Answer:
[693,38:725,62]hs
Step-by-step explanation:
Hello!
Your study variable is X: the lifespan of a light bulb. This variable is said to have an approximately normal distribution.
X≈N(μ;σ²)
Were
μ= populatiom mean
σ= 45 hs standard deviation
A sample of n= 35
To estimate the population mean with a confidence interval you have to use the Z statistic:
X{bar} ± [tex]Z_{1-\alpha /2}[/tex]*(σ/√n)
[tex]Z_{1-\alpha /2} = Z_{0.98} = 2.054[/tex]
710 ± 2.054*(45/√35
[693,38:725,62]hs
I hope you have a SUPER day!
About Here are two relations defined on the set {a, b, c, d}: S = { (a, b), (a, c), (c, d), (c, a) } R = { (b, c), (c, b), (a, d), (d, b) } Write each relation as a set of ordered pairs. (a) S ο R (b) R ο S (c) S ο S
Answer with Step-by-step explanation:
We are given that a set {a,b,c,d}
S={(a,b),(a,c),(c,d),(c,a)}
R={(b,c),(c,b),(a,d),(d,b)]
Composition of relation:Let R and S are two relations on the given set
If ordered pair (a,b) belongs to relation R and (b,c) belongs to S .
Then, SoR={(a,c)}
By using this rule
SoR={(b,d),(b,a)}[/tex]
Because [tex](b,c)\in R[/tex] and [tex](c,d)\in S[/tex].Thus, [tex](b,d)\in SoR[/tex]
[tex](b,c))\in R[/tex] and [tex](c,a)\in S[/tex].Thus, [tex](b,a)\in SoR[/tex]
b.RoS={(a,c),(a,b),(c,b),(c,d)}
Because
[tex](a,b)\in S,(b,c)\in R[/tex] .Therefore, the ordered pair [tex](a,c)\in[/tex] RoS
[tex](a,c)\in S,(c,b)\in R[/tex] .Thus, [tex](a,b)\in RoS[/tex]
[tex](c,d)\in S,(d,b)\in R[/tex].Thus, [tex](c,b)\in RoS[/tex]
[tex](c,a)\in S,(a,d)\in R[/tex].Thus,[tex](c,d)\in RoS[/tex]
c.SoS={(a,d),(a,a),(c,c),(c,b)}
Because
[tex](a,c)\;and\; (c,d)\in S[/tex].Thus, [tex](a,d)\in SoS[/tex]
[tex](c,a),(a,b)\in S[/tex].Thus,[tex](c,b)\in SoS[/tex]
[tex](a,c)\in S[/tex] and [tex](c,a)\in S[/tex].Thus,[tex](a,a)\in SoS[/tex]
[tex](c,a)\in S[/tex] and [tex](a,c)\in S[/tex].Thus ,[tex](c,c)\in SoS[/tex]
Babcock and Marks (2010) reviewed survey data from 2003–2005 and obtained an average of μ = 14 hours per week spent studying by full-time students at four-year colleges in the United States. To determine whether this average has changed in the past 10 years, a researcher selected a sample of n = 64 of today’s college students and obtained an average of M = 12.5 hours. If the standard deviation for the distribution is σ = 4.8 hours per week, does this sample indicate a significant change in the number of hours spent studying?
Answer:
We do not have enough evidence to accept H₀
Step-by-step explanation:
Normal Distribution
size sample = n = 64 (very small sample for evaluating population of 5 years
Standard deviation 4,8
1.- Test hypothesis
H₀ null hypothesis ⇒ μ₀ = 14 and
Hₐ alternative hypothesis ⇒ μ₀ ≠ 14
2.- z(c) we assume α = 0,05 as we are dealing with a two test tail we should consider α/2 = 0.025.
From z table we the z(c) value
z(c) = 1.96 and of course by symmetry z(c) = -1.96
3.- We proceed to compute z(s)
z(s) = [ ( μ - μ₀ ) /( σ/√n) ] ⇒ z(s) = - (1.5)*√64/4.8
z(s) = - 2.5
We compare z(s) and z(c)
z(s) < z(c) -2.5 < -1.96 meaning z(s) is in the rejection zone
we reject H₀ .
From the start we indicate sample size as to small for the experiment nonetheless we found that we dont have enough evidence to accept H₀
Background: Based on the National Center of Health Statistics, the proportion of babies born at low birth weight (below 2,500 grams) in the United States is roughly .078, or 7.8% (based on all the births in the United States in the year 2002). A study was done in order to check whether pregnant women exposed to second hand smoke increases the risk of low birth weight. In other words, the researchers wanted to check whether the proportion of babies born at low birth weight among women who were exposed to second hand smoke during their pregnancy is higher than the proportion in the general population. The researchers followed a sample of 400 women who had been exposed regularly to second hand smoke during their pregnancy and recorded the birth weight of the newborns. Based on the data, the p-value was found to be .119.1.Based on the p-value, what is your conclusion (use .05 significance level)?
Answer:
Step-by-step explanation:
Hello!
Remember, the rule to decide using the p-value is always the same.
If the p-value ≤ α, you reject the null hypothesis.
If the p-value > α, you support the null hypothesis.
The experiment states the hypothesis that the proportion of babies born with low weight is higher if their mothers were exposed to second-hand smoking during pregnancy. Symbolically:
H₀: ρ ≤ 0.078
H₁: ρ > 0.078
α: 0.05
The given p-value is 0.119
Since the p-value (0.119) is greater than the level of significance (α: 0.05) the decision is to not reject the null hypothesis.
The proportion of babies born with low weight among women that were exposed to second-hand smoking during pregnancy is no different from the proportion in the general population.
I hope it helps!
An educator wants to determine whether a new curriculum significantly improves standardized test scores for fourth grade students. She randomly divides 90 fourth dash graders into two groups. Group 1 is taught using the new curriculum, while group 2 is taught using the traditional curriculum. At the end of the school year, both groups are given the standardized test and the mean scores are compared. Determine whether the sampling is dependent or independent. Indicate whether the response variable is qualitative or quantitative.
Answer:
Independent Sampling
Step-by-step explanation:
There are two scenarios for independent sampling .
Testing the mean we get the differences between samples from each population. When both samples are randomly inferences about the populations. can get.
Independent sampling are sample that are selected randomly. Observation does not depend upon value.Many analysis assume that sample are independent.
In this statement 90 dash are divides into two groups Group 1 and Group 2 . Both are standardized that mean both are randomly selected. Means are observed. Observation doesn't depend upon value. So this style of sampling is independent Sample.
The owner of a large car dealership believes that the financial crisis decreased the number of customers visiting her dealership. The dealership has historically had 800 customers per day. The owner takes a sample of 100 days and finds the average number of customers visiting the dealership per day was 750. Assume that the population standard deviation is 350. The value of the test statistic is ____________. Multiple Choice z = –1.429 t99 = 1.429 z = 1.429 t99 = –1.429
Answer:
Z= -1.429
Step-by-step explanation:
Hello!
The owner thinks that the number of customers decreased. The hystorical average per day is 800 customers.
So the variable X: " Customers per day" X~N(μ;σ²)
σ= 350
Symbolically the test hypothesis are:
H₀: μ ≥ 800
H₁: μ < 800
The statistic value is:
Since you are studying the population sample, the study variable has a normal distribution and you know the value of the population variance, the propper statistic to make the test is:
Z= X[bar] - μ = 750 - 800 = -1.4285 ≅ -1.429
σ/√n 350/√100
where X[bar] is the sample mean
μ is the population mean under the null hypothesis
σ is the population standard
n is the sample taken
The value is Z= -1.429
I hope it helps!
8. Suppose you are testing H0 : p = 0.4 versus H1 : p < 0.4. From your data, you calculate your test statistic as z = +1.7. (a) Calculate the p-value for this scenario. (b) Using a significance level of 0.06, what decision should you make?
Answer:
(a) The p-value is 0.9554
(b) We cannot reject the null hypothesis at the significance level of 0.06
Step-by-step explanation:
We are dealing with a lower-tail alternative [tex]H_{1}: p < 0.4[/tex]. Because the test statistic is z = +1.7 which comes from a normal distribution, the p-value is the probability of getting a value as extreme as the already observed.
(a) P(Z < +1.7) = 0.9554
(b) The p-value is very large, and 0.9554 > 0.06, so, we cannot reject the null hypothesis at the significance level of 0.06
A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated from a sample of 49 apples randomly selected from a shipment indicated a mean weight of 122.5 grams per apple. Is there sufficient evidence to reject the company’s claim? (useα = 0.05)
Answer:
There is no sufficient evidence to reject the company's claim at the significance level of 0.05
Step-by-step explanation:
Let [tex]\mu[/tex] be the true mean weight per apple the company ship. We want to test the next hypothesis
[tex]H_{0}:\mu=120[/tex] vs [tex]H_{1}:\mu\neq 120[/tex] (two-tailed test).
Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by
[tex]Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}[/tex] which is normally distributed. The observed value is
[tex]z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583[/tex]. The rejection region for [tex]\alpha = 0.05[/tex] is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.
To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test.
Explanation:To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test. The null hypothesis, H0, is that the mean weight per apple is 120 grams, and the alternative hypothesis, Ha, is that the mean weight per apple is not 120 grams. With a sample of 49 apples, we can calculate the test statistic using the formula:
t = (sample mean - population mean) / (standard deviation / sqrt(sample size))
Once we have the test statistic, we can compare it to the critical value from the t-distribution to determine if we reject or fail to reject the null hypothesis. As per calculation, yhe area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well.
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What is the relationship between probability of type 1 error and probability of type 2 error?
A) As α increases, β decreases
B) As α decreases, β decreases
C) The relationship depends on the context of the hypothesis test
D) Both α and β are independent of one anothe
Answer:
A) As α increases, β decreases
Step-by-step explanation:
I am assuming that the probabilities of both kind of error are positive, otherwise there is no reason to make an hypothesis test.
Both events arent possible simultaneously, so they cant be independent because the probability of both kind of errors is greater than 0.
If the p-value increases, then by definition, the probability of type 1 error increases, in this case, you are more likely to reject the null hypothesis, therefore you are less likely to make the mistake of not rejecting the null hypothesis when you have to. Thus, the probability of type 2 error decreases.
On the other way around, if you are less likely to reject the null hypothesis (hence, the probability of type 1 error decreases), then you will be more likely to not reject the null hypothesis even on weird scenarios, therefore the probability of type 2 error increases.
There is no formula to determine the relatioship between both errors, but the errors are related: when the probability of type 1 error increases, then the probability of type 2 error decreases. The correct answer is A)
Fill in the blank. (Note: The notation x → c+ indicates that x approaches c from the right and x → c− indicates that x approaches c from the left. Enter your answer in radians.) As x → 1−, the value of arcsin(x) → _____ .
Answer:
As x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]
Step-by-step explanation:
To find the value of [tex]arcsin(x)[/tex] when x approaches 1 from the left, you can use the graph of the function [tex]arcsin(x)[/tex]
Examine what happens as x approaches from the left.
As x approaches 1 from the left, the function seems to be approaching [tex]\frac{\pi }{2}[/tex]
Therefore, as x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]
As x approaches 1 from the left, the value of arcsin(x) approaches π/2 radians.
Explanation:The expression x → 1− refers to the limit as x approaches 1 from the left. For the function presented, arcsin(x), we need to consider the value that this function approaches as x gets infinitely close to 1, but is still less than 1. This is asking what angle in radians has a sine of 1, or almost 1 from the left side. The sine of an angle is 1 when that angle is π/2 (or 90° in degree measure), hence as x → 1−, the value of arcsin(x) approaches π/2.
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A student writes an incorrect step while checking if the sum of the measures of the two remote interior angles of triangle ABC below is equal to the measure of the exterior angle.
Step 1: m∠m + m∠n + m∠o = 180 degrees (sum of angles of a triangle)
Step 2: m∠p − m∠o = 90 degrees (alternate interior angles)
Step 3: Therefore, m∠m + m∠n + m∠o = m∠o + m∠p
Step 4: So, m∠m + m∠n = m∠p
In which step did the student first make a mistake and how can it be corrected?
1.) Step 1; it should be m∠m + m∠n + m∠o = 90 degrees (corresponding angles)
2.) Step 1; it should be m∠m + m∠n + m∠o = 90 degrees (adjacent angles)
3.) Step 2; it should be m∠o + m∠p = 180 degrees (alternate exterior angles)
4.) Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles)
Answer:
4.) Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles)
Step-by-step explanation:
o and p are supplementary angles, and therefore add up to 180 degrees.
An upright cylindrical tank with radius 8 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing? Part 1 of 3. If h is the water's height, the volume of the water is V = πr2h. We must find dV/dt. Differentiating both sides of the equation gives Dv/Dt= πr2 Dh/Dt Subsituting for r , this becomes Dv/Dt ____________ π Dh/Dt What goes in the blank ? Thanks !
Answer:
16 goes in the blank
Step-by-step explanation:
V(c) = 2*π*r*h
Differentiating boh sides
DV(c)/Dt = 2πr Dh/Dt now radius is 8 m
DV(c)/Dt = 8π Dh/Dt
That expression gives the relation of changes in V and h
DV(c)/Dt is the speed of growing of the volume
Dh/Dt is the speed of increase in height
so if the cylinder is filling at a rate of 2 m³/min the height will increase at a rate of 16π m/min
Real estate ads suggest that 61 % of homes for sale have garages, 39 % have swimming pools, and 14 % have both features.
a) If a home for sale has a garage, what's the probability that it has a pool, too?
b) Are having a garage and having a pool independent events? Explain.
c) Are having a garage and having a pool mutually exclusive? Explain.
Answer: a) 0.23, b) No, c) No.
Step-by-step explanation:
Since we have given that
Probability of homes for sale have garages = 61%
Probability of homes having swimming pool = 39%
Probability of homes having both = 14%
a) If a home for sale has a garage, what's the probability that it has a pool, too?
Using "Conditional probability", we get that
[tex]P(P|G)=\dfrac{P(P\cap G)}{P(G)}=\dfrac{0.14}{0.61}=0.23[/tex]
b) Are having a garage and having a pool independent events? Explain.
P(P).P(S) = 0.61×0.39=0.24≠P(P∩S)=0.14
Hence, they are not independent events.
c) Are having a garage and having a pool mutually exclusive? Explain.
Since P(P∩S)≠0
So, they are not mutually exclusive.
Hence, a) 0.23, b) No, c) No.
Final answer:
To calculate the probability of a home having a pool given it has a garage, one uses the given percentages to find that it's approximately 23%. Examining whether having a garage and having a pool are independent or mutually exclusive reveals they are neither independent (because the probabilities differ when one condition is known) nor mutually exclusive (because homes can and do have both features).
Explanation:
Given information implies that 61% of homes have garages (G), 39% have pools (P), and 14% have both a garage and a pool (G∩P).
a) The probability that a home for sale has a pool given it has a garage can be calculated using the formula P(P|G) = P(G∩P) / P(G). Substituting the given values, we get P(P|G) = 0.14 / 0.61, which approximately equals 0.23 or 23%. This means if a home has a garage, there's a 23% chance it also has a pool.
b) Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. To check if G and P are independent, we calculate P(P) and compare it to P(P|G). Since P(P) = 0.39 and P(P|G) = 0.23, they are not equal; thus, having a garage and having a pool are not independent events. This is evident because knowing that a home has a garage changes the probability that it also has a pool.
c) Two events are mutually exclusive if they cannot occur at the same time. Given that 14% of homes have both a garage and a pool, it's clear that having a garage and having a pool are not mutually exclusive, as a home can have both features simultaneously.
A bicyclist is training for a race on a hilly path. Their bike keeps track of their speed at any time, but not the distance traveled. Their speed traveling up a hill is 3mph, 8mph when traveling down a hill, and 5mph when traveling along a flat portion. Part A. Construct linear models that describe their distance, D in miles, on a particular portion of the path in terms of the time, t in hours, spent on that part of the path.
Answer:
[tex]D_{up}(t)=3t[/tex]
[tex]D_{down}(t)=8t[/tex]
[tex]D_{flat}(t)=5t[/tex]
Explanation:
To construct a linear model of their distance, in miles, as a function of time in hours, since we were given their traveling speed in miles per hour, simply multiply the traveling speed in mph by the time, t, in hours.
When traveling up a hill:
[tex]D_{up}(t)=3t[/tex]
When traveling down a hill:
[tex]D_{down}(t)=8t[/tex]
When traveling along a flat portion:
[tex]D_{flat}(t)=5t[/tex]
Final answer:
To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance: D = V * T, where D is the distance, V is the velocity, and T is the time.
Explanation:
To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance:
D = V * T
where D is the distance, V is the velocity, and T is the time.
For the uphill portion, the speed is 3 mph, so the linear model would be:
D = 3T
For the downhill portion, the speed is 8 mph, so the linear model would be:
D = 8T
For the flat portion, the speed is 5 mph, so the linear model would be:
D = 5T
MR is the angle bisector of
Answer:
∠NMP, so m∠1 = m∠2.
Step-by-step explanation:
pls need help with this problem
Answer:
36x^2
Step-by-step explanation:
Since area is base*height:
8*4.5=36
x*x=x^2
The monomial is 36x^2
In a survey of 1118 U.S. adults conducted in 2012 by the Financial Industry Regulatory Authority, 810 said they always pay their credit cards in full each month. Construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month. Round the answers to three decimal places. A 99% confidence ineral for the proportion of U.S. adults who pay their credit cards in full each month is ____________< p < ________________
Answer:
99.8% confidence:
[0.6833, 0.7656]
99% confidence:
0.6902 < p < 0.7588
Step-by-step explanation:
Lets call p the probability that a U.S adult pay its credit catrd in full each month. Lets call Y the random variable that counts the total of persons that paid their credit card in full from a random sample of 1118 adults. Y is a random variable with distribution Y ≈ Bi(1118,p) . The expected value of Y is μ = 1118*p and its variance is σ² = 1118*p(1-p). The proportion of adults that paid their credit card is a random variable, lets call it X, obtained from Y by dividing by 1118. The expected value is p and the variance is p(1-p)/1118. The Central Limit Theorem states that X can be approximated by a random variable with Normal Distribution, with mean μ = p, and standard deviation σ = √(p(1-p)/1118).
The standarization of X, W, is a random variable with distribution (approximately) N(0,1) obtained from X by substracting μ and dividing by σ. Thus
[tex] W = \frac{X - \mu}{\sigma} = \frac{X - p}{\sqrt{\frac{p(1-p)}{1118}}} [/tex]
If we want a 99.8% confidence interval, then we can find a value Z such that P(-Z < W < Z) = 0.998. If we do so, then P(W < Z) = 0.999, therefore, Ф(Z) = 0.999, were Ф is the cummuative distribution function of the standard normal distribution. The values of Ф can be found on the attached file. We can find that Ф(3.08) = 0.999, thus, Z = 3.08.
We have that P(-3.08 < W < 3.08) = 0.998, in other words
[tex] P(-3.08 < \frac{X-p}{\sqrt{\frac{p(1-p)}{1118}}} < 3.08) = 0.998 [/tex]
[tex] P(-3.08 * \sqrt{\frac{p(1-p)}{1118}} < X-p < 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]
[tex] P(-X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < -p < -X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]
Taking out the - sign from the -p and reversing the inequalities, we finally obtain
[tex] P(X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < p < X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]
As a conclusion, replacing p by its approximation X, a 99.8% confidence interval for p is
[tex] [X -3.08 * \sqrt{\frac{X(1-X)}{1118}}\, , \, X + 3.08 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]
replacing X with the proportion of the sample, 810/1118 = 0.7245, we have that our confidence interval is
[tex] [0.7245 -3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}\, , \, 0.7245 + 3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}] [/tex]
By solving the fraction and multypling by 3.08, we have
[0.7245 - 0.0411 < p < 0.7245 + 0.0411]
FInally, the 99.8 % confidence interval for p is
[0.6833, 0.7656]
If p is 0.99 (99% confidence), then we would want Z such that Ф(Z) = 0.995, by looking at the table we have that Z is 2.57, therefore the 99% interval for p is
[tex] [X -2.57 * \sqrt{\frac{X(1-X)}{1118}}\, , \, X + 2.57 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]
and, by replacing X by 0.7245 we have that the 99% confidence interval is
[0.6902, 0.7588]
thus, 0.6902 < p < 0.7588
I hope i could help you!
Final answer:
To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full, we use the sample proportion, the z-value for the confidence level, and the sample size to calculate the margin of error and produce the interval.
Explanation:
To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month from a survey result where 810 out of 1118 respondents paid their credit cards in full, we will use the formula for the confidence interval of a proportion: CI = p ± Z*sqrt((p(1-p))/n). Here, 'p' is the sample proportion, 'Z' is the z-value corresponding to the desired confidence level, and 'n' is the sample size.
The sample proportion (p) is 810/1118. To find the appropriate 'Z' value for a 99.8% confidence interval, we can refer to a standard normal distribution table or use a calculator to find that Z is approximately 3.08.
Step-by-step:
Find the sample proportion: p = 810/1118
Determine the 'Z' value for 99.8% confidence level: Z ≈ 3.08
Calculate the standard error (SE): SE = sqrt(p(1-p)/n)
Calculate the margin of error (ME): ME = Z * SE
Construct the confidence interval: (p - ME, p + ME)
After calculations, we can state that the confidence interval for the proportion of U.S. adults who pay their credit cards in full each month is between two values with three decimal places precision (the exact numbers would need to be calculated).
A set of 20 cards consists of 12 red cards and 8 black cards. The cards are shuffled thoroughly and you choose one at random, observe its color, and replace it in the set. The cards are thoroughly reshuffled, and you again choose a card at random, observe its color, and replace it in the set. This is done a total of six times. Let X be the number of red cards observed in these six trials. The variance of X isA. 6.B. 3.60.C. 2.4.D. 1.44.
Answer:
the answer D=1.44
Step-by-step explanation:
if the X= number of red cards observed in 6 trials , since each card observation is independent from the others and the sampling process is done with replacement ( the card is observed, then returned and reshuffled) , X follows an binomial distribution.
X(x)= n!/((n-x)!*x!) *p^x *(1-p)^(n-x)
where n = number of trials = 6 , x= number of red cards observed , p= probability of obtaining a red card in one try
the probability of obtaining the card in one try is
p = number of red cards / total number of cards = 12/ (12+8) = 0.6
since we know that X has a binomial distribution, the variance of this kind of distribution is
variance = σ² = n * p * (1-p)
therefore the variance of X is
variance = σ² = n * p * (1-p) = 6 * 0.6 * (1-0.6) = 1.44
According to a study conducted in one city, 35% of adults in the city have credit card debts more than $2.000. A simple random sample of n=250 adults is obtained from the city. Describe the sampling distribution of P^, the sample proportion of adults who have credit card debts of more than $2000.(Round to three decimal places when necessary.)Select from one of the 4 answers belowA. approximately- normal; \mu p=0.35, \sigma p=0.030B. approximately- normal; \mu p=0.35, \sigma p=0.001C. exactly- normal; \mu p=0.35, \sigma p=0.030D.Binomial; \mu p=87.5, \sigma p=7.542
Final Answer:
A. approximately-normal; μp = 0.35, σp = 0.030
1. Identify conditions for a normal sampling distribution:
Random sampling: The problem states it's a simple random sample.
Large sample size: n = 250 is greater than 10% of the population (assumed to be large), satisfying the condition.
Success-failure condition: np = 250 * 0.35 = 87.5 and n(1-p) = 250 * 0.65 = 162.5 are both greater than 10, meeting the condition.
2. Calculate the mean and standard deviation of the sampling distribution:
Mean (μp): μp = p = 0.35 (equal to the population proportion)
Standard deviation (σp): σp = sqrt(p(1-p)/n) = sqrt(0.35*0.65/250) ≈ 0.030
Final Answer:
A. approximately-normal; μp = 0.35, σp = 0.030
Explanation:
The sampling distribution of P^ is approximately normal due to the satisfaction of the conditions mentioned above.The mean of the sampling distribution is equal to the population proportion (0.35).The standard deviation of the sampling distribution is calculated as 0.030.When the student summarizes the data, she finds that 42 of the 50 business students and 38 of the 70 nursing students admitted to cheating in their courses. True or false? The counts suggest that the normal model is a good fit for the sampling distribution of sample differences.
Answer:
True
Step-by-step explanation:
The answer is "True" because the data set was picked at random.
a. Suppose that a single card is selected from a standard 52-card deck. What is the probability that the card drawn is a spade?
b. Now suppose that a single card is drawn from a standard 52-card deck, but it is told that the card is black.What is the probability that the card drawn is a spade?
The probability of drawing a spade from a 52-card deck is 0.25. If it's known that the card is black, the probability that it's a spade increases to 0.5.
Explanation:The subject of this question is the calculation of probability in a card game scenario. When a single card is selected from a standard 52-card deck, the probability that the card drawn is a spade is calculated by dividing the total number of spades in the deck (13) by the total number of cards in the deck (52). This gives a probability of 13/52 or 0.25.
However, if a single card is drawn from a standard 52-card deck, and it is known that the card is black, then the total number of possible outcomes is reduced to 26 (the number of black cards in the deck). In this scenario, the probability that the card drawn is a spade becomes 13/26 or 0.5.
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The probability of drawing a spade from a 52-card deck is 1/4. If we already know the card is black, the probability increases to 1/2.
Explanation:This question relates to the field of probability. To address part a, in a standard 52-card deck, there are 13 spades. We therefore have a 13 out of 52 chance to draw a spade, simplifying to a probability of 1/4.
For part b, being told that the card drawn is black means our sample space becomes 26 (13 spades and 13 clubs). Since there are still 13 spades, the probability of drawing a spade given that the card is black is 13 out of 26, or 1/2.
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An angle's initial ray points in the 3-o'clock direction and its terminal ray rotates CCW. Let θ represent the angle's varying measure (in radians).
If θ=0.5θ=0.5 what is the slope of the terminal ray?
If θ=1.78θ=1.78, what is the slope of the terminal ray?
Write an expression (in terms of θθ) that represents the varying slope of the terminal ray.
Answer:
0.546 , -4.71
Step-by-step explanation:
Given:
An angle's initial ray points in the 3-o'clock direction and its terminal ray rotates counter -clock wise.
Here, Slope = tan\theta
If θ = 0.5
Then, Slope = tan(θ) = tan(0.5) = 0.546
If θ = 1.78
Then, Slope = tan(θ) = tan(1.78) = - 4.71
The expression (in terms of θ) that represents the varying slope of the terminal ray.
Slope = m = tanθ, where θ is the varying angle
A) The slope of the terminal ray when θ = 0.5 radians is; 0.5463
B) The slope of the terminal ray when θ = 1.78 radians is; -4.71
C) The expression that will represent the varying slope of the terminal ray is;
tan (Δy/Δx)
We are given that θ represents the angle's varying measure (in radians).
Now, in mathematics, slope is simply the tangent of an angle. Thus;
A) At θ = 0.5 radians ,
Slope of terminal ray = tan θ
Slope = tan 0.5
Using radians calculator, tan 0.5 = 0.5463
Thus, slope = 0.5463
B) At θ = 1.78
Slope of terminal ray = tan θ
Slope = tan 1.78
Using radians calculator, tan 1.78 = -4.71
Thus, slope = -4.71
C) The expression that will represent the varying slope of the terminal ray is;
tan (Δy/Δx)
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A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately normally distributed with variance \sigma ^{2} = 1000(psi) 2. A random sample of 12 specimens has a mean compressive strength of Mean \mu= 3250psi.
a.) Construct a 95% two-sided confidence interval on mean compressive strength.
b.) Construct a 99% two-sided confidence interval on the mean compressive strength. Compare the width of this confidence interval with the width of the one found in a)
Answer:
solution is in the image below
Step-by-step explanation:
Final answer:
To construct confidence intervals for the mean compressive strength of concrete with known variance, formulas involving the z-score are employed. The 95% confidence interval is narrower than the 99% interval, illustrating that higher confidence requires a wider interval to encapsulate the true mean with more assurance.
Explanation:
To construct a confidence interval for the mean compressive strength of concrete, we utilize the formula for the confidence interval of the mean when the population variance (σ2) is known. Given that the variance is 1000 psi2 and the mean (μ) is 3250 psi for a sample of n=12 specimens, the z-score corresponding to a 95% confidence level is approximately 1.96, and for a 99% confidence level, the z-score is approximately 2.576.
95% Confidence Interval:
CI = μ ± z(σ/√n)
= 3250 ± 1.96(/1000/12)
= 3250 ± (1.96)(28.8675)
= 3250 ± 56.61
= (3193.39, 3306.61) psi
99% Confidence Interval:
CI = μ ± z(σ/√n)
= 3250 ± 2.576(/1000/12)
= 3250 ± (2.576)(28.8675)
= 3250 ± 74.36
= (3175.64, 3324.36) psi
Comparing the widths, the 99% confidence interval is wider than the 95% confidence interval, which is a reflection of increased certainty (or confidence level) requiring a wider interval.
A ship leaves port at 12:30 pm and travels Upper N 55 degrees Upper E at the rate of 8 mph. Another ship leaves the same port at 1:30 pm and travels Upper N 30 degrees Upper W at the rate of 10 mph. How far apart in miles are the ships at 2:00 pm ?
Answer:
d=0.0167mille
Step-by-step explanation:
we must determine its position (X-Y).
ship 1
[tex]v_{1}=8\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-12.5=1.5h\\d_{s1}=v_{1}*t_{t1}=8\frac{m}{h}*1.5h=12m[/tex]
ship 2
[tex]v_{2}=10\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-13.5=0.5h\\d_{s2}=v_{1}*t_{t2}=10\frac{m}{h}*0.5h=5m[/tex]
component x-y
[tex]sin55=\frac{y_{1}}{h};y_{1}=12*sin55=9.82\\ cos55=\frac{x_{1} }{h};x_{1}=5*cos55=6,88[/tex]
[tex]sin150=\frac{y_{2}}{h};y_{2}=5*sin150=2.5\\cos150=\frac{x_{2} }{h};x_{2}=5*cos150=-4.33[/tex]
[tex]s_{1}=(6.88,9.82); s_{2}=(-4.33,2,5)[/tex]
we must find the distance S1-S2
[tex]d_{s1-s2}=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}=\sqrt{(2.5-9.8)^{2}+(-4.33-6.88)^{2}}\\=\sqrt{(24.5)^{2}+(-11.21)^{2}}=26.94m[/tex]
but the units are requested to be miles
d=26.94m*mille/1,609.34m=0.0167mille
A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. A random sample of 25 bottles is selected, and the contents are measured. The sample yielded a mean content of 11.88 ounces, with a standard deviation of 0.24 ounces. With a 0.05 level of significance, test to see if the machine is in perfect adjustment. Assume the distribution of the population is normal.
Answer:
we reject H₀
Step-by-step explanation:
Normal Distribution
sample size n = 25 degees of fredom = 25 - 1 df = 24
sample standard deviation = s = 0,24
sample mean 11.88
We have a one tail test (left) investigation
1.-Test hypothesis
H₀ ⇒ null hypothesis μ₀ = 12
Hₐ ⇒ Alternative hypothesis μ₀ < 12
2.-Significance level 0,05 t(c) = - 1.7109
3.-Compute of t(s)
t(s) = ( μ - μ₀ )/s/√n ⇒ t(s) =[ ( 11.88 - 12 )*√25 ]/0.24
t(s) = - 0.12*5/0.24
t(s) = - 2.5
4.-We compare t(s) with t(c)
In this case t(s) < t(c) - 2.5 < -1.71
5.-t(s) is in the rejection region, we reject H₀
The machine is not adjusted
The owner of a meat market has an assistant who has determined that the weights of roasts are normally distributed, with a mean of 3.2 pounds and a standard deviation of 0.8 pounds. If a sample of 25 roasts yields a mean of 3.6 pounds, what is the Z-score for this ample mean?
Answer:
2.5
Step-by-step explanation:
We are given that
Mean=[tex]\mu=3.2[/tex] pounds
Standard deviation=[tex]\sigma=0.8[/tex] pounds
n=25
We have to find the Z-score if the mean of a sample of 25 roasts is 3.6 pounds.
We know that Z-score formula
[tex]Z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
We have [tex]\bar X=3.6[/tex]
Substitute the values then we get
Z- score=[tex]\frac{3.6-3.2}{\frac{0.8}{\sqrt{25}}}[/tex]
Z- score=[tex]\frac{0.4}{\frac{0.8}{5}}=\frac{0.4\times 5}{0.8}=2.5[/tex]
Hence, the Z-score value for the given sample mean=2.5
The Z-score for a sample mean of 3.6 pounds for 25 roasts, given a population mean of 3.2 pounds and a population standard deviation of 0.8 pounds, is calculated through the formula 'Z = (X - μ) / (σ / √n)' and is determined to be 2.5. The fact that the Z-score is positive indicates that the sample mean is above the population mean.
Explanation:The question presented involves calculation of a Z-score in a situation involving weights of roasts. The Z-score is a measure used in statistics that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. Here, we can use the formula for calculating Z-score from a sample mean:
Z = (X - μ) / (σ / √n)
where 'X' is the sample mean, 'μ' is the population mean, 'σ' is the population standard deviation, and 'n' is the size of the sample. Plugging in the values given in the problem, we get:
Z = (3.6 - 3.2) / (0.8 / √25) = 0.4 / 0.16 = 2.5
So, the Z-score for the sample mean of 25 roasts weighing 3.6 pounds each is 2.5.
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A random sample of 24 items is drawn from a population whose standard deviation is unknown. The sample mean is x⎯⎯ = 880 and the sample standard deviation is s = 5. Use Appendix D to find the values of Student’s t. (a) Construct an interval estimate of μ with 99% confidence. (Round your answers to 3 decimal places.) The 99% confidence interval is from 877.371 to (b) Construct an interval estimate of μ with 99% confidence, assuming that s = 10. (Round your answers to 3 decimal places.) The 99% confidence interval is from 874.742 to (c) Construct an interval estimate of μ with 99% confidence, assuming that s = 20. (Round your answers to 3 decimal places.) The 99% confidence interval is from 869.484 to (d) Describe how the confidence interval changes as s increases.
To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. The 99% confidence interval for the given sample mean (x = 880) and sample standard deviation (s = 5) is (877.371, 882.629). Assuming a different sample standard deviation, the 99% confidence intervals are (874.24, 885.76) for s = 10 and (868.48, 891.52) for s = 20. The confidence interval width increases as the sample standard deviation increases.
Explanation:To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. First, we find the value of the t-statistic for the given confidence level and degrees of freedom. For a 99% confidence level and a sample size of 24, the degrees of freedom is 23. Using Appendix D or a t-table, we find the critical t-value to be approximately 2.819.
(a) To construct a confidence interval with the given sample mean (x = 880) and sample standard deviation (s = 5), the margin of error is calculated as: E = t * (s / sqrt(n)) = 2.819 * (5 / sqrt(24)) = 2.819 * 1.021 = 2.88 (rounded to 3 decimal places). The 99% confidence interval is then calculated as: (x - E, x + E) = (880 - 2.88, 880 + 2.88) = (877.12, 882.88), rounded to 3 decimal places as (877.371, 882.629).
(b) Using the same method, but assuming a sample standard deviation of s = 10, the margin of error is calculated as: E = 2.819 * (10 / sqrt(24)) = 5.76 (rounded to 3 decimal places). The 99% confidence interval is then (874.24, 885.76).
(c) Assuming s = 20, the margin of error is calculated as: E = 2.819 * (20 / sqrt(24)) = 11.52 (rounded to 3 decimal places). The 99% confidence interval is then (868.48, 891.52).
(d) As the sample standard deviation (s) increases, the margin of error (E) and confidence interval width will also increase. This means that as s increases, we become less certain about the true population mean, resulting in a wider range of values in the confidence interval.
A polymer is manufactured in a batch chemical process. Viscosity measurements are normally made on each batch, and long experience with the process has indicated that the variability in the process is fairly stable with o 20. Fifteen batch
viscosity measurements are given as follows:
724, 718, 776, 760, 745, 759, 795, 756, 742, 740, 761, 749, 739, 747, 742
A process change that involves switching the type of catalyst used in the process is made. Following the process change, eight batch viscosity measurements are taken:
735, 775, 729, 755, 783, 760, 738, 780
Assume that process variability is unaffected by the catalyst change. If the difference in mean batch viscosity is 10 or less, the manufacturer would like to detect it with a high probability.
(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions? Find the P-value.
(b) Find a 9096 confidence interval on the difference in mean batch viscosity resulting from the process change.
(c) Compare the results of parts (a) and (b) and discuss your findings.
Answer:
a) Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]
Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]
[tex]p_v =P(Z<-1.904)=0.0284[/tex]
Comparing the p value with the significance level [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 the difference between the two groups is significantly lower than 10.
b) The 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]
c) If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{s}=750.2[/tex] represent the mean for the sample standard process
[tex]\bar X_{n}=756.875[/tex] represent the mean for the sample with the new process
[tex]s_{s}=19.128[/tex] represent the sample standard deviation for the sample standard process
[tex]s_{n}=21.283[/tex] represent the sample standard deviation for the new process
[tex]\sigma_s=\sigma_n=\sigma=20[/tex] represent the population standard deviation for both samples.
[tex]n_{s}=15[/tex] sample size for the group Cincinnati
[tex]n_{n}=8[/tex] sample size for the group Pittsburgh
z would represent the statistic (variable of interest)
Concepts and formulas to use
(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions?
We need to conduct a hypothesis in order to check if the difference in mean batch viscosity is 10 or less system of hypothesis would be:
Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]
Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]
We have the population standard deviation, so for this case is better apply a z test to compare means, and the statistic is given by:
[tex]z=\frac{(\bar X_{s}-\bar X_{n})-\Delta}{\sqrt{\frac{\sigma^2_{s}}{n_{s}}+\frac{\sigma^2_{n}}{n_{n}}}}[/tex] (1)
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
With the info given we can replace in formula (1) like this:
[tex]z=\frac{(750.2-756.875)-10}{\sqrt{\frac{20^2}{15}+\frac{20^2}{8}}}}=-1.904[/tex]
Statistical decision
Since is a one tail left test the p value would be:
[tex]p_v =P(Z<-1.904)=0.0284[/tex]
Comparing the p value with the significance level [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 the difference between the two groups is significantly lower than 10.
(b) Find a 90% confidence interval on the difference in mean batch viscosity resulting from the process change.
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_s -\bar X_n) \pm z_{\alpha/2}\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_s})}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_s -\bar X_n =750.2-756.875=-6.675[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_n})}[/tex]
And replacing we have:
[tex]SE=\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=8.756[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]-6.675-1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=-21.035[/tex]
[tex]-6.675+1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=7.685[/tex]
So on this case the 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]
(c) Compare the results of parts (a) and (b) and discuss your findings.
If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.
Final answer:
The viscosity analysis involves a hypothesis test to check for significant changes in mean batch viscosity and constructing a confidence interval to estimate the range of this difference following a process change in the manufacturing of a polymer.
Explanation:
The question involves conducting a hypothesis test and constructing a confidence interval to analyze the change in mean batch viscosity of a polymer before and after a process change. A hypothesis test will be used to determine if the mean batch viscosity has changed significantly, and the confidence interval will provide a range in which the true difference in means likely falls.
Hypothesis Test:
Calculate the mean of both sets of viscosity measurements.
Set up the null hypothesis (H0) and the alternative hypothesis (H1): H0: μ1 - μ2 <= 10 (no significant change), H1: μ1 - μ2 > 10 (significant change).
Determine the test statistic using a t-test for two independent samples.
Calculate the p-value associated with the test statistic.
Compare the p-value with the significance level (α = 0.10). If p <= α, reject H0; otherwise, fail to reject H0.
Calculate the standard deviation of both sets of measurements.
Determine the standard error of the difference in means.
Find the appropriate t-distribution critical value based on the given confidence level (90%) and degrees of freedom.
Calculate the 90% confidence interval using the difference in means ± the margin of error.
Compare the results of the hypothesis test and the confidence interval. If the hypothesis test leads to rejection of the null hypothesis while the confidence interval for the difference in means includes values greater than 10, there is evidence suggesting a significant change in mean batch viscosity. However, if the hypothesis test does not lead to rejection of H0 and the confidence interval includes 10, it suggests that the process change did not have a significant effect on mean batch viscosity.
A rectangular box is to have a square base and a volume of 20 ft3. If the material for the base costs $0.35 per square foot, the material for the sides costs $0.10 per square foot, and the material for the top costs $0.15 per square foot, determine the dimensions of the box that can be constructed at minimum cost.
A. length ft
B. width ft
C. height ft.
Answer:
Length = 2 ft
Width = 2 ft
Height = 5 ft
Step-by-step explanation:
Let the square base of the box has one side = x ft
Therefore, area of the base = x² ft²
Cost of the material to prepare the base = $0.35 per square feet
Cost to prepare the base = $0.35x²
Let the height of the box = y ft
Then the volume of the box = x²y ft³ = 20
[tex]y=\frac{20}{x^{3} }[/tex] -----(1)
Cost of the material for the sides = $0.10 per square feet
Area of the sides = 4xy
Cost to prepare the sides of the box = $0.10 × 4xy
= $0.40xy
Cost of the material to prepare the top = $0.15 per square feet
Cost to prepare the top = $0.15x²
Total cost of the box = 0.35x² + 0.40xy + 0.15x²
From equation (1),
Total cost [tex]C=0.35x^{2}+(0.40x)\times \frac{20}{x^{2} }+0.15x^{2}[/tex]
[tex]C=0.35x^{2}+\frac{8}{x}+0.15x^{2}[/tex]
[tex]C=0.5x^{2}+\frac{8}{x}[/tex]
Now we take the derivative of C with respect to x and equate it to zero,
[tex]C'=0.5(2x)-\frac{8}{x^{2}}[/tex] = 0
[tex]x-\frac{8}{x^{2}}=0[/tex]
[tex]x=\frac{8}{x^{2} }[/tex]
[tex]x^{3}=8[/tex]
x = 2 ft.
From equation (1),
[tex](2)^{2}y=20[/tex]
4y = 20
y = 5 ft
Therefore, Length and width of the box should be 2 ft and height of the box should be 5 ft for the minimum cost to construct the rectangular box.
The volume of a box is the amount of space in it.
The dimensions that minimize cost are:
[tex]\mathbf{Length = 2ft}[/tex]
[tex]\mathbf{Width = 2ft}[/tex]
[tex]\mathbf{Height= 5ft}[/tex]
The volume of the box is:
[tex]\mathbf{V = 20}[/tex]
Assume the dimension of the base is x, and the height is h.
The volume of the box will be:
[tex]\mathbf{V = x^2h}[/tex]
So, we have:
[tex]\mathbf{x^2h = 20}[/tex]
The area of the sides is:
[tex]\mathbf{A_1 = 4xh}[/tex]
The cost of the side material is 0.10.
So, the cost is:
[tex]\mathbf{C_1 = 0.10 \times 4xh}[/tex]
[tex]\mathbf{C_1 = 0.4xh}[/tex]
The area of the base is:
[tex]\mathbf{A_2 = x^2}[/tex]
The cost of the base material is 0.35.
So, the cost is:
[tex]\mathbf{C_2 = 0.35x^2}[/tex]
The area of the top is:
[tex]\mathbf{A_3 = x^2}[/tex]
The cost of the top material is 0.15.
So, the cost is:
[tex]\mathbf{C_3 = 0.15x^2}[/tex]
So, the total cost is:
[tex]\mathbf{C = 0.4xh + 0.35x^2 + 0.15x^2}[/tex]
[tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]
Recall that: [tex]\mathbf{x^2h = 20}[/tex]
Make h the subject
[tex]\mathbf{h = \frac{20}{x^2}}[/tex]
Substitute [tex]\mathbf{h = \frac{20}{x^2}}[/tex] in [tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]
[tex]\mathbf{C = 0.4x \times \frac{20}{x^2} + 0.5x^2}[/tex]
[tex]\mathbf{C = \frac{8}{x} + 0.5x^2}[/tex]
Differentiate
[tex]\mathbf{C' = -\frac{8}{x^2} + x}[/tex]
Set to 0
[tex]\mathbf{-\frac{8}{x^2} + x = 0}[/tex]
Rewrite as:
[tex]\mathbf{x = \frac{8}{x^2}}[/tex]
Multiply both sides by x^2
[tex]\mathbf{x^3 = 8}[/tex]
Take cube roots of both sides
[tex]\mathbf{x = 2}[/tex]
Recall that:
[tex]\mathbf{h = \frac{20}{x^2}}[/tex]
So, we have:
[tex]\mathbf{h = \frac{20}{2^2}}[/tex]
[tex]\mathbf{h = 5}[/tex]
Hence, the dimensions that minimize cost are:
[tex]\mathbf{Length = 2ft}[/tex]
[tex]\mathbf{Width = 2ft}[/tex]
[tex]\mathbf{Height= 5ft}[/tex]
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In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 81.0 km/h , to his enemy's car, which is going 109 km/h . The enemy's car is 13.4 m in front of the Indy's when he lets go of the grenade.
A. If Indy throws the grenade so its initial velocity relative to him is at an angle of 45 above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance.
B. Find the magnitude of the velocity relative to the earth.
Answer:
A. 18.21 m/s or 65.6 km/h
B. 135.5 km/h
Step-by-step explanation:
A. The vertical speed of the grenade is ...
vv = v·sin(45°) = v/√2
The time in air is given by the equation for ballistic motion:
h = -4.9t² +vv·t = 0
t(vv -4.9t) = 0 . . . . . factor out t
v/√2 = 4.9t
v = t·4.9√2 . . . . . . solve for v
__
The horizontal distance the grenade travels in t seconds is ...
d = vh·t = v·cos(45°)·t = vt/√2 = (t·4.9√2)(t/√2) = 4.9t² . . . . meters
1 m/s = 3.6 km/h, so the horizontal distance the target travels is ...
x = (109 - 81.0)/3.6·t + 13.4 = 70t/9 +13.4 . . . . meters in t seconds
We want the target distance to be the same as the grenade's distance, so the time required to hit the target is ...
d = x
4.9t² = (70/9)t +13.4
4.9(t² -(100/63)t) = 13.4
4.9(t -50/63)² = 13.4 +4.9(50/63)² ≈ 16.486
t = 50/63 + √(16.486/4.9) ≈ 2.62793 . . . seconds
This corresponds to a launch speed of ...
v = (4.9√2)t ≈ 18.21 . . . . meters/second
__
B. As we said, 1 m/s = 3.6 km/h, so the launch speed is about 65.558 km/h at an angle of 45° relative to the direction of travel. The magnitude of the velocity relative to the earth (ve) will be this vector added to 81.0 km/h in the direction of travel. The (horizontal, vertical) components of that sum are ...
(81.0, 0) + 65.558(cos(45°), sin(45°)) ≈ (127.357, 46.357) km/h
The magnitude is the Pythagorean sum:
ve = √(127.357² +46.357²) ≈ 135.5 . . . . km/h
The initial velocity Indy must throw the grenade at is found using the equations of motion, and it must be around 14.5 m/s. The magnitude of the velocity of the grenade relative to earth can be found by adding this initial velocity to the velocity of Indy's car, resulting in approximately 37 m/s.
Explanation:This is a physics problem involving the concept of relative velocity. We know that when Indy lets go of the grenade, it will move at the same speed as his car (81.0 km/h) relative to the ground plus the additional speed he gives it. Let's assume Indy throws the grenade when the enemy car is exactly at this distance.
A. If the grenade is thrown at an angle of 45 degrees above the horizontal, we can apply the equations of motion to find the initial velocity of the grenade. Using the equation, Range = [v^2 * sin(2*angle)]/g, we can solve for v, which should be around 14.5 m/s.
B. The magnitude of the velocity of the grenade relative to earth would be the vector sum of the initial velocity of the grenade and the velocity of Indy's car. Converting the car's speed to m/s (81.0 km/h = 22.5 m/s) and using vector addition, we can add 22.5 m/s to the initial speed of the grenade (14.5 m/s) to receive a total of approximately 37 m/s.
Learn more about Relative Velocity here:https://brainly.com/question/34025828
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[tex]97.28 \div 19[/tex]