 (total number of individuals),k = 3 (number of individuals with group A blood that we want to find the probability for),p = 0.40,q = 0.60.Plugging these values into the binomial formula gives:
[tex]P(X = 3) = C(6, 3) * (0.40)^3 * (0.60)^3[/tex]First, calculate the combination:
C(6, 3) = 6! / (3! * (6-3)!) = 20
Then, calculate the probability:
[tex]P(X = 3) = 20 * (0.40)^3 * (0.60)^3 = 20 * 0.064 * 0.216 = 0.27648[/tex]
Hence, the probability that exactly three of the individuals have group A blood is approximately 0.27648 or 27.648%.
We play a card game where we receive 13 cards at the beginning out of the deck of 52. we play 50 games one evening. for each of the the following random variable identify the name and parameters of the distribution: a) The number of aces I get in the first game b) The number of games in which I recieve at least one ace during the evening c) The number of games in which all my cards are from the same suit d) The number of spades I receive in 5th game
The answer & explanation for this question is given in the attachment below.
The number of aces in the first game and the number of spades in the 5th game follow a Hypergeometric Distribution while the number of games receiving at least one ace can be modeled by a Binomial distribution. The event of all cards being from the same suit can be thought of as a Uniform distribution.
Explanation:a) The number of aces you get in the first game follows a Hypergeometric Distribution. In such a distribution, you are drawing cards without replacement. The parameters are N=52 (the population size), K=4 (the number of success states in the population i.e., the number of aces in a deck), and n=13 (the number of draws).
b) The number of games in which you receive at least one ace can be modeled by a Binomial distribution. Each game you play (out of 50) is a single trial, with the probability of success (getting at least one ace) being the same for every trial. The parameters are n=50 (the number of trials/games) and p (the probability of getting at least one ace).
c) The likelihood of all your cards being from the same suit in a game is heavily reliant on chance, can be modeled as a Uniform distribution given its rare occurrence. Essentially, the parameters would be minimum = 0 and maximum = 1. However, determining the parameters would require calculation of the specific probabilities, which is complex due to the nature of the game.
d) The number of spades you receive in the 5th game also follows a Hypergeometric distribution, similar to the situation in the first game. The parameters in this case are N=52, K=13 (number of spades in a deck), and n=13 (the number of drawn cards).
Learn more about Statistical Distributions here:https://brainly.com/question/31861365
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A certain type of plywood consists of 5 layers. The thickness of the layers follows a normal distribution with mean 5 mm and standard deviation 0.2 mm. Find the probability that the plywood is less than 24mm thick.
The resulting probability is approximately 100%.
To find the probability that the plywood is less than 24mm thick, we need to calculate the z-score for 24mm using the given mean and standard deviation. The z-score formula is:
z = (x - mean) / standard deviation
Substituting the values, we get:
z = (24 - 5) / 0.2 = 95
We can then look up the z-score in a standard normal distribution table to find the corresponding probability.
In this case, the probability is practically 1 (or 100%) since 24mm is significantly greater than the mean.
Therefore, the probability that the plywood is less than 24mm thick is approximately 100%.
Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the curves x=4y2−y3 and x=0 about the x-axis.
Answer = 321.7
Explanation:
radius = y
height = [tex]4y^{2} - y^{3}[/tex]
area of cylinder = 2π*r*h
Integrate the area to calculate the volume:
[tex]= \int\limits^0_4 {2\pi(4y^{3} -y^{4}) } \, dy[/tex]
[tex]= 2\pi (\int\limits^0_4 {4y^{3}dy -\int\limits^0_4 y^{4} dy )} \,[/tex]
[tex]= 2\pi (256-\frac{1024}{5} )[/tex]
[tex]=\frac{512\pi }{5}[/tex]
[tex]= 321.7[/tex]
Final answer:
To compute the volume of the solid formed by rotating the region bounded by the curves x = 4y² − y³ and x = 0 around the x-axis, utilize the cylindrical shells method with the relevant volume formula for cylindrical shells and integrate over the intersecting interval of y-values.
Explanation:
Finding the Volume of a Solid of Revolution
To find the volume of the solid obtained by rotating the region bounded by the curves x = 4y2 − y3 and x = 0 about the x-axis using the method of cylindrical shells, we follow these steps:
Identify the range of y-values where the two curves intersect, which will give the limits of integration.Write down the formula for the volume of a cylindrical shell: dV = 2πry • dx, where r is the radius (function of y), and dx is the shell's thickness.Insert the given function into the formula to represent r as the function 4y2 − y3 and integrate with respect to y over the interval from step 1.The relevant formulas for the volume and surface area of a sphere are (4/3)πr3 and 4πr2, respectively. It's important to note that the volume depends on the cube of the radius R3, while the surface area is a function of the square of the radius R2.
The weight of a certain type of brick has an expectation of 1.12 kilograms with a variance of 0.0009 kilograms2. How many bricks would need to be selected so that the average weight has a standard deviation of no more than 0.005 kilograms
Answer:
[tex]n \geq 36[/tex]
Step-by-step explanation:
For this case we know the mean and the deviation:
[tex] \mu = 1.12 kg , \sigma= 0.0009[/tex]
The mean is given by this:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
If we find the expected value for the sample mean we have:
[tex] E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n}) =\frac{1}{n} \sum_{i=1}^n E(X_i)[/tex]
Since each observation in the sample [tex] X_1, X_2,...., X_{25}[/tex] have an expectation [tex] E(X_i) = \mu , i =1,2,...,25[/tex] so then we have that:
[tex] E(\bar X) = \frac{1}{n} n\mu = \mu[/tex]
Now for the variance of the sample mean we have this:
[tex]Var (\bar X) = Var(\frac{\sum_{i=1}^n X_i}{n})= \frac{1}{n^2} \sum_{i=1}^n Var(X_i)[/tex]
And again each observation have a variance [tex] \sigma^2_i = 0.0009 , i =1,2,...,25[/tex] then we have:
[tex]Var (\bar X) =\frac{1}{n^2} n(\sigma^2) =\frac{\sigma^2}{n}[/tex]
And then the standard deviation would be:
[tex] Sd(\bar X) =\sqrt{\frac{\sigma^2}{n}}= \frac{\sigma}{\sqrt{n}}[/tex]
And we want that the standard deviation for the sample no more than 0.005 so we have this condition:
[tex]\frac{\sigma}{\sqrt{n}} \leq 0.005[/tex]
And since we know the value of [tex] \sigma= \sqrt{0.0009}=0.03[/tex] we can solve for the value of n like this:
[tex] \frac{0.03}{0.005} \leq \sqrt{n}[/tex]
[tex] 6 \leq \sqrt{n}[/tex]
And if we square both sides we got:
[tex] n\geq 36[/tex]
Final answer:
To achieve an average weight standard deviation of no more than 0.005 kilograms for a certain type of brick with a variance of 0.0009 kilograms², 36 bricks need to be selected.
Explanation:
The question concerns the calculation of the number of bricks needed so that the average weight of these bricks has a standard deviation of no more than 0.005 kilograms. Given that the weight of a certain type of brick has an expectation (mean) of 1.12 kilograms and a variance of 0.0009 kilograms2, the first step is to understand that the variance of the average weight of n bricks is equal to the variance of a single brick divided by n.
To achieve a standard deviation of the average weight of no more than 0.005 kilograms, we use the formula for the standard deviation of the mean, which is √(variance/n). The variance given is 0.0009 kilograms2. Setting the equation √(0.0009/n) <= 0.005 and solving for n gives:
0.0009/n = 0.0052
0.0009/n = 0.000025
n = 0.0009 / 0.000025 = 36
Therefore, to ensure the standard deviation of the average weight is no more than 0.005 kilograms, 36 bricks would need to be selected.
A box is formed by cutting squares from the four corners of a 9"-wide by 12"-long sheet of paper and folding up the sides.
Let x represent the length of the side of the square cutout (in inches), and let V represent the volume of the box (in cubic inches).
Write a formula that expresses V in terms of x.
The volume of a box is the amount of space in it.
The expression that represents volume is: [tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]
The dimension of the cardboard is given as:
[tex]\mathbf{Length = 12}[/tex]
[tex]\mathbf{Width = 9}[/tex]
Assume the cut-out is x.
So, the dimension of the box is:
[tex]\mathbf{Length =12-2x}[/tex]
[tex]\mathbf{Width =9 - 2x}[/tex]
[tex]\mathbf{Height = x}[/tex]
The volume of the box is:
[tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]
Hence, the expression that represents volume is:
[tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]
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The volume of a box formed by cutting a square of side x from a 9" by 12" sheet and folding the sides is given by the formula V = x(12 - 2x)(9 - 2x).
Explanation:The volume of a box is formed by multiplying its length, width, and height. In this case, if you cut a square of side x from each corner of the sheet, the new dimensions of your box would be:
Length = (12 - 2x) Width = (9 - 2x) Height = x
Therefore, the formula to express the volume V in terms of x is:
V = x(12 - 2x)(9 - 2x)
.
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Donna de paul is raising money for the homeless. She discovers that each church group requires 2 hours of let her writing and 1 hour of follow-up while for each labor union she needs 2 hours of letter written and 3 hours of follow-up. Donna can raise $150 from each church group and $200 from each Union Local and she has a maximum of 20 hours of letters written and a maximum of 16 hours of follow-up available per month. Determine the most profitable mixture of group she should contact in the most money she can lose in a month.
z=()x1+()x2
Answer:
x = 7
y = 3
z (max) = 4950/3 = 1650
Step-by-step explanation:
Let call
x numbers of church goup and
y numbers of Union Local
Then
First contraint
2*x + 2*y ≤ 20
Second one
1*x + 3*y ≤ 16
Objective Function
z = 150*x + 200*y
Then the system is
z = 150*x + 200*y To maximize
Subject to:
2*x + 2*y ≤ 20
1*x + 3*y ≤ 16
x ≥ 0 y ≥ 0
We will solve by using the Simplex method
z - 150 *x - 200*y = 0
2*x + 2*y + s₁ = 20
1*x + 3*y + 0s₁ + s₂ = 16
First Table
z x y s₁ s₂ Cte
1 -150 -200 0 0 = 0
0 2 2 1 0 = 20
0 1 3 0 1 = 16
First iteration:
Column pivot ( y column ) row pivot (third row) pivot 3
Second table
z x y s₁ s₂ Cte
1 -250/3 0 0 200/3 = 3200/3
0 - 4/3 0 -1 2/3 = -20/3
0 1/3 1 0 1/3 = -20/3
Second iteration:
Column pivot ( x column ) row pivot (second row) pivot -4/3
Third table
z x y s₁ s₂ Cte
1 0 0 750/12 700/6 = 4950/3
0 1 0 3/4 -1/2 = 7
0 0 1 -1/4 1/2 = 9/3
Donna should contact four church groups and four labor unions to maximize her fundraising, which would yield $1400. The maximum money she could 'lose' (or not earn) is $0 if she does not contact any group.
Explanation:We need to set up linear inequalities to represent the constraints described by Donna's situation. We will let x represent the number of church groups and y represent the number of labor unions she contacts.
The time required for letter writing can be described by the inequality: 2x + 2y <= 20, and the time for follow-ups is x + 3y <= 16.
We want to maximize the sum z = 150x + 200y which represents the total money she can raise.
To find the maximum, one method is to graph the feasible region defined by the constraints and find the vertices. Then evaluate the objective function at the vertices. In this case, the vertices are (0,0), (0,5), (8,0) and (4,4).
Upon calculation, contacting four church groups and four labor unions yields the highest amount of $1400.
As for the loss, given the nature of the task, the most money she could 'lose' in a month is just not raising any money at all, which would be $0 if she doesn't contact any group.
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Find the missing lengths and angle measures in kite ABCD
Answer:
Part 1) [tex]AC=40\ units[/tex]
Part 2) [tex]DC=29\ units[/tex]
Part 3) [tex]m\angle ABE=39^o[/tex]
Part 4) [tex]m\angle BCE=51^o[/tex]
Step-by-step explanation:
we know that
A kite has two pairs of consecutive, congruent sides the diagonals are perpendicular and the non-vertex angles are congruent
Part 1) Find AC
we know that
BD is the axis of symmetry, bisects the diagonal AC
so
[tex]AE=EC[/tex]
we have
[tex]EC=20\ units[/tex]
[tex]AC=AE+EC[/tex] ----> by segment addition postulate
therefore
[tex]AC=20+20=40\ units[/tex]
Part 2) Find CD
we know that
CDE is a right triangle (the diagonals are perpendicular)
so
Applying Pythagorean Theorem
[tex]DC^2=EC^2+ED^2[/tex]
substitute the values
[tex]DC^2=20^2+21^2[/tex]
[tex]DC^2=841\\DC=29\ units[/tex]
Part 3) Find m∠ABE
we know that
In the right triangle ABE
[tex]51^o+m\angle ABE=90^o[/tex] ----> by complementary angles
[tex]m\angle ABE=90^o-51^o=39^o[/tex]
Part 4) Find m∠BCE
we know that
[tex]m\angle BCE=m\angle BAE[/tex] ----> diagonal BD is the axis of symmetry
we have
[tex]m\angle BAE=51^o[/tex]
therefore
[tex]m\angle BCE=51^o[/tex]
The measure of AC is 40 units.
The measure of the length DC is 29 units.
The measure of the angle ABE is 39 degrees.
The measure of the angle BCE is 51 degrees.
We have to determineThe missing lengths and angle measures in kite ABCD.
According to the question
The rhombus is a four-sided quadrilateral with all its four sides equal in length.
Rhombus is a kite with all its four sides congruent.
A kite is a special quadrilateral with two pairs of equal adjacent sides.
1. The measure of the length of AC is;
In the figure, BD is the axis of symmetry, bisects the diagonal AC.
Then,
[tex]\rm AE = EC[/tex]
And the measure of AC is,
[tex]\rm AC = AE+EC \\ \\ AC = 20+20 \\ \\ AC = 40 \ units[/tex]
The measure of AC is 40 units.
2. In the figure, CDE is a right triangle (the diagonals are perpendicular)
Then,
By applying the Pythagoras Theorem
[tex]\rm DC^2=EC^2+ED^2\\ \\ DC^2=20^2+21^2\\\\ DC^2 = 400+441 \\ \\ DC^2 = 841\\ \\ DC = 29 \ \rm units[/tex]
The measure of the length DC is 29 units.
3. In the right triangle ABE by the complementary angles;
[tex]\rm 51+ \angle ABE = 90\\ \\ \angle ABE=90-51\\ \\ \angle ABE=39 \ degrees[/tex]
The measure of the angle ABE is 39 degrees.
4. By the axis of symmetry the diagonal BD is;
[tex]\rm m \angle BCE = m \angle BAE\\\\ m \angle BCE = m \angle BAE = 51 \ degrees[/tex]
The measure of the angle BCE is 51 degrees.
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A large disaster cleaning company estimates that 30 percent of the jobs it bids on are finished within the bid time. Looking at a random sample of 8 jobs that it has contracted, calculate the mean number of jobs completed within the bid time.
Answer: 2.4
Step-by-step explanation:
Given : A large disaster cleaning company estimates that 30 percent of the jobs it bids on are finished within the bid time.
i..e The proportion of the jobs it bids on are finished within the bid time: p = 30%= 0.30
Sample size of jobs that it has contracted : n= 8
Then , the mean number of jobs completed within the bid time : [tex]\mu=np[/tex]
[tex]= 8\times0.30=2.4[/tex]
Hence, the mean number of jobs completed within the bid time is 2.4 .
An train station has determined that the relationship between the number of passengers on a train and the total weight of luggage stored in the baggage compartment can be estimated by the least squares regression equation y=103+30x. Predict the weight of luggage for a flight with 86 passengers.
Answer:
2683
Step-by-step explanation:
Using the linear regression equation that predict the relationship between the weight of the luggage and the total number of passenger y = 103 + 30x, we can plug in the number of passenger x = 86 to predict the weight of the luggage on a flight:
y = 103 + 30*86 = 2683
July 15: Hire part-time helper to be paid $12 per hour. Pay periods are the 1st through the 15th and 16th through the end of the month, with paydays being the 20th for the first pay period and the 5th of the following month for the second pay period. (No entry is required on this date; it is here for informational purposes only.)
Answer:
The information above is just a part of a long question.The main question is found in the attached as well as the answer following a step by step methodical approach.
Step-by-step explanation:
Take a good at the full question before venturing into the statements prepared so as to have a thorough understanding of the requirements and how the answers provided have touched upon the requirements.
A researcher wants to determine if the nicotine content of a cigarette is related to "tar". A collection of data (in milligrams) on 29 cigarettes produced the accompanying scatterplot, residuals plot, and regression analysis. Complete parts a and b below. ) Explain the meaning of Upper R squared in this context. A. The linear model on tar content accounts for 92.4% of the variability in nicotine content. B. The predicted nicotine content is equal to some constant plus 92.4% of the tar content. C. Around 92.4% of the data points have a residual with magnitude less than the constant coefficient. D. Around 92.4% of the data points fit the linear model.
Answer:
Option A The linear model on tar content accounts for 92.4% of the variability in nicotine content.
Step-by-step explanation:
R-square also known as coefficient of determination measures the variability in dependent variable explained by the linear relationship with independent variable.
The given scenario demonstrates that nicotine content is a dependent variable while tar content is an independent variable. So, the given R-square value 92.4% describes that 92.4% of variability in nicotine content is explained by the linear relationship with tar content. We can also write this as "The linear model on tar content accounts for 92.4% of the variability in nicotine content".
The Upper R squared or the coefficient of determination here represents the percentage of the variability in the nicotine content that can be explained by the tar content in the regression model, which in this case is 92.4%.
Explanation:In this context, the meaning of Upper R squared is represented by option A. The linear model on tar content accounts for 92.4% of the variability in nicotine content. This indicates that 92.4% of the change in nicotine content can be explained by the amount of tar content based on the linear regression model used. This measure is also known as the coefficient of determination. Meanwhile, options B, C, and D are not correct interpretations of the R squared in this context. Both B and D wrongly relate the percentage to the predictability of the data points and option C incorrectly associates this percentage with the residual magnitude.
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Umbrella Corporation purchases a raw material in 55-gallon drums from a supplier. Records for the supplier indicate that the impurity level in the material per drum has a normal distribution with a mean of 3% and a standard deviation of 0.4%. An impurity level of 4% or more in any shipment requires that Umbrella return the entire drum to the supplier.
What is the probability that Umbrella has to return any given shipment?
Answer:
The probability that Umbrella returns any shipment is 0.0062.
Step-by-step explanation:
Let X = the impurity level in the material per drum
Then it is provided that,
[tex]X\sim N(\mu = 0.03,\ \sigma=0.004)[/tex]
Also if the impurity level is more than or equal to 4% or 0.04 in any shipment, then Umbrella returns the entire drum to the supplier.
Compute the probability that Umbrella returns any shipment as:
[tex]P(X\geq 0.04)=P(\frac{X-\mu}{\sigma}\geq \frac{0.04-0.03}{0.004}) \\=P(Z\geq 2.5)\\=1-P(Z<2.5)\\=1-0.99379\\=0.00621\\\approx0.0062[/tex]
Use the z-table fro left z-scores to determine the probability.
Thus, the probability that Umbrella returns any shipment is 0.0062.
At a college, 71% of courses have final exams and 43% of courses require research papers. Suppose that 26% of courses have a research paper and a final exam. Part (a) Find the probability that a course has a final exam or a research paper.
Answer:
There is an 88% probability that a course has a final exam or a research paper.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
E is the probability that a course has final exam.
P is the probability that a course requires research paper.
We have that:
[tex]E = e + (E \cap P)[/tex]
In which e is the probability that a course has final exam but does not require research paper and [tex]E \cap P[/tex] is the probability that a course has both of these things.
By the same logic, we have that:
[tex]P = p + (E \cap P)[/tex]
(a) Find the probability that a course has a final exam or a research paper.
This is
[tex]Pr = e + p + (E \cap P)[/tex]
Suppose that 26% of courses have a research paper and a final exam.
This means that
[tex]E \cap P = 0.26[/tex]
43% of courses require research papers.
So [tex]P = 0.43[/tex]
[tex]P = p + (E \cap P)[/tex]
[tex]0.43 = p + 0.26[/tex]
[tex]p = 0.17[/tex]
71% of courses have final exams
So [tex]E = 0.71[/tex]
[tex]E = e + (E \cap P)[/tex]
[tex]0.71 = e + 0.26[/tex]
[tex]e = 0.45[/tex]
The probability is
[tex]Pr = e + p + (E \cap P) = 0.45 + 0.17 + 0.26 = 0.88[/tex]
There is an 88% probability that a course has a final exam or a research paper.
Convert the following numbers to Mayan notation. Show your calculations used to get your answers. 135?
Answer:
The answer to your question is below
Step-by-step explanation:
In Mayan notation there are only 3 symbols
dot . means one
line --- means five
snail means zero
And the are three levels, the lower means x 1
the first means x 20
the second means x 400
the higher means x 8000
Then, 135 is lower than 400, so we must start in the first level and need to divide 135 by 20. 6
20 135
15
Finally place 6 in the fist level and 15 in the lower lever, like this
And 120 + 15 = 135.
Final answer:
To convert 135 to Mayan notation, one must work with a base-20 system. In this system, 135 is broken down to (6 x 20) + 15, which is written in Mayan as a bar and three dots on top, and below that, six dots.
Explanation:
The Maya used a vigesimal (base-20) numeral system for their calculations, which is different from our familiar base-10 system. To convert the number 135 to Mayan notation, you need to express it in base-20.
135 in base-20 is calculated as:
135 divided by 20 gives 6 as the quotient and 15 as the remainder.Therefore, 135 can be written as (6 x 20) + 15.In Mayan numerals, 6 is represented by six dots and 15 by a bar (which equals 5) and three dots on top of it.In Mayan notation, numbers are written vertically with the largest value at the top, so 135 would be represented with a bar and three dots on top (for 15), and under that, six dots (for 6).
If the equation of a circle is (x - 2)2 + (y - 6)2 = 4, it passes through point (5,6). True or false
Answer: False
Step-by-step explanation:
To know if the circle passes through the given point, we simply insert the coordinates of the given point (5,6) into the general equation Of the Circle. If left hand side of the equation balances with the right hand side, then the circle passed through the point.
The equation of the circle is (x-2)² + (y-6)² = 4
By inserting the given coordinates (5,6) we have.
(5-2)² + (6-6)² = 3² =9
Since the answer gotten is not 4, then the circle does not pass through the point (5,6)
An urn contains n + m balls, of which n are red and m are black. They are withdrawn from the urn, one at a time and without replacement. Let X be the number of red balls removed before the first black ball is chosen. We are interested in determining E[X]. To obtain this quantity, number the red balls from 1 to n. Now define the random variables
if red ball i is taken before any black ball is chosen
Otherwise
a) Express X in terms of the
b) Find E[X]
The answer is a) [tex]X=X_1 +X_2 +X_3+..X_n[/tex] b) The expression is
[tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex].
Given:
An urn contains n+m balls of which n is red and m is black
a)
Expressing X in terms of the defined random variables:
Let be the indicator random variable for the event that red ball i is taken before any black ball is chosen. It takes the value 1 if this event occurs and 0 otherwise.
Now, the value of X is the number of red balls removed before the first black ball is chosen. This can be expressed as the sum of the individual indicator random variables for each red ball:
[tex]X=X_1 +X_2 +X_3+..X_n[/tex]
b)
Find E[X] (expected value of X):
The expected value of a sum of random variables is the sum of the expected values of those random variables. To find E[X],
Find the expected value of each indicator random variable and then sum them up.
Therefore, the expected value [tex]X_i[/tex] is:
[tex]E[X_i] = 1. \dfrac{1}{n+m-i+1} + 0\cdot \dfrac{m}{n+m-i+1}[/tex]
[tex]= \dfrac{1}{n+m-i+1}[/tex]
Now, to find E[X], we sum up the expected values of the indicator random variables for all red balls:
[tex]E[X] = E[X_1]+E[X_2]+E[X_3]+.....+E[X_n][/tex]
Substitute the values of [tex]E[X_i][/tex] into the sum and simplify:
[tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex]
a) [tex]X=X_1 +X_2 +X_3+..X_n[/tex] b) [tex]E[X] = \dfrac{1}{n+m} +\dfrac{1}{n+m-1} +....+ \dfrac{1}{m+1}[/tex]
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X can be expressed as X = R_1 + 2R_2 + 3R_3 + ... + nR_n. To find E[X], use the linearity of expectation and the probabilities of R_i.
Explanation:To determine E[X], we need to express X in terms of the random variables and then find the expected value. Let R_i denote the event that red ball i is taken before any black ball is chosen. The expression for X is:
To find the expected value, we use the linearity of expectation. Since the probabilities for each R_i are the same, we have:
Now, we need to determine the probabilities of R_i. Since the balls are drawn without replacement, the probability that red ball i is drawn before any black ball is chosen is:
Substituting this probability into the expression for E[X], we get:
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Discuss several ways to use fractions in everyday life
Answer:
Step-by-step explanation:
Fractions are used in telling time.
Fractions are used to determine discounts when there is sales going on.
Fraction is used to represent part of whole .
Fractions are used in baking to tell how much of an ingredient to use.
A real estate agent has 14 properties that she shows. She feels that there is a 50% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling more than 4 properties in one week.
To find the probability of selling more than 4 properties in one week, use the binomial probability formula for each number of sales above 4 and sum them, or subtract the sum of probabilities of 4 or fewer sales from 1.
Explanation:To compute the probability of selling more than 4 properties in one week, given that the chance of selling any one property is 50% and is independent of selling another property, we can use the binomial probability formula. In our scenario, we have a total of 14 properties, and the random variable X represents the number of properties sold in a week.
The binomial probability formula is:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
n is the number of trials (in this case, 14 properties)k represents the number of successes (properties sold)p is the probability of success on any given trial (50% or 0.5)However, instead of calculating the probability of selling exactly k properties, we are interested in selling more than 4 properties. This means we need to calculate the probabilities for selling 5, 6, ..., up to 14 properties and sum them up. Alternately, we can calculate 1 minus the probability of selling 4 or fewer properties to save time, since the probabilities must sum to 1.
A ball is dropped from a state of rest at time t=0.The distance traveled after t seconds is s(t)=16t^2 ft.
(a) How far does the ball travel during the time interval [3,3.5] ?Δs=____ft(b) Compute the average velocity over [3,3.5] .Δs/Δt= ____ ft/sec(c) Compute the average velocity over time intervals [3, 3.01] , [3, 3.001] , [3, 3.0001] , [2.9999, 3] , [2.999, 3] , [2.99, 3] .
Use this to estimate the object's instantaneous velocity at t=3 .V(3)= ____ ft/sec
The ball travels a distance of 52 ft during the time interval [3,3.5]. The average velocity over this interval is 104 ft/sec. When we calculate over increasingly smaller time intervals, it appears the instantaneous velocity at t=3 is 96 ft/sec.
Explanation:(a) We need to find the displacement Δs over the time interval [3,3.5]. For that, we subtract the position at time 3 from the position at time 3.5:
Δs = s(3.5) - s(3) = 16(3.5^2) - 16(3^2) = 196 - 144 = 52 ft.
(b) We calculate the average velocity by dividing Δs by Δt. That is Δs/Δt = 52/0.5 = 104 ft/sec.
(c) To estimate the instantaneous velocity at t=3, we have to compute the average velocities over smaller and smaller intervals centered at t=3. Let's calculate the average velocities for [3,3.01],[2.999,3]:
For [3,3.01]: Δs/Δt = [16*(3.01)^2 - 16*3^2]/0.01 = 96.12 ft/sec. For [2.999,3]: Δs/Δt = [16*3^2 - 16*(2.999)^2]/0.001 = 96 ft/sec.
Instantaneous velocity at t=3, v(3) can be estimated as the limit of these average velocities, which seems to be approaching about 96 ft/sec.
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(a) The ball travel Δs= 52 ft during the time interval [3,3.5].
(b) The average velocity over [3,3.5] is Δs/Δt= 104 ft/sec
(c) The object's instantaneous velocity at t=3 .V(3)= 96 ft/sec
Let's solve the problem step-by-step.
Given:
The distance traveled by the ball after t seconds is given by the function: [tex]s(t) = 16t^2[/tex]
(a) To find the distance traveled, we need to calculate the change in the distance function, which is
[tex]\Delta s = s(3.5) - s(3)[/tex]
First, compute [tex]s(3)[/tex]:
[tex]s(3) = 16(3)^2 = 16 \times 9 = 144[/tex] ft
Next, compute [tex]s(3.5)[/tex]:
[tex]s(3.5) = 16(3.5)^2 = 16 \times 12.25 = 196[/tex] ft
So,
[tex]\Delta s = 196 - 144 = 52[/tex] ft
(b) Average velocity is calculated by dividing the change in distance by the change in time, which is
[tex]\frac{\Delta s}{\Delta t} = \frac{52}{0.5} = 104[/tex] ft/sec
(c) For the time intervals, we compute each average velocity:
[3, 3.01]:
[tex]\Delta t = 0.01[/tex]
[tex]s(3.01) = 16(3.01)^2 = 16 \times 9.0601 = 144.9616[/tex] ft
[tex]\Delta s = 144.9616 - 144 = 0.9616[/tex] ft
[tex]\text{Average velocity} = \frac{0.9616}{0.01} = 96.16[/tex] ft/sec
[3, 3.001]:
[tex]\Delta t = 0.001[/tex]
[tex]s(3.001) = 16(3.001)^2 = 16 \times 9.006001 = 144.096016[/tex] ft
[tex]\Delta s = 144.096016 - 144 = 0.096016[/tex] ft
[tex]\text{Average velocity} = \frac{0.096016}{0.001} = 96.016[/tex] ft/sec
[3, 3.0001]:
[tex]\Delta t = 0.0001[/tex]
[tex]s(3.0001) = 16(3.0001)^2 = 16 \times 9.00060001 = 144.00960016[/tex] ft
[tex]\Delta s = 144.00960016 - 144 = 0.00960016[/tex] ft
[tex]\text{Average velocity} = \frac{0.00960016}{0.0001} = 96.0016[/tex] ft/sec
[2.9999, 3]:
[tex]\Delta t = 0.0001[/tex]
[tex]s(2.9999) = 16(2.9999)^2 = 16 \times 8.99940001 = 143.99040016[/tex] ft
[tex]\Delta s = 144 - 143.99040016 = 0.00959984[/tex] ft
[tex]\text{Average velocity} = \frac{0.00959984}{0.0001} = 95.9984[/tex] ft/sec
[2.999, 3]:
[tex]\Delta t = 0.001[/tex]
[tex]s(2.999) = 16(2.999)^2 = 16 \times 8.994001 = 143.904016[/tex] ft
[tex]\Delta s = 144 - 143.904016 = 0.095984[/tex] ft
[tex]\text{Average velocity} = \frac{0.095984}{0.001} = 95.984[/tex] ft/sec
[2.99, 3]:
[tex]\Delta t = 0.01[/tex]
[tex]s(2.99) = 16(2.99)^2 = 16 \times 8.9401 = 143.0416[/tex] ft
[tex]\Delta s = 144 - 143.0416 = 0.9584[/tex] ft
[tex]\text{Average velocity} = \frac{0.9584}{0.01} = 95.84[/tex] ft/sec
From these values, we can see that as the interval gets smaller, the average velocities are approaching a specific value. This helps us estimate the instantaneous velocity at [tex]t = 3[/tex].