Give as good a big-O estimate as possible for each of these functions. For the function g in your estimate f(x) ∈ O(g(x)), the function should be much simpler than f. (a) f(n) = (n log n + n 2 )(n 3 + 2)
Answers
The second and third function are missing and they are;
The second function; ((2^n) + n^(2))(n^(3) + 3^(n))
The third function is;
(n^(n) + n^(2n) + 5^(n))(n! + 5^(n))
Answer:
A) O((n^(3)) logn)
B) O(6^(n))
C) O(n^(n)(n!))
Step-by-step explanation:
I've attached explanation of the 3 answers.
The big-O estimate for the function f(n) = [tex]f(n) = (n log n + n^2)(n^3 + 2)[/tex] is [tex]O(n^5)[/tex] , as the product of the highest order terms [tex]n^2[/tex] and [tex]n^3[/tex] is [tex]n^5[/tex], which dominates the function's growth.
To find a big-O estimate for the function [tex]f(n) = (n log n + n^2)(n^3 + 2)[/tex], we must look for the term with the highest growth rate as n goes to infinity. This function is the product of two terms, n log [tex]n log n + n^2[/tex] and [tex]n^3 + 2[/tex]. The term with the highest growth rate in the first part is n2, and in the second part, it is n3. Therefore, the product of the two highest order terms will give us the term that grows the fastest: [tex]n^2 \times n^3 = n^5[/tex].
Thus, the given function is in big-O of [tex]n^5[/tex], which means [tex]f(n) \in O(n^5)[/tex]. The higher order term [tex]n^5[/tex]completely dominates the growth of the function, making the other terms and constants irrelevant for big-O notation. Remember that big-O notation provides an upper bound and is used to describe the worst-case scenario for the growth rate of the function.
Learn more about big-O here:
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Related Questions
The Operations Manager likes your answer and thanks you for the analyses. He then asks you to do the ABC analysis. The manager adds that the company typically does ABC analysis such that the A segment (all A items togehter) accounts for at least 70% of value, B segment for at least 15% and C segment for at most 10% of value.
SKU Number of Items Value ($/item) Weight (lb/item)
A 220 67 52
B 225 3 59
C 245 21 60
D 145 7 48
E 230 26 42
F 240 83 49
Imagine, you want to first do the ABC analysis based on SKU value.
What percentage of the total number of items is accounted for by SKUs in segment B?
What percentage of the total value of inventory is accounted for by SKUs in segment B?
What percentage of the total weight of inventory is accounted for by SKUs in segment B?
Answers
Answer:
Requirements are computed below.
Step-by-step explanation:
For ABC analysis:
Step1: Calculate the total value of each item.
SKU Number of Items Value ($/item) Value of the product ($)
A 220 67 14740
B 225 3 675
C 245 21 5145
D 145 7 1015
E 230 26 5980
F 240 83 19920
Step2: Reorganize the table on the basis of total value in descending order, & calculate % share of total value of investment and cumulative % share of total value of investment.
SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment
SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment
F 240 83 19920 41.96 41.96 A
A 220 67 14740 31.05 73.01 A
E 230 26 5980 12.60 85.60 B
C 245 21 5145 10.84 96.44 B
D 145 7 1015 2.14 98.58 C
B 225 3 675 1.42 100.00 C
----------
% of total number of items in segment B = (230+245)/1305 = 36.4%
% of total value of inventory in segment B = (12.6 + 10.4)/100 = 23.43%
% of the total weight of inventory in segment B = (9660+14700)/67795 = 35.93%
SKU Number of Items Value ($/item) Weight (lb/item) Total weight
A 220 67 52 11440
B 225 3 59 13275
C 245 21 60 14700
D 145 7 48 6960
E 230 26 42 9660
F 240 83 49 11760
Translate the statement into a confidence interval for p. Approximate the level of confidence.
In a survey of 1000 U.S. adults, 19% are concerned that their taxes will be audited by the Internal Revenue Service. The survey's margin of error is plus or minus 3%. (Source: Rasmussen Reports)
Approximate the level of confidence.
The interval is approximately a _______% confidence interval.
Answers
Answer:
The interval is approximately a 98% confidence interval
Step-by-step explanation:
From the question : Error, E= 3% = 0.03, Total population, n=1000, number of people that are concerned that their taxes will be audited, p = 19% = 0.19
E^{2}=z_{\alpha/2}^{2}\cdot \frac{p(1-p)}{n}
0.03^{2}=z_{\alpha/2}^{2}\cdot \frac{0.19(1-0.19)}{1000}
z_{\alpha/2}^{2}=5.848
z_{\alpha/2}=2.418
Area right to 2.418 is 0.0078. So
\alpha/2=0.0078
Therefore \alpha=0.0156\approx 0.02
Thus, the interval is approximately a 98% confidence interval.
Answer:
98 percent
Step-by-step explanation:
plato
In automobile mileage and gasoline-consumption testing, 13 automobiles were road tested for 300 miles in both city and highway driving conditions. The following data were recorded for miles-per-gallon performance. City 16.2 16.7 15.9 14.4 13.2 15.3 16.8 16.0 16.1 15.3 15.2 15.3 16.2 Highway 19.1 20.3 18.0 18.3 18.9 17.1 16.9 18.3 18.7 20.8 19.1 18.2 18.4 Use the mean, median, and mode to make a statement about the difference in performance for city and highway driving. Compute the mean for the miles-per-gallon performance for city and highway driving. (Round your answers to two decimal places.) city 15.9 miles per gallon highway 15.3 miles per gallon Is the mean mileage better on the highway than in the city? Yes No Compute the median for the miles-per-gallon performance for city and highway driving. city miles per gallon highway miles per gallon Is the median mileage better on the highway than in the city? Yes No Compute the mode(s) for the miles-per-gallon performance for city and highway driving. (Enter your answers as a comma-separated list.) city miles per gallon highway miles per gallon Is the modal mileage better on the highway than in the city? Yes No Make a statement about the difference in performance for city and highway driving conditions. Miles-per-gallon performance is in highway driving condition than in city driving conditions.
Answers
Answer:
A. No
City driving mean requires less miles than Highway driving mean.
B. No
City driving median requires less miles than Highway driving median.
C. No
City driving mode requires less miles than Highway driving mode
D.
The driving conditions in the City is better than the driving conditions in the Highway because all measures of centers are lesser than those of Highway.
Step-by-step explanation:
If z= 0.65, then the raw score is 0.65 standard deviations above the mean. True or false? explain
Answers
Answer:
True, this is the meaning of the z-score. Z = 0.65 means that the raw score is 0.65 standard deviations above the mean.
Step-by-step explanation:
The Z-score measures how many standard deviations a raw score is from the mean.
For example, a z-score of -2 means that the raw score is 2 standard deviations below the mean.
Another example, a z-score of 2 means that the raw score is 2 standard deviations above the mean.
If z= 0.65, then the raw score is 0.65 standard deviations above the mean.
True, this is the meaning of the z-score. Z = 0.65 means that the raw score is 0.65 standard deviations above the mean.
On July 1, 2015, Frank Corp. Purchased $100,000 of 8% bonds at face value. Interest is paid annually on June 30. If the accounting year for Frank ends at December 31, 2015, what will be reported with respect to the bonds on that date
Answers
Answer:
interest income in amount of $4000 will be accrued
Step-by-step explanation:
given data
principal = $100,000
rate = 8 percent = 0.08
time period = 6 month ( July - December ) = [tex]\frac{1}{2}[/tex] year
solution
we get here interest at December 31, 2015 that is express as
interest = principal × rate × time ..................1
put here value and we will get
interest = $100,000 × 0.08 × [tex]\frac{1}{2}[/tex]
interest = $100,000 × 0.08 × 0.5
interest = $4000
so interest income in amount of $4000 will be accrued
The nicotine content in cigarettes of a certain brand is known to be right-skewed with mean (in milligrams) μ and a known standard deviation σ = 0.17. The brand advertises that the mean nicotine content of its cigarettes is 1.5, but measurements on a random sample of 100 cigarettes of this brand give a sample mean of x ¯ = 1.53.
Is this evidence that the mean nicotine content is actually higher than advertised? To answer this, test the hypotheses H₀: μ = 1.5, Hₐ: μ > 1.5 at the 0.05 level of significance.
What do you decide?
Answers
Answer:
There is enough evidence to support the claim the mean nicotine content is actually higher than advertised.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 1.5
Sample mean, [tex]\bar{x}[/tex] = 1.53
Sample size, n = 100
Alpha, α = 0.05
Population standard deviation, σ = 0.17
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 1.5\\H_A: \mu > 1.5[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{1.53 - 1.5}{\frac{0.17}{\sqrt{100}} } = 1.7647[/tex]
Now, we calculate the p-value from the standard normal table.
P-value = 0.038807
Since the p-value is less than the significance level we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim the mean nicotine content is actually higher than advertised.
Answer:
Yes, the mean nicotine content is actually higher than advertised.
Step-by-step explanation:
We are given that the nicotine content in cigarettes of a certain brand is known to be right-skewed with mean (in milligrams) μ and a known standard deviation σ = 0.17.
Also, the brand advertises that the mean nicotine content of its cigarettes, [tex]\mu[/tex] is 1.5, but measurements on a random sample of 100 cigarettes of this brand give a sample mean of x bar = 1.53.
Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.5
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 1.5
The test statistics used here will be;
T.S. = [tex]\frac{xbar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
Here, n = sample size = 100
So, test statistics = [tex]\frac{1.53 - 1.5}{\frac{0.17}{\sqrt{100} } }[/tex] = 1.765
Now, at 0.05 level of significance, the standard z table gives critical value of 1.6449. Since our test statistics is more than the critical value as 1.765 > 1.6449 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region.
Therefore, we conclude that the mean nicotine content is actually higher than advertised.
A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin. a. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P(A). (Round to two decimal places)
Answers
Answer:
P(A) = 3/20
Step-by-step explanation:
P(A)=P(blue)P(head)=(3/10)(1/2)=3/20
as there are 10 cards in total, out of which 3 are blue so the probability to get the blue card is, P(blue) = 3/10. and the probability of getting a head when a coin is tossed is P(head) = 1/2.
So in total
P(A) = P(blue)*p(head) = (3/10)*(1/2) = 3/20 = 0.15
Final answer:
P(A) = 0.15.
Explanation:
To find P(A), the probability of event A happening, we need to consider two separate events: picking a blue card and tossing a head on a coin.
First, the probability of picking a blue card from the special deck is 3 out of 10, because there are 3 blue cards among the 10 total cards.
Next, the probability of tossing a head on a coin is 1 out of 2, since a coin has two sides and both outcomes (heads and tails) are equally likely.
To find the combined probability P(A) of the two independent events, we multiply the probabilities:
[tex]P(A) = P(Blue Card) \(\times\) P(Head Coin Toss)[/tex]
Thus:
[tex]P(A) = (3/10) \(\times\) (1/2) = 3/20 = 0.15[/tex]
Rounded to two decimal places, P(A) = 0.15.
New York City is one of the most expensive cities in the US for lodging. The mean hotel room rate is $244.00 per night; assume that room rates are normally distributed, with the standard deviation of $55.00 What is the probability that a hotel room costs between $250.00 and $285.00?
Answers
Answer:
22.96% probability that a hotel room costs between $250.00 and $285.00
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 = 244, \sigma = 55[/tex]
What is the probability that a hotel room costs between $250.00 and $285.00?
This is the pvalue of Z when X = 285 subtracted by the pvalue of Z when X = 250. So
X = 285
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{285 - 244}{55}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a pvalue of 0.7734
X = 250
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{250 - 244}{55}[/tex]
[tex]Z = 0.11[/tex]
[tex]Z = 0.11[/tex] has a pvalue of 0.5438
0.7734 - 0.5438 = 0.2296
22.96% probability that a hotel room costs between $250.00 and $285.00
A professor planned to give an examination in a large class on the Monday before Thanksgiving vacation. Some students asked whether he could change the date because so many of their classmates had at least one other exam on that date. They speculated that at least 40% of the class had this problem. The professor agreed to poll the class, and if there was convincing evidence that the proportion with at least one other exam on that date was greater than .40, he would change the date. Of the 250 students in the class, 109 reported that they had another exam on that date. What proportion of the class reported that they had another exam on the date?
Answers
Answer:
0.436
Step-by-step explanation:
Given that a professor planned to give an examination in a large class on the Monday before Thanksgiving vacation
Some students asked whether he could change the date because so many of their classmates had at least one other exam on that date.
They speculated that at least 40% of the class had this problem.
The professor agreed to poll the class, and if there was convincing evidence that the proportion with at least one other exam on that date was greater than .40, he would change the date.
No of students in total = 250
Reported they had another exam = 109
proportion of the class reported that they had another exam on the date
= [tex]\frac{109}{250} =0.436[/tex]
A can in the shape of a right circular cylinder is required to have a volume of 700 cubic centimeters. The top and bottom are made up of a material that costs 8� per square centimeter, while the sides are made of material that costs 5� per square centimeter. Find a function that describes the total cost of the material as a function of the radius r of the cylinder
Answers
Answer:
the cost function is Cost=7000 m*$ /R + 50.265 $/m² * R²
Step-by-step explanation:
then the cost function is
Cost= cost of side area+ cost of top + cost of bottom = 2*π*R*L * 5$/m² +
π*R² * 8$/m² + π*R² * 8$/m²
since the volume V is
V=π*R²*L → V/(π*R²)=L
then
Cost=2*π*R*V/(π*R²) * 5$/m² + π*R² * 8$/m² + π*R² * 8$/m²
replacing values
Cost=2*700 m³ /R * 5$/m² + π*R² * 16$/m² = 7000 m*$ /R + 50.265 $/m² * R²
thus the cost function is
Cost=7000 m*$ /R + 50.265 $/m² * R²
Answer: 50.24r² + 7000/r
Step-by-step explanation:
The formula for determining the volume of a cylinder is expressed as
Volume = πr²h
Since the volume of the can is 700cm³, then
πr²h = 700
h = 700/πr²
The formula for determining the total surface area of a cylinder is expressed as
Total surface area = 2πr² + 2πrh
The surface area of the top and bottom of the can is 2πr².
Since top and bottom are made up of a material that costs $8 per square centimeter, then the cost is
2πr² × 8 = 16πr²
Since π = 3.14, the surface area of the top and bottom of the cylindrical can is
16 × 3.14 × r² = 50.24r²
The surface area of the side of the can is
2πrh = 2πr × 700/πr²
= 1400/r
Since the the sides are made of material that costs $5 per square centimeter, then the cost is
1400/r × 5 = 7000/r
The total cost of the material as a function of the radius, r of the cylinder is
50.24r² + 7000/r
A bacteria colony increases in size at a rate of 4.0565e1.3t bacteria per hour. If the initial population is 54 bacteria, find the population four hours later. (Round your answer to the nearest whole number.) bacteria
Answers
Answer:
Population of bacteria four hours later [tex]A=616.5187[/tex] or nearest whole number
[tex]A=616[/tex]
Step-by-step explanation:
Given,
Let amount of bacteria four hours later is=A
Rate of increase of bacteria per hour [tex]=\frac{\partial x}{\partial t}=4.0565e^{1.3t}[/tex]
Initial population of bacteria is =54
Time 4 hours
Find the population of bacteria 4 hours later
Solution,
[tex]\int_{54}^{A}dx=\int_{0}^{4}4.0565e^{1.3t}dt[/tex]
[tex]\left [ x \right ]_{54}^{A}=4.0565/1.3\left [e^{1.3t} \right ]_{0}^{4}[/tex]
[tex]A-54=4.0565/1.3\left ( e^{1.3\times 4} -1\right )[/tex]
[tex]A-54=562.5187301[/tex]
[tex]A=562.5187301+54[/tex]
[tex]A=616.5187[/tex]
Population of bacteria 4 hours later is[tex]A=616[/tex]
The population of the bacteria four hours later was 39708
What is an exponential function?
An exponential function is in the form:
y = abˣ
Where a is the initial value of y and b is the multiplication factor.
Given that the initial population is 54 bacteria and it increases in size at a rate of 4.0565e^(1.3t) bacteria per hour. In 4 hours:
[tex]Bacteria\ population=54 * 4.0565e^{1.3*4}=39708[/tex]
The population of the bacteria four hours later was 39708
Find out more on exponential function at: https://brainly.com/question/12940982
A well-mixed cookie dough will produce cookies with a mean of 66 chocolate chips apiece. What is the probability of getting a cookie with at least 55 chips? Round your answer to four decimal places.
Answers
Answer:
0.7262
Step-by-step explanation:
Let random variable = X
Probability of getting a cookie with at least 55 chips:
P(55≤ X ≤78), where X1 = 55 and X2 = 78
Let mean, U = 66 and standard deviation, S.D. = 10
Let Z = X - U/S.D; Z1 = X1 - U/S.D and Z2 = X2 - U/S.D
Z1 = 55 - 66/10 = - 1.1
Z2 = 78 - 66/10 = 1.2
P(X1≤ X ≤ X2) = P(X1 -U/S.D ≤ X - U/S.D ≤ X2 - U/S.D)
P(Z1 ≤ Z ≤ Z2) = P(0 ≤ Z ≤ Z1) + P(0 ≤ Z ≤ Z2)
Hence, (55 ≤ X ≤ 78) = P( - 1.1 ≤ Z ≤ 1.2)
∴ P(-1.1 ≤ Z ≤ 1.2) = P(0 ≤ Z ≤ 1.1) + (0 ≤ Z ≤ 1.2)
= 0.3413 + 0.3849 = 0.7262
Seventy percent of children who go to the doctor have fevers. Of those with fevers, 30% also have a rash. Of those without fevers, 20% have a rash. What is the probability that a child at the doctor's office with a rash does not have a fever
Answers
Answer:
The probability that a child with a rash does not have a fever is 22%
Step-by-step explanation:
1. Probability of having fever:
[tex]P(fever)=0.70[/tex]
2. Probability of not having fever:
[tex]P(not fever)=1-P(fever)\\P(not fever)=1-0.70\\P(not fever)=0.30[/tex]
3. Probability of fave fevers and a rash:
[tex]P(fever and rash)=(0.70)(0.30)=0.21[/tex]
4. Probability of having a rash but not a fever:
[tex]P(rash and not fever)=(0.30)(0.20)=0.60[/tex]
5. Probability of having a rash:
[tex]P(rash)=P(rash and fever)+P(rash and no fever)\\P(rash)=0.21+0.06=0.27[/tex]
6. Probability a child with a rash does not have a fever
[tex]P=\frac{P(rash and not fever)}{P(rash)} =\frac{0.06}{0.27} =0.22[/tex]
22% of the child at the doctor's office with a rash does not have a fever.
Final answer:
To find the probability that a child with a rash does not have a fever, we analyze the given percentages, calculate how many children have a rash with and without fever, and then find the ratio of those without a fever to the total number with a rash, resulting in a probability of approximately 22.22%.
Explanation:
The question asks: What is the probability that a child at the doctor's office with a rash does not have a fever? To solve this, let's start by analyzing the given percentages.
70% of children who go to the doctor have fevers.
Of those with fevers, 30% also have a rash.
Of those without fevers, 20% have a rash.
To find the probability that a child with a rash does not have a fever, we need to calculate the proportion of children with a rash who are fever-free compared to all children with a rash.
Step-by-step Calculation:
Assuming 100 children visit the doctor: 70 will have fevers, and 30 will not.
Of the 70 with fevers, 21 (30% of 70) have a rash.
Of the 30 without fevers, 6 (20% of 30) have a rash.
In total, 27 children have a rash (21 with fever + 6 without).
The probability a child with a rash does not have a fever is the number of children with a rash but no fever divided by the total number of children with a rash: 6/27.
This calculation shows that the probability of a child having a rash but no fever is 6/27 or approximately 22.22%.
A building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18 \pi18π18, pi meters.
What is the area AAA of the cross section of the column?
Give your answer in terms of pi.
Answers
The area of the cross section of the column is [tex]18 \pi \ m^2[/tex]
Explanation:
Given that a building engineer analyzes a concrete column with a circular cross section.
Also, given that the circumference of the column is [tex]18 \pi[/tex] meters.
We need to determine the area of the cross section of the column.
The area of the cross section of the column can be determined using the formula,
[tex]Area= \pi r^2[/tex]
First, we shall determine the value of the radius r.
Since, given that circumference is [tex]18 \pi[/tex] meters.
We have,
[tex]2 \pi r=18 \pi[/tex]
[tex]r=9[/tex]
Thus, the radius is [tex]r=9[/tex]
Now, substituting the value [tex]r=9[/tex] in the formula [tex]Area= \pi r^2[/tex], we get,
[tex]Area = \pi (9)^2[/tex]
[tex]Area = 81 \pi[/tex]
Thus, the area of the cross section of the column is [tex]18 \pi \ m^2[/tex]
Identify the polynomial.
a 2 + b - cd 3
Answers
Answer: 2a + b - 3cd
Step-by-step explanation:
a 2 + b - cd 3
2a + b - 3cd
Find the largest interval which includes x = 0 for which the given initial-value problem has a unique solution. (Enter your answer using interval notation.) (x − 3)y'' + 4y = x, y(0) = 0, y'(0) = 1
Answers
Answer:
[tex](-\infty,3)[/tex]
Step-by-step explanation:
We are given that
[tex](x-3)y''+4y=x[/tex]
[tex]y''+\frac{4}{x-3}y=\frac{x}{x-3}[/tex]
y(0)=0
y'(0)=1
By comparing with
[tex]y''+p(x)y'+q(x)y=g(x)[/tex]
We get
[tex]p(x)=\frac{4}{x-3}[/tex]
[tex]g(x)=\frac{x}{x-3}[/tex]
q(x)=0
p(x),q(x) and g(x) are continuous for all real values of x except 3.
Interval on which p(x),q(x) and g(x) are continuous
[tex](-\infty,3)[/tex]and (3,[tex]\infty)[/tex]
By unique existence theorem
Largest interval which contains 0=[tex](-\infty,3)[/tex]
Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=[tex](-\infty,3)[/tex]
The largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]
The given parameters are:
(x − 3)y'' + 4y = x,
y(0) = 0
y'(0) = 1
Divide the equation (x − 3)y'' + 4y = x through by (x - 3)
[tex]y'' + \frac{4y}{x - 3} = \frac{x}{x - 3}[/tex]
Compare the above equation to the following equation
y" + p(x) y' + q(x)y = g(x)
Then, we have:
[tex]p(x) = \frac{4y}{x - 3}[/tex]
[tex]q(x) = 0[/tex]
[tex]g(x) = \frac x{x - 3}[/tex]
The domains of functions p(x) and g(x) are all set of real values except 3
This is represented as:
[tex](-\infty, 3)\ u\ (3,\infty)[/tex]
Using the unique existence theorem, we have:
The largest interval that contains x = 0 is [tex](-\infty, 3)[/tex]
Hence, the largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]
Read more about intervals at:
https://brainly.com/question/1399055
During the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two minutes.
What is the expected number of calls in one hour?
Answers
Answer:
Let X the random variable who represent the number of occurences in a period of time for the calls.
For this case we have the following parameter [tex] \lambda = 1 \frac{call}{2 minutes}[/tex]
And we are interested in the expected number of calls in one hour.
We know that 1 hr = 60 mins so then the expected number of calls that arrive in one hour are:
[tex] \lambda = \frac{1 call}{2 minutes} * \frac{60 minutes}{1 hour} = 30 calls per hour[/tex]
Step-by-step explanation:
Definitions and concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]
And the parameter [tex]\lambda[/tex] represent the average ocurrence rate per unit of time.
Solution to the problem
Let X the random variable who represent the number of occurences in a period of time for the calls.
For this case we have the following parameter [tex] \lambda = 1 \frac{call}{2 minutes}[/tex]
And we are interested in the expected number of calls in one hour.
We know that 1 hr = 60 mins so then the expected number of calls that arrive in one hour are:
[tex] \lambda = \frac{1 call}{2 minutes} * \frac{60 minutes}{1 hour} = 30 calls per hour[/tex]
It is known that 70% of all brand A external hard drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that n = 15 drives are randomly selected. Let X = the number of successes in the sample. The statistic X/n is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value X/n is 0.2, corresponding to X = 3. What is the probability of this value (what kind of random variable is X)?] (Round your answers to three decimal places.)
Answers
Answer:
The probability distribution of [tex]\hat p[/tex] is, [tex]N(0.70, 0.032)[/tex].
Step-by-step explanation:
The random variable X is defined as the number of brand A external hard drives work in a satisfactory manner throughout the warranty period.
The sample selected is of size, n = 15.
The probability of selecting a hard drive that works in a satisfactory manner throughout is, p = 0.70.
Every hard drive works independently of the others.
The random variable X follows a Binomial distribution with parameters n = 15 and p = 0.70.
The probability mass function of X, the binomial random variable is:
[tex]P(X=x)={15\choose x}0.70^{x}(1-0.70)^{15-x};\ x=0,1,2,3...[/tex]
The probability distribution of X is shown below.
Now a statistic is defined as:
[tex]\hat p=\frac{X}{n}[/tex]
This statistic is known as the sample proportion.
A Normal approximation to Binomial can be used to approximate the distribution of sample proportion if the following conditions are satisfied:
np ≥ 5n(1 - p) ≥ 5Then the sampling distribution of [tex]\hat p[/tex] follows a Normal; distribution with mean [tex]p=0.70[/tex] and standard deviation [tex]\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.70(1-0.70)}{200}}=0.032[/tex].
Check the conditions:
[tex]np=15\times0.70=10.5\\n(1-p)=15\times (1-0.70)=4.5\approx5[/tex]
So, the probability distribution of [tex]\hat p[/tex] is, [tex]N(0.70, 0.032)[/tex].
The sampling distribution of X/n is approximately normally distributed with a mean of 0.7 and a standard deviation of 0.117.
To determine the sampling distribution of the statistic X/n, we need to consider the probability distribution of the number of successes (X) in the sample of n = 15 drives.
Since the probability of a success (a working hard drive) is p = 0.7 and the probability of a failure (a non-working hard drive) is q = 1 - p = 0.3, we can model X using the binomial distribution with parameters n = 15 and p = 0.7.
The probability of getting exactly x successes (working hard drives) in a sample of n = 15 drives is given by the binomial probability mass function (PMF):
P(X = x) = (nCx) * px * (1-p)^(n-x)
where:
nCx is the binomial coefficient, which represents the number of ways to choose x successes out of n trials.
px is the probability of success (working hard drive) on each trial.
(1-p)^(n-x) is the probability of failure (non-working hard drive) on each trial.
For x = 3 successes (working hard drives), we calculate the probability:
P(X = 3) = (15C3) * (0.7)^3 * (0.3)^12 ≈ 0.193
This means that the probability of observing 3 successes (working hard drives) in a sample of 15 drives is approximately 0.193.
The sample proportion (fraction) of successes, X/n, is a continuous random variable that can take values between 0 and 1. To determine the sampling distribution of X/n, we need to consider the distribution of X and how it transforms to X/n.
The distribution of X/n is approximately normal for large sample sizes (n ≥ 30), but for smaller sample sizes, it can be skewed. In this case, with n = 15, the distribution of X/n is slightly skewed, but it is still close to being normal.
The mean of the sampling distribution of X/n is equal to the population proportion, which is p = 0.7.
The standard deviation of the sampling distribution of X/n is given by:
σ(X/n) = √(p * q / n) ≈ √(0.7 * 0.3 / 15) ≈ 0.117
Therefore, the sampling distribution of X/n is approximately normally distributed with a mean of 0.7 and a standard deviation of 0.117.
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The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The lowest 10 percent of the citizens would need at least how many minutes to complete the form
Answers
Answer:
The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.
Step-by-step explanation:
Problems of normally distributed samples are 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 = 40, \sigma = 10[/tex]
The slowest 10 percent of the citizens would need at least how many minutes to complete the form
This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 40}{10}[/tex]
[tex]X - 40 = 1.28*10[/tex]
[tex]X = 52.8[/tex]
The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.
Final answer:
To find the time needed for the lowest 10 percent of citizens to complete the 2010 U.S. Census 'long' form, we use the z-score formula to calculate the value. The lowest 10 percent of citizens would need at least 26.8 minutes to complete the form.
Explanation:
To find how many minutes would be needed for the lowest 10 percent of citizens to complete the 2010 U.S. Census 'long' form, we need to use the z-score formula.
First, we calculate the z-score using the formula: z = (x - mean) / standard deviation, where x is the value we want to find the z-score for, mean is the mean of the distribution, and the standard deviation is the standard deviation of the distribution.
Next, we use a z-table or a calculator to find the z-score that corresponds to a cumulative probability of 0.1. The z-table gives us a z-score of -1.28. We can rearrange the z-score formula to solve for x: x = (z * standard deviation) + mean. Plugging in the values, we get: x = (-1.28 * 10) + 40 = 26.8. Therefore, the lowest 10 percent of citizens would need at least 26.8 minutes to complete the form.
We examined the relationship between Rotten Tomatoes ratings and Metascore ratings for a sample of 75 popular movies. The scatterplot showed a linear form with strong positive association. Here is are the StatCrunch linear regression results. The r-sq value is about 0.75. What does this tell us? Simple linear regression results: Dependent Variable: Metascore Independent Variable: Rotten Tomatoes Metascore 21.605526+0.62434658 Rotten Tomatoes Sample size: 72 R (correlation coefficient) 0.86590 102 R-sq 0.74978458 Estimate of error standard deviation: 7.7978789a. 75 % of the Metascore ratings are accurately predicted by the regression line. b. Rotten Tomato ratings explain about 75 % of the variation in Metascore ratings.c. For each one point increase in Rotten Tomato ratings, we predict a 0.75 point increase in Metascore ratings.
Answers
Answer:
Therefore Rotten Tomato ratings explain about 75% of the variation in Metascore ratings is the correct answer here.
Step-by-step explanation:
The R2 value for regression or the coefficient of determination represents the proportion of variation in dependent variable that is explained by regression ( the rest of it is residual variation )
The given question is about the analysis of a linear regression model to examine the relationship between Rotten Tomatoes ratings and Metascore ratings for popular movies.
Explanation:The given question pertains to linear regression analysis in statistics. It involves examining the relationship between Rotten Tomatoes ratings and Metascore ratings for a sample of 75 popular movies. The r-squared value of 0.75 indicates that approximately 75% of the variation in Metascore ratings can be explained by the linear relationship with Rotten Tomatoes ratings. Additionally, for every one-point increase in Rotten Tomatoes ratings, the regression model predicts a 0.75-point increase in Metascore ratings.
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A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times and determines that 14 of the plates have blistered.
Does this data provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances?
Use Alpha =0.10.
A. What is the parameter of interest?
B. State the null and alternative hypotheses.
C. Calculate the test statistic.
D. Find the rejection region.
E. Make a decision and interpret.
F. Find a p-value corresponding to the test and compare with your decision in E.
Answers
Answer:
a) Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.
b) Null hypothesis:[tex]p\leq 0.1[/tex]
Alternative hypothesis:[tex]p > 0.1[/tex]
c) [tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]
d) For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:
[tex] z_{critc}= 1.28[/tex]
e) For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1
f) [tex]p_v =P(z>1.33)=0.0917[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
Part a
Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.
X=14 represent the number of the plates have blistered.
[tex]\hat p=\frac{14}{100}=0.14[/tex] estimated proportion of the plates have blistered.
[tex]p_o=0.1[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Part b: Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that more than 10% of all plates blister under such circumstances.:
Null hypothesis:[tex]p\leq 0.1[/tex]
Alternative hypothesis:[tex]p > 0.1[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Part c: Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]
Part d: Rejection region
For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:
[tex] z_{critc}= 1.28[/tex]
Part e
For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1
Part f
Since is a right taild test the p value would be:
[tex]p_v =P(z>1.33)=0.0917[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%
The base of a rectangular tank measures 10 ft by 20 ft. The tank is 16 ft tall, and its top is 10 ft below ground level. The tank is full of water weighing 62.4 lb/ft3. How much work does it take to empty the tank by pumping the water to ground level? Give your answer to the nearest ft ∙ lb.
Answers
Answer:
Step-by-step explanation:
Given a rectangular tank with dimension (10ft by 20ft by 16ft)
Then the volume of the tank is
Volume =length × breadth ×height
Volume=10×20×16=3200ft³
V=3200ft³
Then,
∆F= weight density × volume
∆F= 62.4×3200
∆F= 199,680 lb
Then,
Let the 0-point on the x-axis be at the bottom of the tank, so the level of the water ranges from x = 0 to x = 16ft. (It would just as well to let 0 be ground level and let x range from x = −26ft to x − 0.) Then a slice of water at level x is raised (26− x)ft
Then ∆x=(26-x)ft
Work is given as
W= -∫F∆xdx. From 0 to 16
W= -∫199,680(26-x)dx From 0 to 16
W=-199,680∫26-x dx From 0 to 16
W=-199,680 (26x-x²/2). From 0 to 16
W=-199,680(26×16-0.5×16² -0-0)
W=-199,680(288)
W=-57,507,840J
The work done to empty the tank is 57,507,840J
Answer:
2 X 10∧6 ft.lb
Step-by-step explanation:
Volume of rectangular tank = Base area X Height = (10 X 20 X 16)ft³ = 3200ft³
Mass of tank filled with water = 3200ft³ X 62.4 lb/ft³ = 199680lb or 90706.37kg
Hence, work done due to gravity,Δg = m * g * h, where, m = mass of tank, g = gravity = 9.8m/s² and h = height = 10ft or 3.05 meter(m)
∴ Δg = 90706.37 * 9.8 * 3.05 = 2.711 X10∧6 J or 2 X 10∧6 ft.lb
Consider the set consisting of the first 10 positive whole numbers (that is, 1-10). Complete parts (a) through (c) below.
(a) Determine explicitly the numbers in the set that are at least 4.
(b) Determine explicitly the numbers in the set that are at most 5
(c) Determine explicitly the numbers in the set that are between 4 and 6, inclusive.
Answers
Answer:
a) {4,5,6,7,8,9,10}
b) {1,2,3,4,5}
c) {4,5,6}
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $48 and the estimated standard deviation is about $7.
(a) Consider a random sample of n = 60 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution?
The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.12. The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $7. The sampling distribution of x is not normal. The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.90.
Is it necessary to make any assumption about the x distribution? Explain your answer.
It is necessary to assume that x has an approximately normal distribution. It is not necessary to make any assumption about the x distribution because μ is large. It is necessary to assume that x has a large distribution. It is not necessary to make any assumption about the x distribution because n is large.
(b) What is the probability that x is between $46 and $50? (Round your answer to four decimal places.)
3
(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $46 and $50? (Round your answer to four decimal places.)
Answers
Answer:
Step-by-step explanation:
Hello!
X: amount spent in a supermarket impulse buying in a 10 min unplanned shopping interval by one customer.
It is known that the mean of this variable is μ= $48 and its standard deviation is δ=$7
a.
The Central limit theorem states that if there is a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
X[bar]≈N(μ;σ²/n)
The mean of the sampling distribution is μ= $48
The standard deviation of the sampling distribution is σ/√n= $7/√60= $0.90
It is not necessary to make any assumption about the distribution of X since n=60 is considered large enough, you can directly approximate the sampling distribution to normal regardless of the distribution of X.
b.
To calculate this probability you have to use the approximation of the sampling distribution:
Z= (X[bar]-μ)/(σ/√n)≈N(0;1)
μ= $48
σ/√n= $0.90
P(46≤X[bar]≤50)= P(X[bar]≤50) - P(X[bar]≤46)
P(Z≤(50-48)/0.90) - P(Z≤(46-48)/0.90)
P(Z≤2.22) - P(Z≤-2.22)= 0.987 - 0.013= 0.974
c.
If we assume that X has an approximately normal distribution, then you will use it's a mean and standard deviation to reach the asked probability.
Z= (X-μ)/δ≈N(0;1)
μ= $48
δ= $7
P(46≤X≤50)= P(X≤50) - P(X≤46)
P(Z≤(50-48)/7) - P(Z≤(46-48)/7)
P(Z≤0.29) - P(Z≤-0.29)= 0.61409 - 0.38591= 0.22818≅ 0.2282
I hope it helps!
The Central Limit Theorem ensures that the average amount spent by a sample of 60 customers due to impulse buying is normally distributed with a mean of $48 and a standard error of approximately $0.90.
Explanation:Regarding part (a) of the question, by applying the Central Limit Theorem (CLT), we can state that for the sample of n = 60 customers, the sampling distribution of the sample mean x will be approximately normal due to the large sample size, even if the distribution of x is not normal. The mean of the distribution μx remains the same at $48, and the standard deviation (often referred to as the standard error, σx) can be calculated by dividing the population standard deviation by the square root of the sample size (n), which is σx = $7/√60 ≈ $0.90. Thus, the correct statement is: 'The sampling distribution of x is approximately normal with mean μx = $48 and standard error σx = $0.90.'
determine the solution on the interval [0, 2pi] for the equation
[tex]sec^{2} x-2=0[/tex]
Answers
Step-by-step explanation:
sec² x − 2 = 0
sec² x = 2
cos² x = ½
cos x = ±√½
x = π/4, 3π/4, 5π/4, 7π/4
A tank contains 70 kg of salt and 1000 L of water. A solution of a concentration 0.035 kg of salt per liter enters a tank at the rate 8 L/min. The solution is mixed and drains from the tank at the same rate. (a) What is the concentration of our solution in the tank initially? concentration = (kg/L) (b) Set up an initial value problem for the quantity y, in kg, of salt in the tank at time t minutes. dy/dt = (kg/min) y(0) = (kg) (c) Solve the initial value problem in part (b). y(t) = (d) Find the amount of salt in the tank after 5 hours. amount = (kg) (e) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = (kg/L)
Answers
Answer:
(a) Initial concentration of salt = 0.07 kg/L
(b) [tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]
(c) [tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]
(d)Therefore the amount of salt after 5 hours is =38.18 kg
(e) The concentration of salt in the solution in the tank as time approaches infinity is = 0.035 kg/L
Step-by-step explanation:
Given that,
A tank contains 70 kg of salt and 1000 L of water.
(a)
[tex]\textrm{Concentration of salt }=\frac{\textrm{mass of salt}}{\textrm{volume of water}}[/tex]
[tex]=\frac{70}{1000} kg/L[/tex]
=0.07 kg/L
(b)
Let Y(t) be the amount of salt at any instant time t.
Therefore
[tex]Y'(t)= Y_{in}-Y_{out}[/tex]
[tex]Y_{in}=\textrm{concentration of salt} \times \textrm{rate of enter}[/tex]
=(8×0.035) kg/min
=0.28 kg/min
Since the rate of water in and out are same , the amount of solution remain constant.
[tex]Y_{out}= (\frac{y}{1000}\times 8) kg/min[/tex]
[tex]=\frac{y}{125} kg/L[/tex]
[tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]
(c)
The above equation can be rewrite as
[tex]Y'(t) +\frac{y}{125} =0.28[/tex]
The coefficient of y is p(t) [tex]=\frac{1}{125}[/tex]
The integrating factor of the D.E is[tex]=e^{\int p(t) dt=[/tex] [tex]e^{\int \frac{1}{125} dt[/tex] [tex]=e^{\frac{1}{125} t[/tex]
Multiplying the integrating factor of both sides of D.E
[tex]e^{\frac{1}{125} t} \ \frac{dY}{dt} +e^{\frac{1}{125} t} .\frac{1}{125} Y=0.28 \ e^{\frac{1}{125} t}[/tex]
Integrating both sides
[tex]\int e^{\frac{1}{125} t} \ \ dY+\int e^{\frac{1}{125} t} .\frac{1}{125} Y \ dt=\int0.28 \ e^{\frac{1}{125} t}\ dt[/tex]
[tex]\Rightarrow e^{\frac{1}{125} t} \ Y= \frac{0.28e^{\frac{1}{125}t} }{\frac{1}{125}} +C[/tex]
[tex]\Rightarrow Y=35+Ce^{-\frac{1}{125}t}[/tex]
At initial when t= 0, y =70
Therefore
[tex]70=35+Ce^0[/tex]
⇒C= 70-35
⇒C=35
Therefore
[tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]
(d)
When t= 5 hour = 300 min
To find the amount of salt after 5 hour , we need to put the value of t in the general solution of D.E
Therefore the amount of salt after 5 hours is = [tex]Y(300)=35+35e^{-\frac{300}{125} }[/tex]
= 38.18 Kg
(e)
When t=∞
[tex]Y(\infty )= 35 +e^{-\infty}[/tex]
= 35 Kg [tex][e^{-\infty}=0][/tex]
Since the amount of water is remain same i.e 1000 L
Therefore the concentration of salt is [tex]=\frac{35}{1000}kg/L[/tex]
=0.035 kg/L
This problem involves using differential equations to model the concentration of salt in a tank over time. The initial concentration is determined, an initial value problem is set up, the initial value problem is solved, the amount of salt after a specific time is found, and the approaching concentration as time goes to infinity is identified.
Explanation:
(a) The initial concentration of the solution can be found by dividing the amount of salt by the volume of water. So, 70 kg / 1000 L = 0.07 kg/L.
(b) Let y be the amount of salt in the tank at time t. Then, dy/dt = 0.035 kg/L * 8 L/min - 0.07 kg/L * 8 L/min since salt is being added at a rate of 0.035 kg/L *8 L/min and leaving at a concentration of 0.07 kg/L * 8 L/min. Thus, dy/dt = 0.28 kg/min - 0.56 kg/min = -0.28 kg/min. The initial condition is y(0) = 70 kg.
(c) To solve this differential equation, we use the method of separation of variables. We get y(t) = 70 - 0.28t in kg.
(d) After 5 hours, or 300 minutes, the amount of salt in the tank is y(300) = 70 - 0.28*300 = 16 kg.
(e) As time approaches infinity, the concentration of salt will approach the incoming salt concentration, which is 0.035 kg/L since the solution is mixed and leaves at that rate.
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Elise’s garden is below. Tomato plants require 2 square feet of space.
(Shape of garden is a Rectangle 6.3 feet by 12.6 feet )
How many tomato plants can Elise fit in her garden?
Answers
Answer:
Step-by-step explanation:
You should try going too jeeska
Answer:
Step-by-step explanation:
Elise’s garden is rectangular. The formula for determining the area of a rectangle is expressed as
Area = length × width
From the information given,
Length of garden = 6 feet
Width of garden = 12.6 feet
Area of garden = 6 × 12.6 = 75.6 square feet
Tomato plants require 2 square feet of space. This means that the number of tomato plants that Elise can fit in her garden is
75.6/2 = 37.8
Since the number of tomato plants must be whole number, then the number of tomato plants that Elise can fit in her garden is 37
when Madiha was born, her aunt opened a certificate of deposit in her honor to help send her to college someday. now at the age of 18, there is $31,066 in Madiha's account. how much did her aunt originally invest if the interest rate was 6.5%? (round to the nearest dollar)
A.$375
B.$12,500
C.$10,000
D.$8,000
Answers
Answer: C.$10,000
Step-by-step explanation:
Assuming the interest was compounded annually, then we would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
A = $31066
r = 6.5% = 6.5/100 = 0.065
n = 1 because it was compounded once in a year.
t = 18 years
Therefore,
31066 = P(1+ 0.065/1)^1 × 18
31066 = P(1.065)^18
31066 = 3.12P
P = 31066/3.12
P = $9957.1
Approximately $10000 to the nearest dollar
Read the paragraph and answer the question below: Results from two CNN/USA Today/Gallup polls, one conducted in March 2003 and one in November 2003, were recently presented online. Both polls involved samples of 1001 adults, aged 18 years and older. In the March sample, 45% of those sampled claimed to be fans of professional baseball whereas 51% of those polled in November claimed to be fans. Construct a 99% confidence interval for the proportion of adults who professed to be baseball fans in November 2003 (after the World Series). Interpret this interval.
Answers
Answer:
99% confidence interval for the proportion of adults who professed to be baseball fans in November 2003 is (46.9%, 55.1%)
This interval means that the lower limit of the proportion of adults who professed to be baseball fans is 46.9% and the upper limit of the proportion is 55.1%
Step-by-step explanation:
Confidence interval = P' + or - t×sqrt[P'(1-P') ÷ n]
P' is sample proportion = 51% = 0.51
n = 1001
confidence level = 99%
t-value corresponding to 99% confidence interval and infinity degree of freedom is 2.576
t × sqrt[P'(1-P') ÷ n] = 2.576 × sqrt[0.51(1-0.51) ÷ 1001] = 2.576 × 0.0158 = 0.041
Lower limit = P' - 0.041 = 0.51 - 0.041 = 0.469 = 46.9%
Upper limit = P' + 0.041 = 0.51 + 0.041 = 0.551 = 55.1%
99% Confidence interval is (46.9%, 55.1%)
The interval means that the proportion of adults who professed to be baseball fans in November 2003 is between 46.9% and 55.1%
Find the parametric equations for the line through the point P(2,4,4) that is perpendicular to the plane −1x+1y−4z=1. a) Use the variable t and write these equations so that t=0 corresponds to the point P.
Answers
Answer:
[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]
Step-by-step explanation:
Given that a line passes through P(2,4,4)
Also the line is perpendicular to the plane
[tex]-1x+1y-4z=1.[/tex]
From the equation of the plane we can say that normal to the plane has direction ratios as (-1,1,-4)
Since the required line is also perpendicular to the plane, the direction ratios of the required line is
(-1,1,4)
It passes through (2,4,4)
If Q(x,y,z) are general points on the line then
Direction ratios of PQ are = (x-2, y-4, z-4)
These are proportional to (-1,1,4)
So parametric form of the line is
[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]
Whem t=0 we get the point P.