Answer:
(c)approximately Normal, mean 112, standard deviation 1.414.
Step-by-step explanation:
To solve this problem, we have to understand the Central Limit Theorem
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
[tex]\mu = 112, \sigma = 20, n = 200[/tex]
Using the Central Limit Theorem
The distribution of the sample mean IQ is approximately Normal.
With mean 112
With standard deviation [tex]s = \frac{20}{\sqrt{200}} = 1.414[/tex]
So the correct answer is:
(c)approximately Normal, mean 112, standard deviation 1.414.
The distribution of sample means for 200 adults is approximately normal due to the Central Limit Theorem, with a mean of 112 and a standard deviation of 1.414, making option (e) the correct choice.
Explanation:The question asks about the distribution of the sample mean IQ of 200 randomly selected adults from a large population where the mean IQ is 112 with a standard deviation of 20. According to the Central Limit Theorem, when the sample size is large, the distribution of the sample means will be approximately normal (normal by the Central Limit Theorem), even if the source population itself is not perfectly normal.
The mean of the sampling distribution of the sample mean will be the same as the mean of the population, so the mean will be 112.
However, the standard deviation of the sampling distribution (often called the standard error) is equal to the standard deviation of the population divided by the square root of the sample size. So, the standard deviation for the sample mean for 200 adults would be 20 / √200, which is about 1.414. Therefore, the correct choice is (e) exactly Normal, mean 112, standard deviation 1.414.
An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip is drawn.
What is the probability of:
(A) a white chip on the first and a red on the second?
(B) two green chips being drawn?
(C) a red chip on the second, given that a white chip was drawn on the first?
Answer:
(A) 0.04
(B) 0.25
(C) 0.40
Step-by-step explanation:
Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.
Given:
R = 8, G = 10 and W = 2.
Total number of chips = 8 + 10 + 2 = 20
[tex]P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)= \frac{10}{20}=\frac{1}{2}\\P(W)= \frac{2}{20}=\frac{1}{10}[/tex]
As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.
(A)
Compute the probability of selecting a white chip on the first and a red on the second as follows:
[tex]P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04[/tex]
Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.
(B)
Compute the probability of selecting 2 green chips:
[tex]P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25[/tex]
Thus, the probability of selecting 2 green chips is 0.25.
(C)
Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:
[tex]P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40[/tex]
Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.
The probability of drawing a white chip on the first draw and a red chip on the second draw with replacement is 0.04. The probability of drawing two green chips is 0.25. The probability of drawing a red chip on the second draw, given that a white chip was drawn on the first draw, is 4.
Explanation:To find the probability of different events, we need to use the concept of probability. Probability is the likelihood of an event occurring. In this case, we have an urn with 8 red chips, 10 green chips, and 2 white chips.
(A) The probability of drawing a white chip on the first draw and a red chip on the second draw with replacement is:
P(white first and red second) = P(white first) * P(red second) = (2/20) * (8/20) = 16/400 = 0.04.
(B) The probability of drawing two green chips is:
P(green first and green second) = P(green first) * P(green second) = (10/20) * (10/20) = 100/400 = 0.25.
(C) The probability of drawing a red chip on the second draw, given that a white chip was drawn on the first draw, is:
P(red second | white first) = P(red second and white first) / P(white first) = (8/20) / (2/20) = 8/2 = 4.
A differential equation is given. Classify it as an ordinary differential equation (ODE) or a partial differential equation (PDE), give the order, and indicate the independent and dependent variables. If the equation is an ordinary differential equation, indicate whether the equation is linear or nonlinear.
5 (d^2x/dt^2) + 4 (dx/dt) + 9x = 2 Cos 3t
Answer:
the equation[tex]5(\frac{d^{2}x }{dt^{2} }) +4(\frac{dx}{dt})+9x=2cos3t[/tex] is a partial differential equation(PDE) because it contains unknown multi variables and their derivatives. This is a PDE of order 2.
The independent variable is x while the dependent variable is t.
The PDE is Linear.
Step-by-step explanation:
Partial Differential Equation (PDE): This is a differential equation that contains multi variables and their derivatives.
Ordinary Differential Equation (ODE): this is a differential equation containing a function of one independent variable and its derivatives.
Employment data at a large company reveal that 59 % of the workers are married, that 43 % are college graduates, and that 1/3 of the college graduates are married. What is the probability that a randomly chosen worker is: a) neither married nor a college graduate? Answer = % b) married but not a college graduate? Answer = % c) married or a college graduate?
Answer:
a. 74.63%
b. 33.63%
c. 87.67%
Step-by-step explanation:
If 59% (0.59) of the workers are married, then It means (100-59 = 41%) of the workers are not married.
If 43% (0.43) of the workers are College graduates, then it means (100-43= 57%) of the workers are not college graduates.
If 1/3 of college graduates are married, it means portion of graduate that are married = 1/3 * 43% = 1/3 * 0.43 = 0.1433.
For question a, Probability that the worker is neither married nor a college graduate becomes:
= (probability of not married) + (probability of not a graduate) - (probability of not married * not a graduate)
= 0.41 + 0.57 - (0.41*0.57) = 0.98 - 0.2337
= 0.7463 = 74.63%
For question b, probability that the worker is married but not a college graduate becomes:
=(probability of married) * (probability of not a graduate.)
= 0.59 * 0.57
= 0.3363 = 33.63%
For question c, probability that the worker is either married or a college graduate becomes:
=probability of marriage + probability of graduate - (probability of married and graduate)
= 0.59 + 0.43 - (0.1433)
= 0.8767. = 87.67%
The probability that a randomly chosen worker is neither married nor a college graduate is 0%. The probability that a worker is married but not a college graduate is approximately 34%. The probability that a worker is married or a college graduate is approximately 75%.
Explanation:To calculate the probabilities, we need to use the formulas for conditional probability. Let's solve each part:
a) To find the probability that a randomly chosen worker is neither married nor a college graduate, we can subtract the probability that the worker is married and the probability that the worker is a college graduate from 1. The probability of being married is 59%, and the probability of being a college graduate is 43%. Using the formula, 1 - 0.59 - 0.43 = 0.
b) To find the probability that a worker is married but not a college graduate, we can multiply the probability of being married and the probability of not being a college graduate. The probability of being married is 59%, and the probability of not being a college graduate is 57% (100% - 43%). Using the formula, 0.59 * 0.57 = 0.3363 (approximately 34%).
c) To find the probability that a worker is married or a college graduate, we can add the probabilities of being married and being a college graduate, and then subtract the probability of being both married and a college graduate to avoid double counting. The probability of being married is 59%, the probability of being a college graduate is 43%, and the probability of being both married and a college graduate is 1/3 of the college graduates (1/3 * 43%). Using the formula, 0.59 + 0.43 - (1/3 * 0.43) = 0.745 (approximately 75%).
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You are about to take the road test for your driver's license. You hear that only 38% of candidates pass the test the first time, but the percentage rises to 76% on subsequent requests. Estimate the average number of tests drivers take in order to get a license. Your simulation should use at least 20 runs.
Answer:
Answers may vary but will most likely be close to 2.
Step-by-step explanation
Given:first test:38%
second test:76%
SIMULATION FIRST TEST
Randomly select a 2-digit number.
If the digit is between 00 and 35 then you passed the test,else you did not pass the test.
SIMULATION SECOND TEST
Randomly select a 2-digit number.
if the digit is between 00 and 75 then you passed the test,else you did not pass the test.
SIMULATION TRIAL
Perform the simulation of the first test.if you did not pass the first test then perform the simulation of the second test.
Record the number of trials needed to pass the first or second test.
Repeat 20 times and take the average of the 20 recorded number of trials
(what is the sum of recorded values divided by 20).
Note:you will most likely obtain a result of about two trials needed.
The estimated average number of tests drivers take to get a license is 1.9
To estimate the average number of tests drivers take to get a license, we can simulate the process using a simple random experiment. We will assume that each test attempt is independent of the others and that the probabilities of passing are 38% on the first attempt and 76% on subsequent attempts. Here's how we can perform the simulation:
1. For each driver, simulate the number of tests they take by generating a random number between 0 and 1. If the number is less than or equal to 0.38 (38%), the driver passes on the first attempt. Otherwise, they proceed to a second attempt.
2. For the second attempt and any subsequent attempts, generate another random number. If the number is less than or equal to 0.76 (76%), the driver passes.
3. Repeat this process until the driver passes the test. Record the number of attempts it took.
4. Perform this simulation for 20 drivers to get a sample size that will help estimate the average number of tests taken.
5. Calculate the average number of tests by summing the number of tests taken by all drivers and dividing by the number of drivers (20).
Let's perform the simulation:
1. For the first attempt, generate a random number for each of the 20 drivers and check if it's less than or equal to 0.38.
2. For any subsequent attempts, generate a random number for each driver who did not pass on the first attempt and check if it's less than or equal to 0.76.
3. Record the number of attempts for each driver.
4. Sum the total number of attempts and divide by 20 to find the average.
Now, let's calculate the average number of tests based on the simulation: Assuming we have performed the simulation and recorded the number of attempts for each driver, we would have a list of numbers. For example, it might look something like this:
Driver 1: 1 attempt
Driver 2: 2 attempts
Driver 3: 1 attempt
Driver 20: 3 attempts
Let's say after performing the simulation, we have the following counts for each number of attempts:
1 attempt: 8 drivers
2 attempts: 7 drivers
3 attempts: 4 drivers
4 attempts: 1 driver
Now, we calculate the average number of tests taken by these 20 drivers:
Average = (8 * 1 + 7 * 2 + 4 * 3 + 1 * 4) / 20
Average = (8 + 14 + 12 + 4) / 20
Average = 38 / 20
Average = 1.9
Find the radian measure of an angle at the center of a circle with radius 70.0 cm that intercepts an arc length of 127 cm. The measure of the angle is nothing.g
Answer:
1.8413 is the radian measure of angle at the center of circle.
Step-by-step explanation:
We are given the following in the question:
Radius,r = 70.0 cm
Arc length,s = 127 cm
Formula:
[tex]\theta = \dfrac{s}{r}[/tex]
where [tex]\theta[/tex] is the angle measure in radians, s is the intercepted arc and r is the radius of the circle.
Putting the values, we get,
[tex]\theta = \dfrac{127}{70} = 1.8143[/tex]
Thus, 1.8413 is the radian measure of angle at the center of circle.
The radian measure of an angle in a circle, given an arc length of 127 cm and radius of 70 cm, is approximately 1.814. This is computed by dividing the arc length by the radius.
Explanation:In mathematics, the measure of an angle in radians in a circle is given by the ratio of the length of the arc that it subtends and the radius of the circle in which this occurs. This is represented by the formula: θ = s / r where s is the arc length and r is the radius.
Given the radius r = 70 cm and arc length s = 127 cm, you can substitute these values into the formula to get the radian measure of the angle. Therefore, by substitution we get: θ = 127 cm / 70 cm which simplifies to approximately 1.814 radians.
To note, a radian is the standard unit of angular measure, used in many areas of mathematics. An angle's measurement in radians is numerically equal to the length of a corresponding arc of a unit circle, hence why this method works.
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The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?
Answer:
92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 30.05, \sigma = 0.2[/tex]
What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?
This is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 29.75
X = 30.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30.5 - 30.05}{0.2}[/tex]
[tex]Z = 2.25[/tex]
[tex]Z = 2.25[/tex] has a pvalue of 0.9878
X = 29.75
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{29.75 - 30.05}{0.2}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
So there is a 0.9878 - 0.0668 = 0.9210 = 92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.
A parcel delivery company delivered 103,000 packages last year, when its average employment was 84 drivers. This year the firm handled 112,000 deliveries with 96 drivers. What was the percentage change in productivity over the past two years
Answer: 4.85%
Step-by-step explanation: productivity = parcel delivered / number of drivers.
At when parcel delivered was 103, 000 the average number of drivers was 84 thus making the productivity to be
103,000/ 84 = 1226.19
At when the parcel delivered was 112, 800 the average number of drivers was 96 thus making the productivity to be
112, 000/ 96 = 1166.67
Thus the change in productivity = 1226.19 - 1166.67 = 59.52.
Initial productivity = 1226.19
final productivity = 1166.67
% change in productivity = change in productivity / initial productivity * 100
% change in productivity = 59.52 /1226 * 100
% change in productivity = 4.85%
A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random. (a) When he flips the coin, what is the probability that it will show heads? (b) The coin shows heads. Now what is the probability that it is the fair coin?
Answer:
a) probability of choosing heads= 1/2 (50%)
b) probability of choosing the fair coin knowing that it showed heads is= 1/3 (33.33%)
Step-by-step explanation:
Since the unfair coin can have 2 heads or 2 tails , and assuming both are equally possible . then
probability of choosing the fair coin (named A)= 1/2
probability of choosing an unfair coin with 2 heads (named B)= (1-1/2)*1/2= 1/4
probability of choosing an unfair coin with 2 tails (named C)= (1-1/2)*(1-1/2)= 1/4
then
probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B + probability of choosing C * probability of getting heads from C =
1/2*1/2 + 1/4*1 + 1/4*0 = 2/4 = 1/2
the probability of choosing the fair coin knowing that it showed heads is
P(A/B) = P(A∩B)/P(B)
denoting event A= the coin is fair and event B= the result is heads
P(A∩B) = 1/2*1/2 = 1/4
but since we know now that that the unfair coin is not possible , the probability of choosing heads is altered:
P(B)=probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B = 1/2*1/2+1/2*1 = 3/4
then
P(A/B) = P(A∩B)/P(B) = (1/4)/(3/4) = 1/3
then the probability is 1/3
The original price of a toy boat was $50. The boat is marked up 15% before it’s sold. What is the selling price of the boat?
Answer: the selling price of the boat is $57.5
Step-by-step explanation:
The original price of a toy boat was $50. The boat is marked up 15% before it’s sold. This means that the amount by which the original price of the toy boat was increased would be
15/100 × 50 = 0.15 × 50 = $7.5
The selling price of the toy boat would be the sum of its original price and the amount by which it was marked up. It becomes.
50 + 7.5 = $57.5
In F, < CFD = < DFE, m< BFA = 4x, m< AFE = 3x + 12, and BE and AC are diameters.
Answer:
m arc DC=48°
Step-by-step explanation:
Angles in a Circle
We know [tex]\overline{BE}[/tex] and [tex]\overline{AC}[/tex] are diameters, so
[tex]m\angle BFA+m\angle AFE=180^o[/tex].
Since [tex]m\angle AFE=3x+12[/tex] and [tex]m\angle BFA=4x[/tex]
We set the equation
[tex]3x+12+4x=180^o[/tex]
Solving
[tex]7x=180-12=168[/tex]
[tex]x=24^o[/tex]
Thus
[tex]m\angle BFA=4(24)=96^o[/tex]
We also know
[tex]m\angle CFD\cong m\angle DFE[/tex]
Being [tex]\angle CFE[/tex] opposite to [tex]\angle BFA[/tex]
Then [tex]m\angle CFE=96^o[/tex]
It's divided in two equal angles [tex]\angle CFD\ and\ \angle DFE[/tex], thus m arc DC is half of 96°:
m arc DC=48°
First option
John works at a restaurant and gets paid $80 per week plus $5 per tip. This week, John wants to earn at least $300. How many tips, x, must he make to reach his goal?
Answer: John must make at least 44 tips to reach his goal.
Step-by-step explanation:
Let x represent the number of tips that John must make this week to reach his goal.
John works at a restaurant and gets paid $80 per week plus $5 per tip. This means that the total amount that John would earn in a week if he makes x tips would be
80 + 5x
This week, John wants to earn at least $300. This is expressed as
80 + 5x ≥ 300
5x ≥ 300 - 80
5x ≥ 220
x ≥ 220/5
x ≥ 44
student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The population's standard deviation is 2.10 hours per week. Based on a sample of 49 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.
A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?
B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.
Answer:
a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.
Step-by-step explanation:
To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
[tex]\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3[/tex]
A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?
This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.
By the Central Limit Theorem, the formula for Z is:
[tex]Z = \frac{X - \mu}{s}[/tex]
X = 7.1
[tex]Z = \frac{7.1 - 6.3}{0.3}[/tex]
[tex]Z = 2.67[/tex]
[tex]Z = 2.67[/tex] has a pvalue of 0.9962
X = 5.5
[tex]Z = \frac{5.5 - 6.3}{0.3}[/tex]
[tex]Z = -2.67[/tex]
[tex]Z = -2.67[/tex] has a pvalue of 0.0038
So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.
This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9
X = 6.7
[tex]Z = \frac{6.7 - 6.3}{0.3}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
X = 5.9
[tex]Z = \frac{5.9 - 6.3}{0.3}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918.
So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.
Complete the proof proving YUZ~VWZ
Answer:
1. ΔXYZ is a right Δ with altitude YU.
Given
2. ΔXYZ ~ ΔYUZ
Right Triangle Altitude Similarity Theorem
3. VW || XY
Given
4. ∠VWZ ≅ ∠XYZ
Corresponding angles
5. ∠Z ≅ ∠Z
Reflexive property of congruence
6. ΔXYZ ~ ΔVWZ
AA Similarity postulate
7. ΔYUZ ~ ΔVWZ
Transitive property of similar triangles
Step-by-step explanation:
The first statement is given in the problem. Since we know the altitude of a right triangle, we can use the Right Triangle Altitude Similarity Theorem to say that the triangles formed by the altitude are similar to each other and the original triangle.
Next, we are given in the problem statement that the lines VW and XY are parallel. Therefore, ∠VWZ and ∠XYZ are corresponding angles, which makes them congruent. And since ∠Z is equal to itself (by reflexive property), we can use AA similarity to say ΔXYZ and ΔVWZ are similar.
Finally, combining statements 2 and 6, we can use transitive property to say that ΔYUZ and ΔVWZ are similar.
According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110. How can both a function and its complement be 1 for the same input combination? What's wrong here?
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
The error originates from an incorrect application of DeMorgan's theorem. The correct complement of (W · X) + (Y · Z) is (W' + X') · (Y' + Z'). This corrects the discrepancy seen for WXYZ = 1110.
Explanation:The confusion here likely originates from a mistake in the DeMorgan's theorem application. According to DeMorgan's Theorem, the complement of (W · X) + (Y · Z) is given by (W' + X') · (Y' + Z'), not W' + X' · Y' + Z'.
So, if we have WXYZ = 1110, the given function (W · X) + (Y · Z) equals 1 because we have (1 · 1) + (1 · 0) = 1 + 0 = 1. Whereas, the correct complement function, (W' + X') · (Y' + Z') equals 0 because we have (0 + 0) · (0 + 1) = 0 · 1 = 0.
This explains why we were seeing both original function and the incorrectly applied complement function evaluating to 1 for the same input combination.
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The mean price for new homes from a sample of houses is $155,000 with a standard deviation of $10,000. Assume that the data set has a symmetric and bell-shaped distribution.
(a) Between what two values do about 95% of the data fall?
(b) Estimate the percentage of new homes priced between $135,000 and $165,000?
Answer:
a) 95% of the data falls between $135,000 and $175,000.
b) 81.5% of new homes priced between $135,000 and $165,000.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviations of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
We also have that:
50% of the measures are below the mean and 50% of the measures are above the mean.
34% of the measures are between 1 standard deviation below the mean and the mean, and 34% of the measures are between the mean and 1 standard deviations above the mean.
47.5% of the measures are between 2 standard deviations below the mean and the mean, and 47.5% of the measures are between the mean and 2 standard deviations above the mean.
49.85% of the measures are between 3 standard deviations below the mean and the mean, and 49.85% of the measures are between the mean and 3 standard deviations above the mean.
In this problem, we have that:
Mean = $155,000.
Standard deviation = $10,000.
(a) Between what two values do about 95% of the data fall?
By the Empirical Rule, 95% of the values fall within 2 standard deviations of the mean.
So
155000 - 2*10000 = 135,000
155000 + 2*10000 = 175,000
95% of the data falls between $135,000 and $175,000.
(b) Estimate the percentage of new homes priced between $135,000 and $165,000?
We have to find how many fall between $135,000 and the mean($155,000) and how many fall between the mean and $165,000
$135,000 and the mean
$135,000 is two standard deviations below the mean.
By the empirical rule, 47.5% of the measures are between 2 standard deviations below the mean and the mean.
So 47.5% of the measures are between $135,000 and the mean
Mean and $165,000
$165,000 is one standard deviation above the mean.
By the empirical rule, 34% of the measures are between the mean and 1 standard deviations above the mean.
So 34% of the measures are between the mean and $165,000.
$135,000 and $165,000
47.5% + 34% = 81.5% of new homes priced between $135,000 and $165,000.
Final answer:
The answer explains how to determine the range of values where 95% of data falls based on mean and standard deviation and estimates the percentage of new homes within a specific price range.
Explanation:
The mean price for new homes from a sample of houses is $155,000 with a standard deviation of $10,000.
(a) Between what two values do about 95% of the data fall?
Approximately 95% of the data falls within two standard deviations from the mean.The range would be $155,000 ± 2($10,000) = $155,000 ± $20,000 = $135,000 to $175,000.(b) Estimate the percentage of new homes priced between $135,000 and $165,000?
This range is $20,000 wide, which is equivalent to 2 standard deviations.Since the data is normally distributed, approximately 95% of the values are within 2 standard deviations of the mean.Therefore, we can estimate that around 95% of new homes are priced between $135,000 and $165,000.Suppose c and y vary together such that y= 4x + 8. a. Suppose x varies from x= 2 to x= 7.5. i. Over this interval, how much does x change by? ii. Over this interval, how much does y change by? iii. Over this interval, the change in y is how many times as large as the change in x? b. Suppose x varies from x= -5.1 to x= -5.1. i. Over this interval, how much does x change by? ii. Over this interval, how much does y change by? iii. Over this interval, the change in y is how many times as large as the change in x?
Answer:
ai) 5.5
aii) 22
aiii) 4
bi) 0
bii) 0
biii) undefined
Step-by-step explanation:
ai) The change in x is x2 -x1 = 7.5 -2 = 5.5 . . . . change in x
__
aii) The change in y is (4(7.5) +8) -(4(2) +8) = 4(7.5-2) = 22.0 . . . . change in y
__
aiii) The ratio of changes is (change in y)/(change in x) = 22/5.5 = 4
The change in y is 4 times the change in x.
___
bi) The difference is -5.1 -(-5.1) = 0 . . . . change in x
__
bii) The difference is (4(-5.1)+8) -(4(-5.1)+8) = 0 . . . . change in y
__
biii) The ratio of changes is (change in y)/(change in x) = 0/0 = undefined.
The multiplier of the change in x to get the change in y is undefined.
_____
Comment on part B
We know that the relative rates of change for x and y in this linear function are 1 : 4. However, we cannot compute that ratio directly when the change in x is 0. (The ratio holds for vanishingly small values of change in x, so is 4 in the limit as Δx → 0. That isn't what the problem asks.)
Final answer:
Over the given interval, x changes by 5.5 and y changes by 38. The ratio of the change in y to the change in x is approximately 6.91. Over the given interval, x remains constant at -5.1 and y does not change. Therefore, the ratio of the change in y to the change in x is undefined.
Explanation:
i. To find the change in x, we subtract the initial value from the final value: 7.5 - 2 = 5.5.
ii. To find the change in y, we substitute the initial and final values of x into the equation y = 4x + 8 and subtract: y_final - y_initial = (4 * 7.5 + 8) - (4 * 2 + 8) = 38.
iii. To find the ratio of the change in y to the change in x, we divide the change in y by the change in x: 38 / 5.5 ≈ 6.91.
iv. To find the change in x, we subtract the initial value from the final value: -5.1 - -5.1 = 0.
v. To find the change in y, we substitute the initial and final values of x into the equation y = 4x + 8 and subtract: y_final - y_initial = (4 * -5.1 + 8) - (4 * -5.1 + 8) = 0.
vi. To find the ratio of the change in y to the change in x, we divide the change in y by the change in x: 0 / 0.
Denzel bought headphones two months ago, Solo2 Beats by Dre, for $130. He gives them to his little brother and goes online to buy another for himself but they are now $160.
What is the percentage change in the headphone’s price?
a) 23% b) 81% c) 19% d) 21%
Answer:
a) 23%
Step-by-step explanation:
To find the price change as percentage :
Multiply 100 by 160 then divide it by 130
100 × 160 ÷ 130 = 123 approximately which means there's an additional 23% to the 100% price
A manufacturer of doorknobs has a production process that is designed to provide a doorknob with a target diameter of 2.5 inches. In the past, the standard deviation of the diameter has been 0.035 inch. In an effort to reduce the variation in the process, studies have been conducted that have led to a redesigned process. A sample of 25 doorknobs produced under the new process indicates a sample standard deviation of 0.025 inch. Assume diameters to be normally distributed. (a) Calculate a 99% confidence interval estimate for the variance of the redesigned process. (b) Given the results of (a), does the redesigned process result in a reduced variation
What is the critical value at the 0.05 level of significance for a goodness-of-fit test if there are six categories? Select one: a. 3.841 b. 5.991 c. 7.815 d. 11.070
Answer:
The correct option is (d) 11.070
Step-by-step explanation:
The test statistic for Goodness of fit test for k observations is:
[tex]\chi^{2}=\sum\frac{(Observed-Expected)^{2}}{Expected}[/tex]
This statistic follows a Chi-square distribution with (k - 1) degrees of freedom and α level of significance.
Here α = 0.05 and degrees of freedom is, k - 1 = 6 - 1 = 5 d.f.
Use the chi-square table for the critical value.
[tex]\chi^{2}_{(5)}=11.070[/tex]
Thus, the correct option is (d).
Which of the following best describes galena’s cleavage (see animation in placemark)? a. three directions at ~90° b. three directions not at ~90° c. two directions at ~90° d. two directions not at ~90°
Answer:
a)
Step-by-step explanation:
Given problem:
Which of the following best describes galena’s cleavage (see animation in placemark)?
Options:
a)three directions at ~90
b) three directions not at ~90°
c) two directions at ~90°
d) two directions not at ~90°
Solution:
- As per the animation where the place-mark is observed as a point. We can see that corner point is a point of intersection of three planes, left, right and bottom. We can also see that these three planes are at right angles @ 90 degrees to each other. Hence, option A.
In a normally distributed data set of how long customers stay in your store, the mean is 50.3 minutes and the standard deviation is 3.6 minutes.
Within what range would you expect 95% of your customers to stay in your store?
Answer:
You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 50.3 minutes
Standard deviation = 3.6 minutes
Within what range would you expect 95% of your customers to stay in your store?
Within 2 standard deviations of the mean.
So from
50.3 - 2*3.6 = 43.1 minutes
To
50.3 + 2*3.6 = 57.5 minutes
You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.
An astronaut with a mass of 90 kg (including spacesuit and equipment) is drifting away from his spaceship at a speed of 0.20 m/s with respect to the spaceship. The astronaut is equipped only with a 0.50-kg wrench to help him get back to the ship.
A). With what speed must he throw the wrench for his body to acquire a speed of 0.10 m/s?
B). In what direction relative to the spaceship must he throw the wrench? Towards the spaceship or away from the spaceship.
Answer:
Part A:
[tex]v_w=53.9\ m/s[/tex]
Part B:
Wrench is thrown away from the spaceship
Step-by-step explanation:
This the problem related to conservation of momentum.
According to the conservation of momentum:
Initial Momentum=Final Momentum
[tex]m_av_a=m_wv_w+(m_a-m_v)v[/tex]
where:
[tex]m_a[/tex] is the mass of astronaut
[tex]m_w[/tex] is the mass of wrench
[tex]m_a-m_w[/tex] is the mass when wrench is thrown
[tex]v_a[/tex] is the speed of astronaut
[tex]v_w[/tex] is the speed of wrench
v is the speed acquired
Part A:
(+ve sign for away from ship),( -ve sign for towards ship)
v= -0.10 m/s
[tex]90*0.2=0.5*v_w+(90-0.5)(-0.1)\\v_w=53.9\ m/s[/tex]
Part B:
Since velocity is +ve as calculated above and according to conditions:
+ve sign for away from ship -ve sign for towards shipWrench is thrown away from the spaceship
If Sam earns $97.50 for 15 hours of work,How many hours will he need to work to earn $130?
Answer:
Step-by-step explanation:
$97.50=15hours
$x=1hour
$15x= 97.50
X= 97.50/15
X= 6.5$
Therefore
1hour=6.5$
Therefore
130/6.5
=20hours
Suppose that 12 people enter an elevator on the 1st floor of a 24 floor building. Assume that all 12 independently pick a floor (above the first) randomly to get off on. What is the expected number offloors no one gets off on?
Answer:
∑E(x[tex]_{i}[/tex]) = 13.49167 floors
Step-by-step explanation:
The expected number of floors no one get off = ∑E(x[tex]_{i}[/tex]) where i is from 0 to 23
and E(x[tex]_{i}[/tex]) = ∑x[tex]_{i}[/tex]P(x[tex]_{i}[/tex])
here x[tex]_{i}[/tex] is the indicator of floor where no one gets off, its value is 0 when atleast one person get off on its floor and 1 when when no one gets off.
Now,
P(x[tex]_{i}[/tex]=1) = (22/23)¹²
P(x[tex]_{i}[/tex]=0) = [1-(22/23)¹²]
Now,
E(x[tex]_{i}[/tex]) = ∑x[tex]_{i}[/tex]P(x[tex]_{i}[/tex]) = 0* [1-(22/23)¹²] + 1*(22/23)¹² =0.586594704
For total number of floors where no one gets off
∑E(x[tex]_{i}[/tex]) = E(x₁)+E(x₂)+E(x₃)........................+E(x₂₃)
∑E(x[tex]_{i}[/tex]) = 23*0.586594704
∑E(x[tex]_{i}[/tex]) = 13.49167 floors
Final answer:
To answer, we calculate the expectation by considering the probability of a floor being skipped by all 12 people in a 24-floor building. The expected number of floors not chosen by anyone is around 8.
Explanation:
To find this, we can use the concept of probability. For any given floor (except the first), the probability that a single person does not choose it is (23/23) for the first choice, and the probability that they do choose another floor is (22/23), considering there are 23 floors above the first. Since each of the 12 people chooses independently, the probability that a given floor is not chosen by any of the 12 people is ((22/23)¹²).
With 23 possible floors (excluding the first) for people to exit on, the expected number of floors not chosen by anyone is 23 * ((22/23)¹²). A calculation reveals this to be approximately 7.9. Therefore, there's an expectation that around 8 floors will have no one exiting on them, under the given conditions.
When a certain basketball player takes his first shot in a game he succeeds with probability 1/2. If he misses his first shot, he loses confidence and his second shot will go in with probability 1/3. If he misses his first 2 shots then his third shot will go in with probability 1/4. His success probability goes down further to 1/5 after he misses his first 3 shots. If he misses his first 4 shots then the coach will remove him from the game. Assume that the player keeps shooting until he succeeds or he is removed from the game. Let X denote the number of shots he misses until his first success or until he is removed from the game.
a. Calculate the probability mass function of X.
b. Compute the expected value of X.
a. PMF: [tex]\(P(X=k) = \frac{1}{2^k}\) for \(k=0,1,2,3\); \(P(X > 3) = \frac{1}{2^3}\)[/tex]
b. [tex]\(E(X) = \frac{11}{4}\)[/tex]
a. To calculate the probability mass function (PMF) of X, we need to consider the different scenarios leading to the player's success or removal from the game. Let P(X = k) be the probability that the player misses \(k\) shots before succeeding or getting removed.
[tex]\[P(X = 0) = \frac{1}{2}\][/tex]
The player succeeds on the first shot.
[tex]\[P(X = 1) = \frac{1}{2} \cdot \frac{1}{3}\][/tex]
(The player misses the first shot, then succeeds on the second.)
[tex]\[P(X = 2) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4}\][/tex]
(The player misses the first two shots, then succeeds on the third.)
[tex]\[P(X = 3) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{1}{5}\][/tex]
(The player misses the first three shots, then succeeds on the fourth.)
[tex]\[P(X > 3) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5}\][/tex]
(The player misses the first four shots and is removed.)
b. To compute the expected value of X, we use the formula:
[tex]\[E(X) = \sum_{k=0}^{\infty} k \cdot P(X = k)\][/tex]
[tex]\[E(X) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) + \sum_{k=4}^{\infty} k \cdot P(X > 3)\][/tex]
Now plug in the values from part (a) into this formula and compute the sum. The result will give you the expected value of X.
The median weight of a boy whose age is between 0 and 36 months can be approximated by the function w (t )equals 8.38 plus 1.51 t minus 0.0069 t squared plus 0.000254 t cubed, where t is measured in months and w is measured in pounds. Use this approximation to find the following for a boy with median weight in parts a) through c) below. a) The rate of change of weight with respect to time.
Answer:
a) W'(t) = 1.51 -0.0138t + 0.000762t² pounds/day
Step-by-step explanation:
The median weight of a boy with age between 0 and 36 months is given by:
[tex]W(t) = 8.38+1.51t-0.0069t^2 + 0.000254t^3[/tex]
To find the rate of change of weight with respect to time, that is, the change in weight measured in pounds caused by a unit change in time, measured in days, simply derivate the weight function over time:
[tex]\frac{d(W(t)}{dt}= W'(t) = 1.51-0.0138t + 0.000762t^2[/tex]
The rate of change is 1.51 -0.0138t + 0.000762t² pounds/day.
One percent of a certain model of television have defective speakers. Suppose 500 televisions of this model are ready to ship. Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers (round off to second decimal place).
Answer:
49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
Step-by-step explanation:
For each television, there are only two possible outcomes. Either they have defective speakers, or they do not. The probabilities of each television having defective speakers are independent from each other. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]p = 0.01, n = 500[/tex]
Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
This is
[tex]P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{500,5}.(0.01)^{5}.(0.99)^{495} = 0.1764[/tex]
[tex]P(X = 6) = C_{500,6}.(0.01)^{6}.(0.99)^{494} = 0.1470[/tex]
[tex]P(X = 7) = C_{500,7}.(0.01)^{7}.(0.99)^{493} = 0.1048[/tex]
[tex]P(X = 8) = C_{500,8}.(0.01)^{8}.(0.99)^{492} = 0.0652[/tex]
[tex]P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1764 + 0.1470 + 0.1048 + 0.0652 = 0.4934[/tex]
49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers
The approximate probability that 5 to 8 out of 500 televisions have defective speakers is 0.49. This was calculated using the binomial distribution with n = 500 and p = 0.01. We summed P(X = 5), P(X = 6), P(X = 7), and P(X = 8) to find the result.
To find the approximate probability that 5 to 8 televisions out of 500 have defective speakers, we can use the binomial distribution.
Let X be the random variable representing the number of defective televisions in the shipment. Since 1% of the televisions are defective, the probability of a television being defective is p = 0.01, and the sample size is n = 500.The probability mass function for a binomial distribution is given by:
[tex]P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)[/tex]
where C(n, k) is the binomial coefficient.
We need to calculate P(5 ≤ X ≤ 8), which is P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8).
Using a binomial calculator or statistical software, we get:
P(X = 5) ≈ 0.1755P(X = 6) ≈ 0.1447P(X = 7) ≈ 0.1040P(X = 8) ≈ 0.0681Summing these probabilities gives:
P(5 ≤ X ≤ 8) ≈ 0.1755 + 0.1447 + 0.1040 + 0.0681
≈ 0.4923
Therefore, the approximate probability that 5 to 8 televisions have defective speakers is 0.49 (rounded to two decimal places)
Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad non-wetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found 727 that were judged defective, inspector B found 756 such joints, and 940 of the joints were judged defective by at least one of the inspectors.
Suppose that one of the 10,000 joints is randomly selected.
(a) What is the probability that the selected joint was judged to be defective by neither of the two inspectors? (Enter your answer to four decimal places.)
(b) What is the probability that the selected joint was judged to be defective by inspector B but not by inspector A? (Enter your answer to four decimal places.)
Answer:
(a) The probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.
(b) The probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.
Step-by-step explanation:
The sample of joints randomly selected is, n = 10,000.
Number of joints judged defective by inspector A is, n (A) = 727.
The probability that a joint is judged defective by inspector A is:
[tex]P(A)=\frac{n(A)}{n} =\frac{727}{10000} =0.0727[/tex]
Number of joints judged defective by inspector B is, n (B) = 756.
The probability that a joint is judged defective by inspector B is:
[tex]P(B)=\frac{n(B)}{n} =\frac{756}{10000} =0.0756[/tex]
Number of joints judged defective by at least one of the inspectors is,
n (At least 1) = 940.
The probability that a joint is judged defective by at least one of the inspectors is:
[tex]P(At\ least\ 1)=\frac{n(At\ least\ 1)}{n} =\frac{940}{10000}=0.094[/tex]
(a)
Compute the probability that a selected joint was judged to be defective by neither of the two inspectors as follows:
P (At least 1) = 1 - P (Less than 1)
= 1 - P (None)
P (None) = 1 - P (At least 1)
[tex]=1-0.094\\=0.906[/tex]
Thus, the probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.
(b)
Compute the probability that a selected joint was judged to be defective by inspector B but not by inspector A as follows:
P (B but not A) = P (At least 1) - P (A)
[tex]=0.094-0.0727\\=0.0213[/tex]
Thus, the probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.
Given the following data, find the weight that represents the 73rd percentile. Weights of Newborn Babies 8.2 6.6 5.6 6.4 7.9 7.1 6.5 6.0 7.8 8.0 6.8 8.8 9.3 7.7 8.8
To find the 73rd percentile of the given data, we need to arrange the weights in ascending order and calculate the rank. Then, we interpolate to find the weight at the desired percentile.
Explanation:To find the weight that represents the 73rd percentile, we need to follow these steps:
Arrange the weights in ascending order: 5.6, 6.0, 6.4, 6.5, 6.6, 6.8, 7.1, 7.7, 7.8, 7.9, 8.0, 8.2, 8.8, 8.8, 9.3Calculate the rank of the percentile by using the formula: rank = (percentile/100) * (n - 1) + 1, where n is the number of data points. In this case, n = 15.The rank of the 73rd percentile is (73/100) * (15 - 1) + 1 = 11.02.Since the rank is not a whole number, we need to interpolate to find the weight. The weight at rank 11 is 8.0 and the weight at rank 12 is 8.2.Use the formula for interpolation: weight = weight at lower rank + (rank - lower rank) * (weight at higher rank - weight at lower rank), which gives: weight = 8.0 + (11.02 - 11) * (8.2 - 8.0) = 8.02.Therefore, the weight that represents the 73rd percentile is 8.02.
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A bee with a velocity vector r ′ ( t ) r′(t) starts out at ( 3 , − 6 , 10 ) (3,−6,10) at t = 0 t=0 and flies around for 9 9 seconds. Where is the bee located at time t = 9 t=9 if ∫ 9 0 r ′ ( u ) d u = 0
The velocity vector of the bee is simply the rate of change of the position of the bee.
The bee is located at [tex]\mathbf{ (3,-6,10)}[/tex] at 9 seconds.
The given parameters are:
[tex]\mathbf{r'(t) = (3,-6,10)}[/tex], when t = 0
A vector is represented as:
[tex]\mathbf{r'(t) = xi + yj + zk}[/tex]
From the question, we have:
[tex]\mathbf{\int\limits^9_0 {r'(u)} \, du = 0}[/tex]
Integrate
[tex]\mathbf{r(u)|\limits^9_0 = 0}[/tex]
Expand
[tex]\mathbf{r(9) - r(0) = 0}[/tex]
Rewrite as:
[tex]\mathbf{r(9) = r(0) }[/tex]
Recall that: [tex]\mathbf{r'(t) = (3,-6,10)}[/tex]
So, we have:
[tex]\mathbf{r(9) = 3i -6j + 10k }[/tex]
Also, we have:
[tex]\mathbf{xi + yj + zk = 3i -6j + 10k }[/tex]
By comparison;
[tex]\mathbf{x = 3}[/tex]
[tex]\mathbf{y = -6}[/tex]
[tex]\mathbf{z = 10}[/tex]
So, the bee is located at [tex]\mathbf{ (3,-6,10)}[/tex] at 9 seconds.
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After 9 seconds, the bee returns to its initial position (3, -6, 10) because the integral of the velocity function from 0 to 9 seconds is zero, indicating that the net displacement of the bee in these 9 seconds is zero.
Explanation:The question is essentially asking where the bee is located based on its velocity vector, initial position, and the change in time. To answer this, we will use the concept of integrals from calculus. The integral of the velocity function, r′(t), over a period of time gives the displacement of the bee. However, it's given that ∫9 0 r′(u) du = 0. This tells us that the net displacement of the bee in that 9 seconds is zero.
Displacement refers to the change in position of an object, so if the net displacement in these 9 seconds is zero, then the bee is at the same position at t=9 as at t=0. In other words, after travelling around for 9 seconds, the bee returns to the same starting point. Therefore, at time t=9, the bee is located at the same initial position, that is, (3, -6, 10).
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