Answer:
Answer explained below
Step-by-step explanation:
Let a be the working hours of plant A.
Let b be the working hours of plant H.
Let c be the working hours of plant M.
We have to minimize , Z = 60a + 92b + 140c
subject to constraint , 40a + 20b + 40c >=350
20a + 40b + 40c >=240
where a>=0 , b>=0 ,c>=0
So ,by solving this , a = 7.67 , b= 2.17 , c = 0
So ,working hours of plant A = 8 hrs
working hours of plant H = 3 hrs
working hours of plant M = 0 hrs
Answer:
A = 8hrs
H = 3hrs
M = 0hrs
Step-by-step explanation:
Let a be the working hours of plant A.
Let b be the working hours of plant H.
Let c be the working hours of plant M.
∴ Hours of plants A,H,M are a,b,c respectively.
We have to minimize the cost of production, Z = 60a + 92b + 140c
Regular ice cream: 40a + 20b + 40c = 350
Deluxe ice cream: 20a + 40b +40c =240
Where a>=0, b>=0, c>=0
So, by solving this, a= 7.67, b= 2.77,c= 0
So working hours of plant A = 8 hrs
Working hours of plant H = 3 hrs
Working hours of plant M = 0 hrs
In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the Select one US National Academy of Sciences, reviewed a study measuring bone quality 10 points and levels of vitamin-D in a random sample from bodies of 675 people who died in good health. 8.5% of the 82 bodies with low vitamin-D levels (below 50 nmol/L) had weak bones. Comparatively, 1% of the 593 bodies with regular vitamin-D levels had weak bones. Is a normal model a good fit for the sampling distribution? A. Yes, there are close to equal numbers in each group. B. O Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group. C. O No, the groups are not the same size. D. O No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.
Answer:
B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
The correct answer is:
B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.
As regards using the normal model, the correct answer is D. No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.
Why can't the normal model be used?In sampling distributions, the normal model can be used if np ≥ 10 and n (1 - p) ≥ 10.
In this case, those with weak bones are:
= 8.5% x 82
= 6.97 people which is less than 10
= 1% x 593
= 5.93 people
We do not have 10 or more people for the sample sizes so the normal model will not be a good fit.
Find out more on the normal model at https://brainly.com/question/15399601.
A professor compared differences in class grades between students in their freshman, sophomore, junior, and senior years of college. If different participants were in each group, then what type of statistical design is appropriate for this study?(a) a two-independent sample t test(b) a one-way between-subjects ANOVA(c) a two-way between-subjects ANOVA(d) both a two-independent sample t test and a one-way between-subjects ANOVA
Answer:
(b) a one-way between-subjects ANOVA
That's the correct option since we have one factor (class grade) and we have more than two groups.
Step-by-step explanation:
(a) a two-independent sample t test
We can't apply a two independnet t test since we are comparing more than two groups (Freshman, sophomore, Junior and senior). And for this case when we have more than two groups, the most powerful method is the one way ANOVA between subjects.
(b) a one-way between-subjects ANOVA
That's the correct option since we have one factor (class grade) and we have more than two groups.
One way Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
(c) a two-way between-subjects ANOVA
We can't apply a two way ANOVA since we have just one factor (or variable of interest) the class grades measured with a score. So then is not appropiate use this method for this case.
(d) both a two-independent sample t test and a one-way between-subjects ANOVA
False since we can't apply the two way ANOVA, that's not correct.
A BINGO card is a 5 × 5 grid. The center square is a free space and has no number. The first column is filled with five distinct numbers from 1 to 15, the second with five numbers from 16 to 30, the middle column with four numbers from 31 to 45, the fourth with five numbers from 46 to 60, and the fifth with five numbers from61 to 75.
Since the object of the game is to get five in a row horizontally, vertically, or diagonally, the order is important.
How many BINGO cards are there?
Answer:
[tex]5.52*10^{26}[/tex]
Step-by-step explanation:
For the columns with 5 slots, there are 15 distinct options to fill in. The number of ways to fill them in is
15 * 14 * 13 * 12 * 11 = 360360 ways (order matters)
For the column with 4 slots and 15 options. The number of ways to fill them in is
15 * 14 * 13 * 12 = 32760 ways (order matters)
Since a bingo card has 4 columns of 5 slots and 1 column with 4 slots, the total number of combination there is:
[tex]360360^4*32760 \approx 5.52*10^{26}[/tex] (order matters)
Park officials make predictions of times to the next eruption of a particular geyser, and collect data for the errors (minutes) in those predictions. The display from technology available below results from using the prediction errors to test the claim that the mean prediction error is equal to zero. Comment on the accuracy of the predictions. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Answer:
a) Null hypothesis: [tex]\mu =0[/tex]
Alternative hypothesis: [tex]\mu \neq 0[/tex]
b) t =-7.44
c) [tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]
d) Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.
We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.
Step-by-step explanation:
Assuming this output:
[tex]t_{difference}=-0.395[/tex]
t(observed value) =-7.44
t(Critical value )= 1.984
DF = 98
p value (two tailed) < 0.0001
[tex]\alpha =0.05[/tex]
a) What are the null and alternative hypothesis
Null hypothesis: [tex]\mu =0[/tex]
Alternative hypothesis: [tex]\mu \neq 0[/tex]
The reason is because the output says a bilateral test so for this case w eselect this option.
b) Identify the statistic
The correct formula for the statistic is given by:
[tex]t=\frac{\bar X_1 -\bar X_2 -0}{\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}}[/tex]
Based on the output the calculated value its t =-7.44
c) Identify the p value
We have the degrees of freedom given 98
Based on the alternative hypothesis the p value is given by:
[tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]
d) State the final conclusion that addresses the original claim
Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.
We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.
Compute the upper Riemann sum for the given function f(x)=x2 over the interval x∈[−1,1] with respect to the partition P=[−1,−14,14,34,1].
Answer:
Upper Riemann Sum is 9/16
Step-by-step explanation:
Final answer:
To compute the upper Riemann sum for the function f(x) = x^2 over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1], we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval. The upper Riemann sum is obtained by summing up all these values.
Explanation:
To compute the upper Riemann sum, we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval.
Given the function f(x) = x^2 and the partition P = [-1, -1/4, 1/4, 3/4, 1], we can calculate the upper Riemann sum as follows:
Calculate the width of each subinterval: Δx = (1 - (-1)) / 4 = 1/2.
Find the value of the function at each partition point:
Multiply the value of the function at each partition point by the width of the corresponding subinterval:
Sum up all the values obtained: 1/2 + 1/32 + 1/32 + 9/32 + 1/2 = 17/16
Therefore, the upper Riemann sum for the given function over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1] is 17/16.
The yield stress of a random sample of 25 pieces of steel was measured, yielding a mean of 52,800 psi. and an estimated standard deviation of s = 4,600 psi. a. What is the probability that the population mean is less than 50,000 psi? b. What is the estimated fraction of pieces with yield strength less than 50,000 psi? c. Is this sampling procedure sampling-by-attributes or sampling-by-variable?
Answer:
Step-by-step explanation:
Given than n = 25 , mean = 52800 sd = 4600
1) P(X<50000) , so please keep the z tables ready
we must first convert this into a z score so that we can look for probability values in the z table
using the formula
Z = (X-Mean)/SD
(50000- 52800)/4600 = -0.608
checkng the value in the z table
P ( Z>−0.608 )=P ( Z<0.608 )=0.7291
b)
again using the same formula and converting to z score
we need to calculate P(X<50000)
Z = (X-Mean)/SD
(50000- 52800)/4600 = -0.608
P ( Z<−0.608 )=1−P ( Z<0.608 )=1−0.7291=0.2709
proportion is 27% approx
c)
When your data points are measurements on a numerical scale you have variables data , here yield stress is numeric in nature , hence its a sampling by variable plan
The term "between-subjects" refers to
a. observing the same participants in each group
b. observing different participants one time in each group
c. the type of post hoc test conducted
d. the type of effect size estimate measured
Answer:
b. observing different participants one time in each group
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
If we analyze the formula for the sum of squares between we see that we are subtracting the mean for each group minus the grand mean. But in order to find the mean of each group we just need to observe just one time the dependent variable of interest for each group.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
Solution to the problem
If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
If we analyze the formula for the sum of squares between we see that we are subtracting the mean for each group minus the grand mean. But in order to find the mean of each group we just need to observe just one time the dependent variable of interest for each group.
For this reason the best option on this case is:
b. observing different participants one time in each group
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property :
[tex]SST=SS_{between}+SS_{within}[/tex]
Answer:
b. observing different participants one time in each group
Step-by-step explanation:
Between-subjects or groups design is a common experimental design used in psychology and other social science fields. Between-subjects is a type of experimental design in which the subjects of an experiment are assigned to different groups or conditions or participants in which each subject is being tested base on only one of the experimental conditions at a time. In Between-subjects experimental study, participants can be part of the treatment group or the control group, but cannot be part of both. A complete new group is required for each group, If more than one treatment is tested. The other way of assigning test to participants is Within-subjects design.
The major difference between Between-subjects design and Within-subjects design is that in Within-subjects design, the same participants test all the conditions of the experiment while in Between-subjects design different participants test each condition of the experiment.
Jake save cash into 5 different bank accounts . What is the probability that Jade save most into account A and save least into account B?
Answer:0.05
Step-by-step explanation:
There are 5 different bank accounts
Probability that A will give maximum benefit i.e. Jake save most in account A
[tex]P_1=\frac{1}{5}[/tex]
so picking 1 bank we are left with 4 banks . So we Probability that he will save least into account B is [tex]P_2=\frac{1}{4}[/tex]
Probability that Jake save most into account A and save least into account B is
[tex]=P_1\times P_2[/tex]
[tex]=\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}[/tex]
It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of 200-500°F. In a test of one type of mask, 12 of 60 masks had lenses pop out at 250°. Construct a 90% upper confidence limit for the true proportion of masks of this type whose lenses would pop out at 250°. (Round your answers to four decimal places.)
The upper 90% confidence limit for the true proportion of firefighter masks of this type whose lenses would pop out at 250° is approximately 0.2949.
Explanation:To construct a 90% upper confidence limit for the true proportion of firefighter's masks whose lenses would pop out at 250°, firstly, we need to calculate the sample proportion (p) which is the ratio of the number of masks that had lenses popping out (12) to the total number of masks tested (60). Thus, the sample proportion is 12/60 = 0.2.
Next, we use the formula for an upper confidence limit for proportions: p + z*sqrt((p*(1-p))/n), where z is the z-score associated with the desired level of confidence (for 90% confidence, z is 1.645), p is the sample proportion, and n is the sample size.
So the upper 90% confidence limit would be calculated as: 0.2 + 1.645*sqrt((0.2*0.8)/60) = 0.2 + 1.645*0.05774 = 0.2949322. Rounding to four decimal places, we get an upper confidence limit of 0.2949.
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Write the formula for Newton's method and use the given initial approximation to compute the approximations x_1 and x_2. f(x) = x^2 + 21, x_0 = -21 x_n + 1 = x_n - (x_n)^2 + 21/2(x_n) x_n + 1 = x_n - (x_n)^2 + 21 x_n + 1 = x_n - 2(x_n)/(x_n)^2 + 21 Use the given initial approximation to compute the approximations x_1 and x_2. x_1 = (Do not round until the final answer. Then round to six decimal places as needed.)
Answer:
[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]
[tex]x_{1} = -10[/tex]
[tex]x_{2} = -3.95[/tex]
Step-by-step explanation:
Generally, the Newton-Raphson method can be used to find the solutions to polynomial equations of different orders. The formula for the solution is:
[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]
We are given that:
f(x) = [tex]x^{2} + 21[/tex]; [tex]x_{0} = -21[/tex]
[tex]f^{'} (x)[/tex] = df(x)/dx = 2x
Therefore, using the formula for Newton-Raphson method to determine [tex]x_{1}[/tex] and [tex]x_{2}[/tex]
[tex]x_{1} = x_{0} - \frac{f(x_{0} )}{f^{'}(x_{0})}[/tex]
[tex]f(x_{0}) = x_{0} ^{2} + 21 = (-21)^{2} + 21 = 462[/tex]
[tex]f^{'}(x_{0}) = 2*(-21) = -42[/tex]
Therefore:
[tex]x_{1} = -21 - \frac{462}{-42} = -21 + 11 = -10[/tex]
Similarly,
[tex]x_{2} = x_{1} - \frac{f(x_{1} )}{f^{'}(x_{1})}[/tex]
[tex]f(x_{1}) = (-10)^{2} + 21 = 100+21 = 121[/tex]
[tex]f^{'}(x_{1}) = 2*(-10) = -20[/tex]
Therefore:
[tex]x_{2} = -10 - \frac{121}{20} = -10+6.05 = -3.95[/tex]
Mr. Kim needs to buy 3 plane tickets for him, his wife & their 2 children at $259 for each person. How much will he spend on all of their plane tickets?
Answer: [tex]\$3108[/tex]
Step-by-step explanation:
Mr. Kim, his wife and two children are 4 persons. Now, he needs to buy 3 plane tickets for each one, which means he has to buy 12 plane tickets:
[tex](3 tickets)(4)=12 tickets[/tex]
On the other hand, we know each ticket costs [tex]\$259[/tex] per person. So, if we have 12 tickets to buy, we will have to multiply [tex]\$259[/tex] by 12:
[tex](12)(\$259)=\$3108[/tex] This is the amount Mr. Kim will spend in the plane tickets
Can the following points be on the graph of the equation x-y = 0? Explain
Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Consider the differences between right and left arm blood pressure measurements.
Right Arm 102 101 94 79 79
Left Arm 175 169 182 146 144
a. Find the values of d and sd (you may use a calculator).
b. Construct a 90% confidence interval for the mean difference between all right and left arm blood pressure measurements.
Answer:
a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]
b) [tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/tex]
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
Solution
Let put some notation
x=value for right arm , y = value for left arm
x: 102, 101,94,79,79
y: 175,169,182,146,144
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: 73, 68, 88, 67, 65
Part a
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]
Part b
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=5-1=4[/tex]
Now we need to calculate the critical value on the t distribution with 4 degrees of freedom. The value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], so we need a quantile that accumulates on each tail of the t distribution 0.05 of the area.
We can use the following excel code to find it:"=T.INV(0.05;4)" or "=T.INV(1-0.05;4)". And we got [tex]t_{\alpha/2}=\pm 2.13[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
Now we have everything in order to replace into formula (1):
[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.331[/tex]
[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.069[/tex]
So on this case the 90% confidence interval would be given by (63.331;81.069).
[tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/tex]
A calculus exam has a mean of µ = 73 and a standard deviation of σ = 4. Trina's score on the exam was 79, giving her a z-score of +1.50. The teacher standardized the exam distribution to a new mean of µ = 70 and standard deviation of σ = 5. What is Trina's z-score for the standardized distribution of the calculus exam?
Answer:
I get z = +1.8
x= 79
m= 73;s= 4;z= (x-m)/s= 1.5
hope it helps you
Trina's new z-score is calculated using the z-score formula z = (x-μ)/σ where the raw score, mean, and standard deviation are 79, 70, and 5, respectively. Therefore, Trina's z-score on the standardized distribution of the calculus exam is +1.80.
Explanation:The subject you are asking about is related to calculating the z-score on a standardized distribution in a calculus exam. The z-score, in simple terms, tells you how many standard deviations a given value is from the mean. It's a way of standardizing scores on different distributions.
Now, we know Trina's original z-score was +1.50 when the mean was 73 with a standard deviation of 4. This score was moved to a new distribution that has a mean of 70 and standard deviation of 5. We can calculate the new z-score using the formula: z = (x-μ)/σ, where x is the raw score, μ is the mean, and σ is the standard deviation.
Plugging Trina's score into the formula we get z = (79-70)/5. Thus, Trina's new z-score on the standardized distribution of the calculus exam is +1.80.
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I need help with 1 and 4 please!
Answer:
Step-by-step explanation:
1) The diagram is a polygon with unequal sides. The number of sides and angles is 5. This means that it is an irregular Pentagon. The formula for the sum of interior angles of a polygon is expressed as
(n - 2)180
Where n is the number if sides of the polygon. Since the number of sides of the given polygon is 5, the sum if the interior angles would be
(5-2)×180 = 540 degrees. Therefore,
10x - 3 + 5x + 2 + 7x - 11 + 13x - 31 + 8x - 19 = 540
43x - 62 = 540
43x = 540 + 62 = 602
x = 602/43 = 14
Angle S = 13 - 31 = 13×14 - 31 = 182 - 31
Angle S = 151 degrees.
4) The diagram is a rectangle. Opposite sides are equal. Triangle JNM is an isosceles triangle. It means that its base angles, Angle NMJ and angle NJM are equal. Therefore
3x + 38 = 7x - 2
7x - 3x = 38 + 2
4x = 40
x = 40/4
x = 10
Angle NMJ = 7x - 2 = 7×10 - 2 = 68 degrees.
Angle JML = 90 degrees ( the four angles in a rectangle are right angles). Therefore,
Angle NML = 90 - 68 = 22 degrees
Best answer gets brainliest!
You must find the horizontal distance between two towers (points A and B) at the same elevation on opposite sides of a wide canyon running east and west. The towers lie directly north and south of each other. You mark off an east/west line CD running perpendicular to AB.
A: From C you measure the angle between the two towers (angle ACB) as 88.60°. Given the distance from C to B is 389 feet, write an equation and solve it to find an expression for the distance AB to the nearest whole foot. (note: AB is perpendicular to CD.)
B: You want to check your work to make sure it’s right.You should be able to both measure and compute the angle at D. Knowing the distance between the two towers from above and the distance BD is 459 feet, what is the angle at D to the nearest hundredth degree?
C: What is angle CAD in radians? Give your answer rounded correctly to 4 decimal places.
A. 15917 ft B.∠D=88.35° C.0.0532 rad
Step-by-step explanation:
A. Given that angle ∠ACB = 88.60° and the distance from C to B is 389 ft then triangle ABC is right ,90° at B. Applying the formula for tangent of an angle which is;
Tan of an angle = opposite side length/adjacent side length
Tan Ф = O/A =AB/389 ft
Tan 88.60°= AB/389
AB=389*tan 88.60° = 389×40.92 =15916.87 ft ⇒15917 ft
B.
The distance from B to D is given as 459 ft and the distance between the towers , AB, is 15917 ft. To get angle ∠D apply the formula for tangent of an angle where ;
Tan ∠D=O/A =15917/459 =34.6775599129
∠D =tan⁻(34.6775599129)
∠D=88.35°
C. To get angle ∠A subtract the sum of angle ∠C and ∠D from 180°. Apply the sum of angles in a triangle theorem
∠A =180° - (88.60°+88.35°)
∠A = 180°-(176.95°)=3.05°
Changing degrees to radians you multiply the degree value with 0.0174533
3.05°=3.05*0.0174533=0.05323254 rad
To 4 decimal places
=0.0532 rad
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Keywords : horizontal distance, elevation, equation, expression, distance
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Answer:
Step-by-step explanation:
An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route canbe justified only if it reduces the mean travel time by more than 10 minutes.a. If u 1 is the mean for the scenic route and u 2 for the nonscenic route. what hypotheses should be tested?b. If u 1 is the mean for the nonscenic route and u Zfor the scenic route, what hypotheses should be tested?
Final answer:
The null and alternative hypotheses for testing the mean travel time on scenic and nonscenic routes.
Explanation:
a. In this case, the hypotheses to be tested are:
Null hypothesis (H0): The mean travel time for the nonscenic route is not shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 ≤ 10.
Alternative hypothesis (Ha): The mean travel time for the nonscenic route is shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 > 10.
b. If u1 is the mean for the nonscenic route and u2 for the scenic route, the hypotheses to be tested would be:
Null hypothesis (H0): The mean travel time for the scenic route is not shorter than 10 minutes compared to the nonscenic route, or in other words, u2 - u1 ≤ 10.
Alternative hypothesis (Ha): The mean travel time for the scenic route is shorter than 10 minutes compared to the non scenic route, or in other words, u2 - u1 > 10.
The correct options are : - (a) The correct hypothesis is 2. [tex]\(\mu_1 - \mu_2 < -10\)[/tex] and (b) The correct hypothesis is 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]
Part (a): [tex]\(\mu_1\)[/tex] is the mean for the scenic route and [tex]\(\mu_2\)[/tex] is the mean for the non-scenic route
The individual decides that the non-scenic route is justified only if it reduces the mean travel time by more than [tex]10[/tex] minutes. In this context, we need to test whether the mean travel time for the scenic route [tex]\(\mu_1\)[/tex] minus the mean travel time for the non-scenic route [tex](\(\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]
Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = -10\)[/tex]
Alternative Hypothesis[tex](\(H_a\)) : \(\mu_1 - \mu_2 < -10\)[/tex]
So, the correct option for (a) is: 2.[tex]\(\mu_1 - \mu_2 < -10\)[/tex]
Part (b): [tex]\(\mu_1\)[/tex] is the mean for the non-scenic route and [tex]\(\mu_2\)[/tex] is the mean for the scenic route
In this case, we need to test whether the mean travel time for the non-scenic route [tex](\(\mu_1\))[/tex] minus the mean travel time for the scenic route [tex](\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]
Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = 10\)[/tex]
Alternative Hypothesis [tex](\(H_a\)) : \(\mu_1 - \mu_2 < 10\)[/tex]
So, the correct option for (b) is: 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]
The complete Question is
An individual can take either a scenic route to work or a non-scenic route. She decides that use of the non-scenic route can be justified only if it reduces the mean travel time by more than 10 minutes.
(a) If μ1 is the mean for the scenic route and μ2 for the non-scenic route, what hypotheses should be tested?
1. μ1 − μ2 = −10
2. μ1 − μ2 < −10μ1 − μ2 = −10
3. μ1 − μ2 > −10 μ1 − μ2 = −10
4. μ1 − μ2 ≠ −10μ1 − μ2 = 10
5. μ1 − μ2 < 10μ1 − μ2 = 10
6. μ1 − μ2 > 10
(b) If μ1 is the mean for the non-scenic route and μ2 for the scenic route, what hypotheses should be tested?
1. μ1 − μ2 = −10
2. μ1 − μ2 < −10μ1 − μ2 = −10
3. μ1 − μ2 > −10 μ1 − μ2 = −10
4. μ1 − μ2 ≠ −10μ1 − μ2 = 10
5. μ1 − μ2 < 10μ1 − μ2 = 10
6. μ1 − μ2 > 10
The seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row. If there are 30 rows altogether, how many seats are there in all? Select one: O
A. 1470
B. 1635
C. 2940
D. 3270
Answer:
B. 1635
Step-by-step explanation:
We have been given that the seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row.
We can see that the seating order is in form of arithmetic sequence, whose first term is 11 and common difference is 3.
Since there are 30 rows altogether, so we need to find the sum of 30 first terms of the sequence using sum formula.
[tex]S_n=\frac{n}{2}[2a+(n-1)d][/tex], where,
[tex]S_n=[/tex] Sum of n terms,
n = Number of terms,
a = First term,
d = Common difference.
Upon substituting our given value is above formula, we will get:
[tex]S_n=\frac{30}{2}[2(11)+(30-1)3][/tex]
[tex]S_n=15[22+(29)3][/tex]
[tex]S_n=15[22+87][/tex]
[tex]S_n=15[109][/tex]
[tex]S_n=1635[/tex]
Therefore, there are 1635 seats in all and option B is the correct choice.
Answer: the total number of seats is 1635
Step-by-step explanation:
For each additional row after the first row, there are 3 more seats than there are in the previous row. This means that the seats in each row is increasing in arithmetic progression with a common difference of 3. The formula for determining the sum of n terms of an arithmetic sequence is expressed as
Sn = n/2[2a + (n - 1)d]
Where
n represents the number of terms in the sequence.
a represents the first term,
d represents the common difference.
From the information given,
a = 11
n = 30
d = 3
Therefore,
S30 = 30/2[2×11 + (30 - 1)3]
S30 = 15[22 + 29×3]
S30 = 15 × 109= 1635
Suppose that you own a store that sells a particular stove for $1,000. You purchase the stoves from the distributor for $800 each. You believe that this stove has a lifetime which can be faithfully modeled as an exponential random variable with a parameter of lambda = 1/10, where the units of time are years. You would like to offer the following extended warranty on this stove: if the stove breaks within r years, you will replace the stove completely (at a cost of $800 to you). If the stove lasts longer than r years, the extended warranty pays nothing. Let $C be the cost you will charge the consumer for this extended warranty. For what pairs of numbers (C,r) will the expected profit you get from this warranty be zero. What do you think are reasonable choices for C and r? Why?
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The pairs (C,r) where the expected profit from a warranty is zero are calculated by balancing the warranty cost to the store for stoves failing within the warranty period with the revenue from selling the warranties. Reasonable values should consider customer appeal, market competition, and business risk.
Explanation:The question involves calculating the expected profit from a warranty which depends on an exponential random variable representing the lifetime of a stove. The lifetime of the stove, modeled as an exponential random variable, is typically used for modeling the lifespan of objects like mechanical or electronic devices whose failure rate is constant over time. This property is also known as the memoryless property. The distribution is defined by the parameter lambda (λ). In this case, λ equals 1/10, which suggests the average lifespan of the stove is 10 years.
The cost to replace the stove is $800, and the price charged for the warranty is denoted as C. The question asks for pairs of (C,r) where the expected profit is zero. This occurs when the cost of the warranty C is equal to the cost of replacing the stove, $800, over the time r in years for which the warranty is valid. It is also important to remember that not every stove will require a replacement, only those that fail within the warranty period r. The probability that a stove fails within the r years is computed using the exponential distribution's cumulative probability function as P(X
The expected profit is zero when the total warranty cost compensates for all the stoves replaced within their warranty periods i.e., C*P(X>r) = $800*P(X
Reasonable choices for C and r would depend on many factors such as the company's risk tolerance, competition in the warranty market, and customers' willingness to pay for extended warranties. A higher warranty price C increases profit but may discourage customers from buying the warranty. A longer warranty period r increases customer appeal but also the company's costs if more stoves fail and need to be replaced.
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The temperature in degrees Celsius on the surface of a metal plate is T(x, y) = 19 − 4x2 − y2 where x and y are measured in centimeters. Estimate the average temperature when x varies between 0 and 2 centimeters and y varies between 0 and 4 centimeters. °C
Answer:
Average value of temperature will be [tex]8.335^{\circ}C[/tex]
Step-by-step explanation:
We have given the temperature in degree Celsius [tex]T(x,y)=19-4x^2-y^2[/tex]
It is given that x varies between 0 to 2
So [tex]0\leq x\leq 2[/tex]
And y varies between 0 and 4
So [tex]0\leq y\leq 4[/tex]
So area between the region A = 4×2 = 8[tex]cm^2[/tex]
Now average temperature is given by
[tex]Average\ value=\frac{1}{A}\int \int T(x,y)dA[/tex]
[tex]Average\ value=\frac{1}{8}\int \int (19-4x^2-y^2)dxdy[/tex]
[tex]=\frac{1}{8}\int \int (19x-4\frac{x^3}{3}-y^2x)dy[/tex]
As limit of x is 0 to 2
[tex]=\frac{1}{8}\int (19\times 2-4\times \frac{2^3}{3}-y^2\times 2)-(19\times 0-4\times \frac{0^3}{3}-y^2\times 0))dy=\frac{1}{8}\int(38-\frac{32}{3}-2y^2)dy[/tex]
=[tex]=\frac{1}{8}(38y-\frac{32}{3}y-\frac{2y^3}{3})[/tex]
As limit of y is 0 to 4
So [tex]Average\ value=\frac{1}{8}(38\times 4-\frac{16}{3}\times 4-\frac{2\times 4^3}{3})-0=8.335^{\circ}C[/tex]
A box with a square base and open top must have a volume of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.(a) Sides of base (cm)(b) Height (cm)
If a box with a square base and open top must have a volume of 32,000 cm3. The dimensions of the box that minimize the amount of material used are:
(a) Sides of the base ≈ 40 cm
(b) Height ≈ 20 cm
What is the dimensions of the box?Let's the sides of the square base be x cm
Height of the box =h cm.
Volume of the box is given by:
Volume (V) = x² * h
We are given that the volume of the box is 32,000 cm³:
x²* h = 32,000
Surface area (A) of the box is the sum of the area of the base and the four sides:
Surface Area (A) = x² + 4xh
Express hin terms of x using the volume equation:
h = 32,000 / x²
Substitute
A = x² + 4x * (32,000 / x²)
A = x² + 128,000 / x
Let find the critical points of the surface area function
dA/dx = 2x - 128,000 / x^2 = 0
Solve for x
2x = 128,000 / x²
x³ = 64,000
x ≈ 40 cm
No, we can find the corresponding height
h = 32,000 / x²
h = 32,000 / (40²)
h ≈ 20 cm
Therefore the dimensions of the box that minimize the amount of material used are: Sides of the base ≈ 40 cm, Height ≈ 20 cm
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To minimize the amount of material used, we need to minimize the surface area of the box while keeping the volume constant. The dimensions of the box that minimize the amount of material used are approximately x ≈ 252.98 cm and h ≈ 32000 / (x²) cm.
Explanation:To find the dimensions of the box that minimize the amount of material used, we need to minimize the surface area of the box while keeping the volume constant. Let's assume the side length of the square base is x and the height of the box is h.
The volume of the box is given by [tex]V = x^2 * h = 32000 cm^3.[/tex]
The surface area of the box is given by [tex]A = x^2 + 4xh.[/tex]
To minimize A, we can use the volume equation to solve for h in terms of x: h = 32000 / (x^2). Substituting this into the surface area equation, we get:
[tex]A(x) = x^2 + 4x(32000 / x^2)[/tex]
To find the critical points of A(x), we differentiate A(x) with respect to x and set the result to 0: [tex]dA(x)/dx = 2x - (128000 / x^2) = 0.[/tex]
Solving this equation gives us x = sqrt(64000) or x ≈ 252.98 cm.
Since we are dealing with real-world dimensions, we can't have a negative value for x. Therefore, the dimensions of the box that minimize the amount of material used are approximately x ≈ 252.98 cm and h ≈ 32000 / (x²) cm.
The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. Each can holds a maximum of 12.50 ounces of soda. Every can that has more than 12.50 ounces of soda poured into it causes a spill and the can must go through a special cleaning process before it can be sold. What is the probability that a randomly selected can will need to go through this process? A) .1587 B) .6587 C) .8413 D) .3413
Answer: A) .1587
Step-by-step explanation:
Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.
i.e. [tex]\mu=12.30[/tex] and [tex]\sigma=0.20[/tex]
Let x denotes the amount of soda in any can.
Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.
Then, the probability that a randomly selected can will need to go through the mentioned process = probability that a randomly selected can has more than 12.50 ounces of soda poured into it =
[tex]P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587[/tex]
Hence, the required probability= A) 0.1587
Final answer:
The probability that a randomly selected can will need a special cleaning process because it holds more than 12.50 ounces of soda is calculated using the Z-score. The correct answer is A) 0.1587.
Explanation:
The question asks us to calculate the probability that a randomly selected can will require a special cleaning process if it holds more than 12.50 ounces of soda, given a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. To find this probability, we can use the standard normal distribution, also known as the Z-score.
First, we calculate the Z-score for 12.50 ounces using the formula: Z = (X - \\mu\) / \\sigma\, where X is the value we are evaluating (12.50 ounces), \mu is the mean (12.30 ounces), and \sigma is the standard deviation (0.20 ounce). Plugging in the values gives us Z = (12.50 - 12.30) / 0.20 = 1. The Z-score represents the number of standard deviations the value X is from the mean.
Next, to find the probability that a can will overflow (have more than 12.50 ounces), we look up the corresponding probability in the standard normal distribution table or use a calculator suitable for such statistical computations. The probability corresponding to a Z-score of 1 is approximately 0.8413. However, since we are interested in the probability of a can having more than 12.50 ounces, we need the area to the right of the Z-score, which is 1 - 0.8413 = 0.1587. Therefore, the probability that a randomly selected can will require a special cleaning process is 0.1587.
The correct answer to the question is A) 0.1587.
A school board has a plan to increase participation in the PTA. Currently only about 15 parents attend meetings. Suppose the school board plan results in logistic growth of attendance. The school board believes their plan can eventually lead to an attendance level of 45 parents. In the absence of limiting factors the school board believes its plan can increase participation by 20% each month. Let m denote the number of months since the participation plan was put in place, and let P be the number of parents attending PTA meetings
(a) What is the carrying capacity K for a logistic model of P versus m? K45
(b) Find the constant b for a logistic model. b15
(c) Find ther value for a logistic model. Round your answer to three decimal places
(d) Find a logistic model for P versus m. Pw
The carrying capacity K for the logistic model is 45. In the context of the problem, the constant b, which refers to the growth rate divided by the carrying capacity, would be approximately 0.00444. The logistic model representing P vs. m under these conditions would be P(m) = 45/ (1+ (45/15-1) * e^(-0.00444m)).
Explanation:In the field of mathematics, specifically in growth modeling, a logistic model incorporates a carrying capacity. The carrying capacity, denoted as K, is the maximum stable value of the population, in this case, the number of parents attending the PTA meeting. The carrying capacity is expected to be 45 in this instance.
The constant b in the logistic model can be found using the initial value (15 parents) and the growth rate (20%). However, the question does not specify whether this rate is relative or absolute. Assuming relative growth, the initial growth rate r is 0.2 and the constant b = r/K would be 0.00444. This is not the traditional definition of b in a logistic equation; typically, b would denote the initial population size, so the question seems to have a specific, non-standard usage in mind that we need to respect.
A good starting value of ther, in order to ensure convergence and stability of the numerical method, can be 15, the initial population size.
The logistic model would thus be represented as P(m) = K/ (1+ (K/P_0-1) * e^(-bm)), where P_0 is the initial number of parents, K is the carrying capacity, b is the constant/ growth rate, m is the number of months, and e is the mathematical constant approximated as 2.718.
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Is this sequence arithmetic, geometric, or neither 45, 59, 65, 70, 85
Answer: It's neither arithmetic or a geometric sequence
Explanation: For an arithmetic sequence, to get the common difference, we subtract the first term from the second, third term from second which should give the same value. While for a geometric sequence, to get the common ratio, divide the second term by the first term, third term by second term and so on, all of which should give the same answer. But in the above sequence, it doesn't follow this pattern
The sequence 45, 59, 65, 70, 85 is neither arithmetic nor geometric.
To determine if a sequence is arithmetic, one must check if the difference between consecutive terms is constant. For this sequence:
- The difference between the second and first terms is 59 - 45 = 14.
- The difference between the third and second terms is 65 - 59 = 6.
- The difference between the fourth and third terms is 70 - 65 = 5.
- The difference between the fifth and fourth terms is 85 - 70 = 15.
Since these differences are not the same, the sequence is not arithmetic.
To determine if a sequence is geometric, one must check if the ratio between consecutive terms is constant. For this sequence:
- The ratio between the second and first terms is 59/45.
- The ratio between the third and second terms is 65/59.
- The ratio between the fourth and third terms is 70/65.
- The ratio between the fifth and fourth terms is 85/70.
Simplifying these ratios:
- 59/45 = 1.3111
- 65/59 = 1.1017
- 70/65 = 1.0769
- 85/70 = 1.2143
Since these ratios are not the same, the sequence is not geometric.
To determine the aptness of the model, which of the following would most likely be performed?
A. Check to see whether the residuals have a constant variance
B. Determine whether the residuals are normally distributed
C. Check to determine whether the regression model meets the assumption of linearity
D. All of the above
Answer:
D. All of the above
Step-by-step explanation:
Linear regression models relate to some assumptions about the distribution of error terms. If they are violated violently, the model is not suitable for drawing conclusions. Therefore, it is important to consider the suitability of the model for information before further analysis can be performed based on this model.
The fit of the model is related to the remaining behavior complying with the basic assumptions for error values in the model. When a regression model is constructed from a series of data, it should be shown that the model responds to the standard statistical assumptions of the linear model because of conducting inference. Residual analysis is an effective tool to investigate hypotheses. This method is used to test the following statistical assumptions for a simple linear regression model:
-Regression function is linear in parameters,
-Error conditions have constant variance,
-Error conditions are normally distributed,
-Error conditions are independent.
If no statistical hypothesis of the model is fulfilled, the model is not suitable for data. The fourth hypothesis (independence of error conditions) relates to the regulation of time series data. It is now used some simple graphical methods to analyze analysis, the feasibility of a model, and some formal statistical tests. In addition, when a model fails to meet these assumptions, some data changes may be made to ensure that the assumptions are reasonable for the modified model.
At time t=0 sec, a tank contains 15 oz of salt dissolved in 50 gallons of water. Then brine containing 88oz of salt per gallon of brine is allowed to enter the tank at a rate of 55 gal/min and the mixed solution is drained from the tank at the same rate.a. How much salt is in the tank at an arbitrary time t? b. How much salt is in the tank after 25 min?
Answer:
a) [tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]
b) [tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]
Step-by-step explanation:
Part a
Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.
For this case we know that the rate of change can be expressed on this way:
[tex]Rate change = In-Out[/tex]
And we can name the rate of change as [tex]\frac{ds}{dt}=rate change[/tex]
And our variable s would represent the amount of salt for any time t.
We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.
For the concentration out we can assume that is [tex]\frac{s}{50gal}[/tex]
And now we can find the expression for the amount of salt after time t like this:
[tex]\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}[/tex]
And we have this differential equation:
[tex]\frac{dS}{dt} +\frac{1}{10} s = 40[/tex]
With the initial conditions y(0)=15 oz
As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:
[tex]\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}[/tex]
And then in order to solve the differential equation we need to multiply with the integrating factor like this:
[tex]e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt[/tex]
[tex]e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C[/tex]
Now we can divide both sides by [tex] e^{\frac{1}{25} t} [/tex] and we got:
[tex]s(t) =400 + C e^{-\frac{1}{10} t}[/tex]
Now we can apply the initial condition in order to solve for the constant C like this:
[tex]15 = 400+C[/tex]
[tex]C=-385[/tex]
And then our function would be given by:
[tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]
Part b
For this case we just need to replace t =25 and see what we got for the value of the concentration:
[tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]
a) The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex].
b) There are 4400 ounces of salt in the tank after 25 minutes.
How to model a dissolution process in a tankIn this question we must model the salt concentration in the tank ([tex]c(t)[/tex]), in ounces per gallon, as a function of time ([tex]t[/tex]), in minutes, in which salt is dissolved into the tank due to a constant inflow rate ([tex]\dot V[/tex]), in gallons. Likewise, an equal outflow rate exists with resulting concentration.
a) The process is modelled mathematically by a non-homogeneous first order differential equation and physically by principle of mass conservation, whose description is shown below:
[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V_{T}}\cdot c(t) = \frac{\dot V}{V_{T}}\cdot c_{in}[/tex] (1)
Where:
[tex]V_{T}[/tex] - Tank volume, in gallons[tex]c_{in}[/tex] - Inflow salt concentration, in ounces per gallonThe solution of this differential equation is:
[tex]c(t) = c_{in} + \left(\frac{m_{T}}{V_{T}}-c_{in} \right)\cdot e^{-\frac{\dot V}{V_{T}}\cdot t }[/tex] (2)
Where [tex]m_{T}[/tex] is the initial salt mass of the tank, in ounces.
And the salt mass in the tank at an arbitrary time [tex]t[/tex] ([tex]m(t)[/tex]), in ounces, is obtained by multiplying (2) by the volume of the tank. That is to say:
[tex]m(t) = c(t)\cdot V_{T}[/tex] (3)
By replacing [tex]c(t)[/tex] in (3) by (2), we have the following expression:
[tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex] (4)
The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]. [tex]\blacksquare[/tex]
b) If we know that [tex]V_{T} = 50\,gal[/tex], [tex]\dot V = 55\,\frac{gal}{min}[/tex], [tex]c_{in} = 88\,\frac{oz}{gal}[/tex], [tex]m_{T} = 15\,oz[/tex] and [tex]t = 25\,min[/tex], then the quantity of salt is:
[tex]m(25) = (50)\cdot (88)+[15-(50)\cdot (88)]\cdot e^{-\left(\frac{55}{50} \right)\cdot (25)}[/tex]
[tex]m(25) = 4400\,oz[/tex]
There are 4400 ounces of salt in the tank after 25 minutes. [tex]\blacksquare[/tex]
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To determine the health benefits of walking, researchers conduct a study in which they compare the cholesterol levels of women who walk at least 10 miles per week to those of women who do not exercise at all. The study finds that the average cholesterol level for the walkers is 198, and that the level for those who don't exercise is 223. Which of the following statements is true?I. This study provides good evidence that walking is effective in controlling cholesterol.II. This is an observational study, not an experiment.III. Although the study was conducted only on women, we can confidently generalize the results to men in the same age group.A. I onlyB. II onlyC. III onlyD. II and III onlyE. I and III only
Answer:
B. II only
Step-by-step explanation:
For this case the correct options would be:
B. II only
We analyze one by one the statements:
I.This study provides good evidence that walking is effective in controlling cholesterol
That's FALSE because there are many other factors that can influence the cholesterol and we just have two possible conditions (women who walk at least 10 miles and women who do not exercise at all), and we have just two averages obtained from two samples that we don't know if is a paired values or independent values. And is not appropiate to conclude this with the lack of info on this case.
II. This is an observational study, not an experiment
Correct we don't have a design for this experiment or factors selected in order to test the hypothesis of interest. So for this case we can conclude that we have an observational study for this case.
III. Although the study was conducted only on women, we can confidently generalize the results to men in the same age group.
False, we can't generalize results from women to men since they are different groups of people and with different characteristics.
In some year, a candy shop produced 100 boxes of candy per working day in January. In each month following this, the shop produced 25 more boxes of candy per working day in addition to the previous month.
How much boxes did the candy shop produce on each working day in October?
Answer: 325 boxes of candy will be produced on each working day in October
Step-by-step explanation:
The initial number of boxes of candy produced is 100. In each month following this, the shop produced 25 more boxes of candy per working day in addition to the previous month. This means that the number of boxes produced each month is increasing in arithmetic progression. The formula for the nth term of an arithmetic sequence is expressed as
Tn = a + (n-1)d
Where
a is the first term of the sequence
n is the number of terms
d is the common difference.
From the given information,
a = 100
d = 25
n = number of months from January to October = 10
Tn = 100 + (10 - 1)25
Tn = 100 + 9×25
Tn = 100 + 225 = 325
A pizza delivery chain advertises that it will deliver your pizza in no more 20 minutes from when the order is placed. Being a skeptic, you decide to test and see if the mean delivery time is actually more than advertised. For the simple random sample of 63 customers who record the amount of time it takes for each of their pizzas to be delivered, the mean is 20.49 minutes with a standard deviation of 1.42 minutes. Perform a hypothesis test using a 0.01 level of significance.
Answer:
[tex]t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738[/tex]
[tex]p_v =P(t_{(62)}>2.738)=0.0040[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .
Step-by-step explanation:
Data given and notation
[tex]\bar X=20.49[/tex] represent the mean time for the sample
[tex]s=1.42[/tex] represent the sample standard deviation for the sample
[tex]n=63[/tex] sample size
[tex]\mu_o =20[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean time is actually higher than 20 min, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 20[/tex]
Alternative hypothesis:[tex]\mu > 20[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=63-1=62[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(62)}>2.738)=0.0040[/tex]
And we can use the following excel code to find it:
"=1-T.DIST(2.738,62,TRUE)"
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .
To perform the hypothesis test, set up null and alternative hypotheses, calculate the test statistic, and compare it to the critical value from the t-distribution table.
Explanation:To perform a hypothesis test, we need to set up the null and alternative hypotheses. The null hypothesis, H0, states that the mean delivery time is 20 minutes or less. The alternative hypothesis, Ha, states that the mean delivery time is more than 20 minutes. We can perform a one-sample t-test using the sample mean, standard deviation, sample size, and the desired level of significance. Calculate the test statistic and compare it to the critical value from the t-distribution table. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis.
Learn more about Hypothesis test here:https://brainly.com/question/34171008
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Which polyhedron is convex?
Answer:
The fourth one
Step-by-step explanation:
Because Convex polygons can only be shapes like hexagons and squares and stuff like that. They cant be curved or zig zagged.
Answer:
The fourth one
Step-by-step explanation:
Convex polygons can be hexagons and stuff like that.