Answer:
1) Number of scoops = 42
2)Number of scoops = 10
Step-by-step explanation:
Volume of the sink = (4000/3)π in³
1)
Diameter = 4 in ∵ Radius = 4/2 = 2 in
Height = 8 in
Volume of the cylindrical cup = [tex]\pi r^{2}h = \pi 2^{2} * 8 = 32\pi[/tex] in³
Number of cups of water that must be scooped out = Volume of the sink/ Volume of the cylindrical cup
= (4000/3)π / 32π ==> 42
Number of scoops = 42
b)
Diameter = 8 in ∵ Radius = 8/2 = 4 in
Height = 8 in
Volume of the cylindrical cup = [tex]\pi r^{2}h = \pi 4^{2} * 8 = 128\pi[/tex] in³
Number of cups of water that must be scooped out = Volume of the sink/ Volume of the cylindrical cup
= (4000/3)π / 128π ==> 10
Number of scoops = 10
A random sample of 1800 NAU students in Flagstaff found 1134 NAU students who love their MAT114 class. Find a 95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class. Express your results to the nearest hundredth of a percent. .
Answer:
95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (60.77% , 65.23%)
Step-by-step explanation:
Among 1800 NAU students, 1134 students love their class. We have to find the 95% confidence interval of students who love their class.
We will use the concept of confidence interval of population proportion for this problem.
The proportion of students who love the class = p = [tex]\frac{1134}{1800}=0.63[/tex]
Proportion of students who do not love the class = q = 1 - p = 1 - 0.63 = 0.37
Total number of students in the sample = n = 1800
Confidence Level = 95%
The z values associated with this confidence level(as seen from z table) = 1.96
The formula to calculate the confidence interval for population proportion is:
[tex](p-z\times\sqrt{\frac{p \times q}{n}},p+z\times\sqrt{\frac{p \times q}{n}})[/tex]
Using the values in this expression gives:
[tex](0.63-1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}}, 0.63+1.96 \times \sqrt{\frac{0.63 \times 0.37}{1800}})\\\\ =(0.6077,0.6523)[/tex]
Thus, 95% confidence interval for the true percent of NAU students in Flagstaff who love their MAT114 class is (0.6077 ,0.6523) or (60.77% , 65.23%
Suppose 10000 people are given a medical test for a disease. About1% of all people have this condition. The test results have a 15% false positive rate and a 10% false negative rate. What percent of the people who tested positive actually have the disease?
Answer:
The percent of the people who tested positive actually have the disease is 38.64%.
Step-by-step explanation:
Denote the events as follows:
X = a person has the disease
P = the test result is positive
N = the test result is negative
Given:
[tex]P(X)=0.01\\P(P|X^{c})=0.15\\P(N|X)=0.10[/tex]
Compute the value of P (P|X) as follows:
[tex]P(P|X)=1-P(P|X^{c})=1-0.15=0.85[/tex]
Compute the probability of a positive test result as follows:
[tex]P(P)=P(P|X)P(X)+P(P|X^{c})P(X^{c})\\=(0.85\times0.10)+(0.15\times0.90)\\=0.22[/tex]
Compute the probability of a person having the disease given that he/she was tested positive as follows:
[tex]P(X|P)=\frac{P(P|X)P(X)}{P(P)}=\frac{0.85\times0.10}{0.22} =0.3864[/tex]
The percentage of people having the disease given that he/she was tested positive is, 0.3864 × 100 = 38.64%.
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night. Assume that room rates are normally distributed with a standard deviation of $55. What is the minimum cost that captures the 20% most expensive hotel rooms in New York City?
Answer:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 250.31.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the room hotel rate of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(204,55)[/tex]
Where [tex]\mu=204[/tex] and [tex]\sigma=55[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.20[/tex] (a)
[tex]P(X<a)=0.80[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.842[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.842[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 250.31.
Suppose that, during the next hour, 20 customers each purchase one cup of coffee. Assume that the customers make their decisions independently, and each customer has a 17% probability of ordering decaffeinated coffee. What is the probability that everyone will order a coffee with caffeine? g
Answer:
0.0241 or 2.41%
Step-by-step explanation:
Since each customer has a 17% probability of ordering decaffeinated coffee, the probability of a customer ordering a coffee with caffeine is:
[tex]P(C) = 1-0.17 = 0.83[/tex]
If customers make their decisions independently, the probability that all 20 order a coffee with caffeine is:
[tex]P(X=20)=P(C)^{20}\\P(X=20) = 0.83^{20}=0.0241=2.41\%[/tex]
The probability is 0.0241 or 2.41%.
The altitude of a triangle is increasing at a rate of 2.5 2.5 centimeters/minute while the area of the triangle is increasing at a rate of 2 2 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.5 7.5 centimeters and the area is 96 96 square centimeters
Answer:
The base of the triangle is decreasing at a rate 8 centimeter per minute.
Step-by-step explanation:
We are given the following in the question:
[tex]\dfrac{dh}{dt} = 2.5\text{ cm per minute}\\\\\dfrac{dA}{dt} = 2\text{ square cm per minute}[/tex]
Instant height = 7.5 cm
Instant area = 96 square centimeters
Area of triangle =
[tex]A=\dfrac{1}{2}\times b\times h[/tex]
where b is the base and h is the height of the triangle.
[tex]96 = \dfrac{1}{2}\times b \times 7.5\\\\b = \dfrac{96\times 2}{7.5} = 25.6[/tex]
Rate of change of area of triangle =
[tex]\dfrac{dA}{dt} = \dfrac{d}{dt}(\dfrac{bh}{2})\\\\\dfrac{dA}{dt} =\dfrac{1}{2}(b\dfrac{dh}{dt} + h\dfrac{db}{dt})[/tex]
Putting values, we get,
[tex]2 = \dfrac{1}{2}(7.5\dfrac{db}{dt}+25.6(2.5))\\\\4 = 7.5\dfrac{db}{dt} + 64\\\\-60 = 7.5\dfrac{db}{dt} \\\\\Rightarrow \dfrac{db}{dt} = \dfrac{-60}{7.5} = -8[/tex]
Thus, the base of the triangle is decreasing at a rate 8 centimeter per minute.
The area of a rectangle is 18 square inches. The length of the rectangle is twice its width. Find the width of the rectangle (in inches).
Let the width = X
Then the length would be 2X ( twice the width).
Area is Length x width, so you have X * 2X = 18
x *2x = 2x^2
Now you have 2x^2 = 18
Divide both sides by 2:
x^2 = 9
Take the square root of both sides:
x = sqrt(9)
x = 3
The width is 3 inches.
The width of the rectangle is determined by establishing an equation from the given area (18 square inches) and the fact that the length is twice the width, which leads to the solution that the width is 3 inches.
Explanation:To find the width of a rectangle when the area is 18 square inches, and the length is twice the width, we need to set up an equation. Let the width be w inches. Then the length would be 2w inches, since it's twice the width. We know that the area (A) of a rectangle is the multiplication of its length and width, so we have:
A = length imes width
18 = 2w imes w
18 = 2w2
Now we solve for w:
w2 = 18 / 2
w2 = 9
w = 3
The width of the rectangle is 3 inches.
ALGEBRA 2 HELP!!!
Find all the linear factors
f(x)=[tex]3x^{4}-7x^{3}-10x^{2} +28x-8[/tex]
Answer:
x + 2
3x − 1
x − 2
Step-by-step explanation:
The order of the polynomial is 4, so there are 4 roots.
Use rational root theorem to find possible rational roots.
±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3
By trial and error, -2, ⅓, and 2 are all roots. The fourth root can't be imaginary (complex roots come in pairs), so one of the real roots must be repeated.
Answer: the factors are (x - 2)(x + 2)(3x - 1)
Step-by-step explanation:
The given polynomial function is expressed as
f(x) = 3x⁴ - 7x³ - 10x² + 28x - 8
The first step is to test for any value of x that satisfies the polynomial when
3x⁴ - 7x³ - 10x² + 28x - 8 = 0
Assuming x = 2, then
3(2)⁴ - 7(2)³ - 10(2)² + 28 × 2 - 8 = 0
48 - 56 - 40 + 56 - 8 = 0
0 = 0
It means that x - 2 is a factor.
To determine the other factors, we would apply the long division method. The steps are shown in the attached photo. Looking at the photo, we would factorize the quadratic equation which is expressed as
3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x + 2) - 1(x + 2)
(3x - 1)(x + 2) = 0
Use Green's Theorem to find the work done by the force F(x, y) = x(x + y) i + xy2 j in moving a particle from the origin along the x-axis to (6, 0), then along the line segment to (0, 6), and then back to the origin along the y-axis.
Final answer:
The question involves using Green's Theorem to calculate work done by a specific force field along a triangular path. Green's Theorem links the line integral around a closed curve to a double integral over the region it encloses. However, direct calculation of line integrals may be more applicable for this problem.
Explanation:
The question asks to use Green's Theorem to calculate the work done by a force F(x, y) = x(x + y) i + xy^2 j moving a particle along a specified path. Green's Theorem relates a line integral around a simple closed curve C, to a double integral over the plane region D bounded by C. It's expressed in the form ∑_C (M dx + N dy) = ∫∫_D (∂N/∂x - ∂M/∂y) dA where M and N are components of a vector field.
To use Green's Theorem, we identify M = x(x + y) and N = xy^2. Thus, ∂N/∂x = y^2 and ∂M/∂y = x. The path described (origin to (6,0), to (0,6), and back to origin) encloses a triangle, the area of which is easily integrated over. However, it is crucial to note that direct integration methods or alternative strategies may sometimes be more straightforward for such paths, especially when they do not form a typical closed curve as in standard Green's Theorem applications.
In the context of this problem, the key would be computing the line integrals directly due to the specific nature of the force field and the path involved. The use of Green's Theorem hints at setting up an integration over the area, but with the given vector field and triangular path, the execution would involve careful calculation of respective line integrals or assessing the area directly since the theorem simplifies to a calculation over the region enclosed by the path.
An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.
a) Write out the hypotheses.
b) Calculate the test statistic.
c) Find the p-value.
d) Do we reject the null hypothesis? Explain.
e) What can you conclude based on this evidence
Answer:
An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.
a) Write out the hypotheses.
Fewer than 75% of claims settled within two months of filing.
b) Calculate the test statistic.
Test statistic = percent of claims settled in two months = 23/35 = 65.7%
c) Find the p-value.
we need to use the z-score with a table
=
standard deviation = s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}, =
√(1/34) [(0.657)(12) + (0.343)(23)] = √0.463911765 = 0.681
let 1 = "solved" and 0 = "unresolved"
thus our mean is 0.657
z = (0.75 - 0.657)/(0.681) - 0.137
p = 0.55172
d) Do we reject the null hypothesis? Explain.
Yes, because our p value is not below 0.05 and is not substantial to prove our null hypothesis
e) What can you conclude based on this evidence?
That further testing is needed.
A salesperson obtained a systematic sample of size 2525 from a list of 500500 clients. To do so, he randomly selected a number from 1 to 2020, obtaining the number 1313. He included in the sample the 1313th client on the list and every 2020th client thereafter. List the numbers that correspond to the 2525 clients selected.
Question
A salesperson obtained a systematic sample of size 25 from a list of 500 clients. To do so, he randomly selected a number from 1 to 20, obtaining the number 13. He included in the sample the 13th client on the list and every 20th client thereafter. List the numbers that correspond to the 25 clients selected.
Answer:
13, 33, 53, ...... 393
Step-by-step explanation:
The Question is making reference to arithmetic progression
Where the first term is 13
And the difference between each interval is 20 (i.e number of clients)
Using the following formula, well obtained the client in any position
Tn = a + (n - 1) d
Where Tn = nth term
a = T1 = 1st term = 13
n = number of terms = 1 to 25
d = common difference = 20
Calculating T2
T2 = 13 + (2 - 1) * 20
T2 = 13 + 20
T2 = 33
Calculating T3
T3 = 13 + (3 - 1) * 20
T3 = 53
.......
Calculating the 20th term
T3 = 13 + (20 - 1) * 20
T20 = 393
The number of views of a page on a Web site follows aPoisson distribution with a mean of 1.5 per minute.a. What is the probability of no views in a minute?b. What is the probability of two or fewer views in10 minutes?c. Does the answer to the previous part depend on whetherthe 10-minute period is an uninterrupted interval? Explain.d. Determine the length of a time interval such that the probabilityof no views in an interval of this length is 0.001.
Answer:
a. P (X=0) = 0.223
b. P(X≤2) = 3.93 * 10^-5
C. No
D. t=4.6 miutes
Step-by-step explanation:
a. P (X=0) = [tex]e^{-1.5}[/tex] = 0.223
b. P(X≤2) =[tex]e^{1.5 * 10} (1 + 1.5*10 + \frac{1.5^{2} * 10^{2} }{2} )[/tex]
= 3.93 * 10^-5
C. No, the answer to the previous part does not depend on whether the 10-minute period is an uninterrupted interval, it only depend on the uniformity of the density of views per minute ad independency of the disjoint time intervals.
D. P (X=0) =0.001
[tex]e^{-1.5t} = 0.001\\\\-1.5t=ln(0.001)\\\\t =\frac{-6.9}{-1.5}[/tex]
t=4.6 miutes
The Poisson distribution is used to calculate the probabilities of certain counts of discrete events occurring in a given time interval. By using the average rate of occurrence, we can determine probabilities relevant to web traffic, call intervals, or urgent care visits.
Explanation:The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. Various scenarios using Poisson distributions involve understanding the probability of a certain number of events occurring over a specified period.
Probability of no views in one minute: This can be computed using the formula for the Poisson distribution where the expected mean number of events (λ) is 1.5. The probability P(X=0) is given by (e^-1.5 * 1.5^0) / 0!.Probability of two or fewer views in 10 minutes: Given that events occur at a rate of 1.5 per minute, over 10 minutes the average rate would be 15. Calculate the cumulative probability for 0, 1, and 2 events and sum them up.Dependence of the interval: The answer to part b does not depend on the interval being uninterrupted, as the Poisson model assumes independence of intervals.Length of interval for a 0.001 probability of no views: Find the interval length 't' such that P(X=0) = 0.001 using the Poisson formula and solving for 't'.In choosing what music to play at a charity fund raising event, Shaun needs to have an equal number of string quartets from Mendelssohn, Beethoven, and Haydn. If he is setting up a schedule of the 66 string quartets to be played, and he has 66 Mendelssohn, 1616 Beethoven, and 6868 Haydn string quartets from which to choose, how many different schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
Answer:
64.69e221
Step-by-step explanation:
When choosing, the combination formula for selection is used. That is when selecting "r" number of items from a possible "n" items, then the number of ways is denoted as:
nCr = n! / (n-r)! × r!
Since 66 string quartet have to be chosen and the 3 genres must be equally represented in the string quartet, then we must have 22 number of each genre in it.
Number of ways to select 22 mendelssohn from possible 66 = 66C22 = 1.82 × 10^17
Number of ways to select 22 Beethoven from possible 1616 = 1616C22 = 2.97 × 10^49
Number of ways to select 22 Haydn from possible 6868 = 6868C22 = 2.2 × 10^63
Total number of ways to arrange these 66 schedules = 66! = 5.44 × 10^92
Number of possible schedule = [1.82 * 10^17] * [2.97*10^49] * [2.2*10^63] * [5.44*10^92]
=64.69 ×10^221. ≈64.69e221
Give (in percents) the two marginal distributions, for marital status and for income. Do each of your two sets of percents add to exactly 100%? If not, why not?
Answer:
Answer is attached
Step-by-step explanation:
6.11
Marginal Distribution of Marital Status
Single 4.1%
Married 93.9%
Divorced 1.5%
Widowed .5%
Marginal Distribution of Job Grade
Job Grade 1 11.6%
Job Grade 2 51.5%
Job Grade 3 30.2%
Job Grade 4 6.7%
Each of the marginal distributions sum up to exactly 100%. Depending on how one rounds the percentages, it is possible that the results will not be exactly 100%.
Imagine that you have been given a dataset of 1,000 documents that have been classified as being about entertainment or education. There are 700 entertainment documents in the dataset and 300 education documents in the dataset. The tables below give the number of documents from each topic that a selection of words occurred in. Word-document counts for the entertainment dataset fun christmas is 695 machine 35 family 400 learning 70 415 Word-document counts for the education dataset is christmas fun 200 machine 120 family 10 learning 105 295a. What target level will a naive Bayes model predict for the following query document: "machine learning is fun"? b. What target level will a naive Bayes model predict for the following query document: "christmas family fun"? c. What target level will a naive Bayes model predict for the query document in part (b) of this question, if Laplace smoothing with k = 10 and a vocabulary size of 6 is used?
Answer:
Please find attached answer.
Step-by-step explanation:
The Naive Bayes model is likely to classify the document 'machine learning is fun' and 'christmas family fun' as 'entertainment', even when using Laplace smoothing.
Explanation:The Naive Bayes model classifies texts based upon the likelihood of a particular term belonging to a given class. To find the target level for the document, you multiply the prior probability of the document being in a given class (either entertainment or education) with the likelihoods of the terms in the query document being in that class and you choose the class with the highest probability.
For the query document "machine learning is fun": Given the word-document counts, the model would predict 'entertainment' due to the higher frequency of 'fun' and 'machine' in entertainment documents than in education documents.For the query document "christmas family fun": The model would probably predict 'entertainment' due to the higher frequency of 'fun' and 'family' in the entertainment dataset than in the education dataset.If Laplace smoothing with k = 10 and a vocabulary size of 6 is used, the model would still predict 'entertainment' for "christmas family fun", as adjusting for unseen words in this way will not change the relative probabilities significantly.Learn more about Naive Bayes model here:https://brainly.com/question/21507963
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how many pounds of chamomile tea that costs 18.20 per pound must be mixed with 12lb of orange tea that costs 12.25 per pound to make a mixture that costs 14.63 per pound
8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
Step-by-step explanation:
We are given the following in the question:
Chamomile tea:
Unit cost = 18.20 per pound
Amount = x pounds
Total cost =
[tex]18.20\times x = 18.20x[/tex]
Orange tea:
Unit cost = 12.25 per pound
Amount = 12 pounds
Total cost =
[tex]12.25\times 12 = 147[/tex]
Total mixture:
Unit cost = 14.63 per pound
Amount = (12+x) pounds
Total cost =
[tex]14.63\times (12+x) = 175.56 + 14.63x[/tex]
We can write the equation that cost of mixture is equal to cost of chamomile tea and orange tea.
[tex]18.20x + 147 = 175.56 + 14.63x\\\Rightarrow 18.20x- 14.63x = 175.56-147\\\Rightarrow 3.57x = 28.56\\\Rightarrow x = 8[/tex]
Thus, 8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
A recent survey of 1010 U.S. adults selected at random showed that 627 consider the occupation of firefighter to have very great prestige. Estimate the probability (to the nearest hundredth) that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.
Answer:
0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Desired outcomes:
Adults who consider the occupation of firefighter to have very great prestige. So the number of desired outcomes is [tex]D = 627[/tex]
Total outcomes.
All adults sampled. So [tex]T = 1010[/tex]
Probability:
[tex]P = \frac{D}{T} = \frac{627}{1010} = 0.62[/tex]
0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.
Consider the infinite series sigma ^infinity _k = 1 1/7^k a. Find the first four terms of the sequence of partial sums. b. Use the results of part (a) to propose a formula for S_n. c. Propose a value of the series. a. Find the first four terms of the sequence of partial sums. S_1 =, S_2 =, S_3 =, S_4 = (Simplify your answers.) b. Use the results of part (a) to propose a formula for S_n. A. S_n = 7^n - 1/6 middot 7^n B. S_n = 7n/7^n C. S_n = n/7^n D. S_n = 6 middot 7^n/7^n - 1 c. Propose a value of the series. A reasonable conjecture for the value of the series is
Answer:
a. First four terms of the sequential partial sums
[tex]S_1=\frac{1}{7}, S_2=\frac{7^2-1}{6*7^2}, S_3 =\frac{7^3-1}{6*7^3}, S_4 =\frac{7^4-1}{6*7^4}[/tex]
b. The formula for Sn is option A = [tex]\frac{7^n-1}{6*7^n}[/tex]
c. Value of the series = [tex]\frac{1}{6}[/tex]
Step-by-step explanation:
a. First four terms of the sequential partial sums
[tex]\sum^{\infty}_{k=1}\\S_1=\frac{1}{7}\\ S_2=\frac{1}{7}+ \frac{1}{7^2}= \frac{8}{49} =\frac{7^2-1}{6*7^2}\\S_3=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3} = \frac{57}{343} =\frac{7^3-1}{6*7^3}\\S_4=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4} = \frac{400}{2401} =\frac{7^4-1}{6*7^4}[/tex]
b. Formula for Sn
The sum of n terms
[tex]S_n=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+ \frac{1}{7^4}+.....+ \frac{1}{7^n}= \frac{7^n-1}{6*7^n}[/tex]
c. Value for the series
This can be given as the Sum of infinite terms;
[tex]S_{\infty}=\frac{1}{7}+ \frac{1}{7^2}+ \frac{1}{7^3}+ \frac{1}{7^4}+....+ \frac{1}{7^n}+ ....= \lim_{n \to \infty} \frac{7^n-1}{6*7^n}= \frac{1}{6}[/tex]
In a study of cereal leaf beetle damage to oats, researchers measured the number of beetle larvae per stem in small plots of oats after applying (or not applying) the pesticide Malathion . Researchers applied Malathion to a random sample of 5 plots, and did not apply the pesticide to an independent sample of 12 plots. A noted scientist claims that Malathion will not make any difference in the mean number of larvae per stem. Test her claim at the .05 level of significance. State H0 and H1. a. H0: Mean Malathion -MeanNoMalathion <= 0 H1: Mean Malathion-Mean Malathion > 0 b. H0: MeanMalathion -MeanNoMalathion >= 0 H1: MeanMalathion-MeanNoMalathion < 0 c. H0: MeanMalathion = MeanNoMalathion H1: MeanMalathion ≠ MeanNoMalathion d. H0: MeanMalathion-MeanNoMalathion = 0 H1: MeanMalathion-MeanNoMalathion ≠ 0
Answer:
At 0.05 level of significance, there is no difference in the mean number of larvae per stem. This supports the scientist's claim.
(d) H0: MeanMalathion - MeanNoMalathion equals 0
H1: MeanMalathion - MeanNoMalathion not equals 0
Step-by-step explanation:
Test statistic (t) = (mean 1 - mean 2) ÷ sqrt[pooled variance (1/n1 + 1/n2)]
Let the difference between the two means be x and the pooled variance be y
n1 = 5, n2 = 12
t = x ÷ sqrt[y(1/5 + 1/12)] = x ÷ sqrt(0.283y) = x ÷ 0.532√y = 1.88x/√y
Assuming the ratio of x to √y is 0.5
t = 1.88×0.5 = 0.94
n1 + n2 = 5 + 12 = 17
degree of freedom = n1 + n2 - 2 = 17 - 2 = 15
significance level = 0.05 = 5%
critical value corresponding to 15 degrees of freedom and 5% confidence interval is 2.131
The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.
The region of no rejection of the null hypothesis lies between -2.131 and 2.131
Conclusion
Fail to reject the null hypothesis because the test statistic 0.94 falls within the region bounded by the critical values.
The scientist's claim is right.
A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.
An alternate hypothesis is also a statement from a population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.
American car travels 32 miles on 1 gallon of gas European car travels 12.7 km on 1 L of gas which car gets better gas mileage explain your reasoning
Answer: the American car gets better gas mileage.
Step-by-step explanation:
American car travels 32 miles on 1 gallon of gas. This means that its gas mileage is 32 miles per gallon.
European car travels 12.7 km on 1 L of gas. We would convert 12.7 km to miles.
1 kilometer = 0.621 mile
12.7 kilometer = 12.7 × 0.621 = 7.8867 miles
We would also convert 1 L to gallons.
1 liter = 0.264 us gallons
Therefore, the gas mileage of the European car in miles per gallon would be
7.8867/0.264 = 29.87 miles per gallon.
Therefore, since 32 miles per gallon is greater than 29.87 miles per gallon, it means that the American car gets better gas mileage.
Rona drove 56 milesto visit a friend she drove 42 miles before stopping for gas what percent of the drive did rona complete before stopping for gas
Answer:
75%
Step-by-step explanation:
Find the percentage corresponding to 42/56: (0.75)(100%) = 75%
Answer:
75%
Step-by-step explanation:
Becauuuuuuseeee if you divide 42 by 56 it would be 0.75 and yeah. so take out the 0 and its 75%
What is 3 1/4 +( 3 1/4 +5 1/5)?
Answer:16.5
Step-by-step explanation:
Answer:
11 7/10
Step-by-step explanation:
Lets make the fractions into common denominators, by mulitplying 4 by 5 and 5 by 4 (see below) and follow PEMDAS
3 1/4(5) +(3 1/4(5) + 5 1/5(4))
3 5/20+ (3 5/20+5 4/20)
Simplify like terms
3 5/20+ (8 9/20)
11 14/20
Simplify fraction
11 7/10
Consider randomly selecting a single individual and having that person test drive 3 different vehicles.
Define events A1, A2, and A3 by A1 = likes vehicle #1 A2 = likes vehicle #2 A3 = likes vehicle #3. Suppose that P(A1) = 0.55, P(A2) = 0.65, P(A3) = 0.70, P(A1 ∪ A2) = 0.80, P(A2 ∩ A3) = 0.50, and P(A1 ∪ A2 ∪ A3) = 0.88.
(a) What is the probability that the individual likes both vehicle #1
and vehicle #2?
(b) Determine and interpret P (A2 |A3 ).
(c) Are A2 and A3 independent events? Answer in two different ways.
(d) If you learn that the individual did not like vehicle #1, what now
is the probability that he/she liked at least one of the other two
vehicles?
Answer:
(a) The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40.
(b) The value of P (A₂ | A₃) is 0.7143.
(c) The events A₂ and A₃ are not independent.
(d) The probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.
Step-by-step explanation:
The events are defined as follows:
A₁ = an individual like vehicle #1
A₂ = an individual like vehicle #2
A₃ = an individual like vehicle #3
The information provided is:
[tex]P(A_{1})=0.55\\P(A_{2})=0.65\\P(A_{3})=0.70\\P(A_{1}\cup A_{2})=0.80\\P(A_{2}\cap A_{3})=0.50\\P(A_{1}\cup A_{2}\cip A_{3})=0.88\\[/tex]
(a)
Compute the probability that the individual likes both vehicle #1 and vehicle #2 as follows:
[tex]P(A_{1}\cap A_{2})=P(A_{1})+P(A_{2})-P(A_{1}\cup A_{2})\\=0.55+0.65-0.80\\=0.40[/tex]
Thus, the probability that the individual likes both vehicle #1 and vehicle #2 is 0.40.
(b)
Compute the value of P (A₂ | A₃) as follows:
[tex]P(A_{2}|A_{3})=\frac{P(A_{2}\cap A_{3})}{P(A_{3}}\\=\frac{0.50}{0.70}\\=0.7143[/tex]
Thus, the value of P (A₂ | A₃) is 0.7143.
(c)
If two events X and Y are independent then,
[tex]P(X\cap Y)=P(X)\times P(Y)\\P(X|Y)=P(X)[/tex]
The value of P (A₂ ∩ A₃) is 0.50.
The product of the probabilities, P (A₂) and P (A₃) is:
[tex]P(A_{2})\times P(A_{3})=0.65\times0.70=0.455[/tex]
Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃)
The value of P (A₂ | A₃) is 0.7143.
The value of P (A₂) is 0.65.
Thus, P (A₂ | A₃) ≠ P (A₂).
The events A₂ and A₃ are not independent.
(d)
Compute that probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ as follows:
[tex]P(A_{2}\cup A_{3}|A_{1}^{c})=\frac{P((A_{2}\cup A_{3})\cap A_{1}^{c})}{P(A_{1}^{c})}\\=\frac{P((A_{2}\cup A_{3}\cup A_{1})-P(A_{1})}{1-P(A_{1})} \\=\frac{0.88-0.55}{1-0.55}\\=0.7333[/tex]
Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.
Final answer:
The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40. Events A2 and A3 are not independent as P(A2 and A3) does not equal P(A2)P(A3). If the individual did not like vehicle #1, the probability of liking at least one of the other two vehicles is 0.73.
Explanation:
Probability of Likes and Independency of Events
To find the probability that the individual likes both vehicle #1 and vehicle #2, which is P(A1 ∩ A2), we use the formula: P(A1 ∩ A2) = P(A1) + P(A2) - P(A1 ∪ A2). Given that P(A1) = 0.55, P(A2) = 0.65, and P(A1 ∪ A2) = 0.80, we can calculate P(A1 ∩ A2):
P(A1 ∩ A2) = 0.55 + 0.65 - 0.80 = 0.40.
The conditional probability P(A2|A3) represents the probability of liking vehicle #2 given that the individual already likes vehicle #3. Using the definition of conditional probability:
P(A2|A3) = P(A2 ∩ A3) / P(A3) = 0.50 / 0.70 = approximately 0.714.
To determine if A2 and A3 are independent events, we can check if P(A2 ∩ A3) equals P(A2)P(A3) which would be 0.65 * 0.70 = 0.455. Since P(A2 ∩ A3) = 0.50, which is not equal to 0.455, A2 and A3 are not independent.
Moreover, if A2 and A3 were independent, then knowing that A3 occurred would not change the probability of A2 happening. However, P(A2|A3) is not equal to P(A2), indicating their dependence. Lastly, when the individual did not like vehicle #1, the probability that they liked at least one of the other two vehicles can be found using the probability of the union of events, subtracting the probability of liking vehicle #1:
P(A2 ∪ A3 | not A1) = P(A2 ∪ A3) - P(A1) + P(A1 ∩ A2 ∩ A3) = 0.88 - 0.55 + 0.40 = 0.73.
Jon is hitting baseballs. When he tosses the ball into the air, his hand is 5 feet above the ground. He hits the ball when it falls back to a height of 4 feet. The height of the ball is given by h = 5 + 25t - 16t^2, where h is measured in feet and t is measured in seconds. How many seconds after throwing the ball does Jon hit it?
Answer:
[tex]t=1.60\ sec[/tex]
Step-by-step explanation:
Functions
We are given the function that relates the height of the ball and the time elapsed t. The function is:
[tex]h = 5 + 25t - 16t^2[/tex]
We need to know the time needed for the height to reach 4 feet, that is h=4
[tex]5 + 25t - 16t^2=4[/tex]
Rearranging
[tex]16t^2-25t -1=0[/tex]
Solving for t
[tex]\boxed{t=1.60\ sec}[/tex]
The other solution is negative and is discarded because t cannot be negative.
Is (2,7) a point on the line y=4x-3?
Given that the equation of the line is [tex]y=4x-3[/tex]
We need to determine the point (2,7) lies on the line.
For a point to lie on the line, substituting the point in the equation of the line makes it valid.
Thus, substituting the point (2,7) in the equation of the line [tex]y=4x-3[/tex], we get;
[tex]7=4(2)-3[/tex]
[tex]7=8-3[/tex]
[tex]7\neq 5[/tex]
Thus, the both sides of the equation are not equal.
Substituting the point (2,7) in the equation makes the equation invalid.
Thus, the point (2,7) does not lie on the line [tex]y=4x-3[/tex]
. You decide you want to start investing for your retirement and you want to have $650,000. If you save $100 a month in an account that averages a 10% annual rate of return, will you have enough money in 25 years? (Hint: this formula was introduced in Section 1)
Answer:
750,0000
Step-by-step explanation:
A state insurance commission estimates that 13% of all motorists in its state areuninsured. Suppose this proportion is valid. Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured. You may assume that the normaldistribution applies.
Answer:
79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
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.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 50, p = 0.13[/tex]
So
[tex]\mu = E(X) = np = 50*0.13 = 6.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.13*0.87} = 2.38[/tex]
Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
This is 1 subtracted by the pvalue of Z when X = 4. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 6.5}{2.38}[/tex]
[tex]Z = -0.84[/tex]
[tex]Z = -0.84[/tex] has a pvalue of 0.2005
1 - 0.2005 = 0.7995
79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
Using the normal approximation to the binomial distribution, the probability that at least 5 out of 50 motorists will be uninsured when 13% of motorists are uninsured is approximately 74.22%.
First, we calculate the mean (μ) and the standard deviation (σ) of the number of uninsured motorists in a sample of 50. The mean number of uninsured motorists is μ = np, where n is the sample size and p is the probability of a motorist being uninsured. So, μ = 50 * 0.13 = 6.5.
The standard d eviation is σ = √(np(1-p)). Therefore, σ = √(50 * 0.13 * (1 - 0.13)) ≈ 2.31.
To find the probability of at least 5 uninsured motorists, we want P(X ≥ 5). We need to standardize this to the standard normal distribution to find the z-score. The z-score is z = (X - μ) / σ.
For X = 5, the z-score is z = (5 - 6.5) / 2.31 ≈ -0.65. We use the z-table or a calculator to find the probability corresponding to z = -0.65.
Since we want at least 5 uninsured, we find P(Z ≥ -0.65), which equals 1 - P(Z < -0.65). The value from the z-table for -0.65 is approximately 0.2578. Therefore, the probability we're looking for is 1 - 0.2578 = 0.7422, or 74.22%.
One mole of nickel (6 1023 atoms) has a mass of 59 grams, and its density is 8.9 grams per cubic centimeter, so the center-to-center distance between atoms is 2.23 10-10 m. You have a long thin bar of nickel, 2.1 m long, with a square cross section, 0.08 cm on a side. You hang the rod vertically and attach a 69 kg mass to the bottom, and you observe that the bar becomes 1.11 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in nickel.
Final answer:
Nickel does not crystallize in a simple cubic structure because the hypothetical density of a simple cubic arrangement is 2.23 g/cm³, significantly less than the actual density of nickel, which is 8.90 g/cm³.
Explanation:
To determine if nickel crystallizes in a simple cubic structure, we can compare the calculated density of a simple cubic arrangement of nickel atoms to the actual density of nickel. Given the edge length of the unit cell for nickel is 0.3524 nm (3.524 x 10⁻⁸ cm) and knowing the properties of a simple cubic lattice, the volume of a single cubic unit cell would be V = a³ = (3.524 x 10⁻⁸ cm)³ = 4.376 x 10⁻²³ cm³. With the mass of a single nickel atom (58.693 g/mol divided by Avogadro's number, 6.022 x 10²³ atoms/mol), if the nickel crystallized in a simple cubic structure, the predicted density would be:
density = mass/volume = (9.746 x 10⁻²³ g) / (4.376 x 10⁻²³ cm³) = 2.23 g/cm³. However, because the actual density of nickel is 8.90 g/cm³, we can conclude that nickel does not form a simple cubic structure.
The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) Find the mean and standard deviation of the sample count X who are married. b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married? c.) What is the probability that we find less than 4 of the seniors are married? d.) What is the probability that we find at least 1 of the seniors are married?
Answer:
a) Mean 6, standard deviation 2.42
b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.
c) 14.85% probability that we find less than 4 of the seniors are married.
d) 99.77% probability that we find at least 1 of the seniors are married
Step-by-step explanation:
For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.
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.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]n = 300, p = 0.02[/tex]
a.) Find the mean and standard deviation of the sample count X who are married.
Mean
[tex]E(X) = np = 300*0.02 = 6[/tex]
Standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42[/tex]
b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?
This is P(X = 8).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040[/tex]
10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.
c.) What is the probability that we find less than 4 of the seniors are married?
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023[/tex]
[tex]P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143[/tex]
[tex]P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436[/tex]
[tex]P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485[/tex]
14.85% probability that we find less than 4 of the seniors are married.
d.) What is the probability that we find at least 1 of the seniors are married?
Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
From c), we have that [tex]P(X = 0) = 0.0023[/tex]. So
[tex]0.0023 + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 0.9977[/tex]
99.77% probability that we find at least 1 of the seniors are married
In this problem, the mean and standard deviation of a binomial distribution, with probability of success 0.02 and sample size 300, are found. Subsequently, the probabilities that 8, less than 4, and at least 1 of the seniors are married are computed using the binomial formula.
Explanation:This problem deals with the Binomial distributions in statistics. Since we know the proportion of high school seniors who are married is 0.02, and the sample size is 300, we can use these values to calculate the mean and the standard deviation.
a.) The mean (mean = np) of the sample count X who are married is 0.02*300=6, and the standard deviation would be sqrt(n*p*(1-p)) = sqrt(300*0.02*0.98) = √5.88≈2.43.
b.) The probability that, in our sample of 300, we find that 8 of the seniors are married is given by the binomial formula P(X=k) = C(n,k)(p^k)(1-p)^(n-k). Plugging n=300, k=8, p=0.02 into the formula, we get the desired probability.
c.) The probability that we find less than 4 of the seniors are married is sum of P(X=k) from k=0 to 3. This could be computed using the aforementioned binomial formula. Remember, you're summing the probabilities for each k.
d.) The probability that we find at least 1 of the seniors are married can be found by subtracting the probability that none of the seniors are married from 1 (i.e., P(X >=1) = 1 - P(X=0)).
Learn more about Binomial Distribution here:https://brainly.com/question/39749902
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Test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.
The null and alternative hypothesis would be:
a. H0:p≥0.9H0:p≥0.9
H1:p<0.9H1:p<0.9
b. H0:p=0.9H0:p=0.9
H1:p≠0.9H1:p≠0.9
c. H0:p≤0.9H0:p≤0.9
H1:p>0.9H1:p>0.9
d. H0:μ=0.9H0:μ=0.9
H1:μ≠0.9H1:μ≠0.9
e. H0:μ≤0.9H0:μ≤0.9
H1:μ>0.9H1:μ>0.9
f. H0:μ≥0.9H0:μ≥0.9
H1:μ<0.9H1:μ<0.9
The test is:
a. left-tailed
b. two-tailed
c. right-tailed
Based on a sample of 100 people, 94% owned cats
The p-value is: (to 2 decimals)
Based on this we:
Fail to reject the null hypothesis
Reject the null hypothesis
Box 1: Select the best answer
Box 2: Select the best answer
Box 3: Enter your answer as an integer or decimal number. Examples: 3, -4, 5.5172
Enter DNE for Does Not Exist, oo for Infinity
Box 4: Select the best answer
Answer:
1. H0 : p = 0.9
H1 : p ≠ 0.9
2. The test is two tailed.
3. Reject the null hypothesis
Step-by-step explanation:
We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.
So, Null Hypothesis, [tex]H_0[/tex] : p = 0.90
Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.9
Here, the test is two tailed because we have given that to test the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.
Now, test statistics is given by;
[tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1) , where, n = sample size = 100
[tex]\hat p[/tex] = 0.94 (given)
So, Test statistics = [tex]\frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } }[/tex] = 1.68
Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)
= 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%
Now, our decision rule is that;
If p-value < significance level ⇒ Reject null hypothesis
If p-value > significance level ⇒ Accept null hypothesis
Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that proportion of people who own cats is significantly different than 90%.
At St. Eligius Hospital, the weights of newborn babies follow a Normal distribution, with a mean of 7.5 pounds and a standard deviation of 1.2 pounds.
If Dr. Cavanero tells a mother that her newborn baby has a weight that is at the third quartile (or Q3), this means the baby must weigh approximately _____.
Answer:
The baby must weigh approximately 8.31 pounds.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 7.5 pounds
Standard Deviation, σ = 1.2 pounds
We are given that the distribution of weights of newborn babies is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the third quartile or [tex]Q_3[/tex]
We have to find the value of x such that the probability is 0.75
[tex]P( X < x) = P( z < \displaystyle\frac{x - 7.5}{1.2})=0.75[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 7.5}{1.2} = 0.674\\\\x = 8.3088 \approx 8.31[/tex]
Thus, the baby must weigh approximately 8.31 pounds.