Answer:
We need to conduct a hypothesis in order to test the claim that the true proportion is equal to 14.6% or not. So we need to use a one proportion z test and the system of hypothesis are:
Null hypothesis:[tex]p=0.146[/tex]
Alternative hypothesis:[tex]p \neq 0.146[/tex]
A. One-proportion z-test
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
And the conditions required are:
1) The data comes from a random sampling
2) Independence condition between observations
3) np>10 and n(1-p)>10
4) The sample size is 10 times lower than the population size.
Step-by-step explanation:
Data given and notation
n=865 represent the random sample taken
X=159 represent the housing units that are vacant
[tex]\hat p=\frac{159}{865}=0.184[/tex] estimated proportion of vacant units
[tex]p_o=0.146[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Solution to the problem
We need to conduct a hypothesis in order to test the claim that the true proportion is equal to 14.6% or not. So we need to use a one proportion z test and the system of hypothesis are:
Null hypothesis:[tex]p=0.146[/tex]
Alternative hypothesis:[tex]p \neq 0.146[/tex]
A. One-proportion z-test
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
And the conditions required are:
1) The data comes from a random sampling
2) Independence condition between observations
3) np>10 and n(1-p)>10
4) The sample size is 10 times lower than the population size.
An incoming college student took her college’s placement exams in French and mathematics. In French, she scored 85, and in math 80. The overall results for both exams are approximately normal. The mean French test score was 72 with a SD of 12, while the mean math test score was 68 with a SD of 8. On which exam did she do better as compared with the other incoming college students? Compute the z-scores and the percentiles for each exam to support your answer
Answer:
In Math, she scored in the 93rd percentile, which is higher than the French percentile. So she did better on the Math exam as compared with the other incoming college students
Step-by-step explanation:
Z-score:
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
On which exam did she do better as compared with the other incoming college students?
On the exam for which she had the higher z-score.
Franch:
Scored 85.
Mean 72, SD = 12. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{85 - 72}{12}[/tex]
[tex]Z = 1.08[/tex]
[tex]Z = 1.08[/tex] has a pvalue of 0.8810, so her French score is in the 88th percentile.
Math:
Scored 80
Mean 68, SD = 8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 68}{8}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332.
In Math, she scored in the 93rd percentile, which is higher than the French percentile. So she did better on the Math exam as compared with the other incoming college students
An object is taken from a freezer at negative 8 degrees Upper C. Let t be the time in hours after the object was taken from the freezer. At time t the average temperature of the object is increasing at the rate of Upper T prime (t )equals 20 e Superscript negative 0.4 t degrees Celsius per hour. Find the temperature of the object at time t.
Answer:
[tex]T(t)=-50e^{-0.4t}+42[/tex]
Step-by-step explanation:
We are given that
[tex]T(0)=-8^{\circ} C[/tex]
[tex]T'(t)=20e^{-0.4t}[/tex]
Integrating on both sides
[tex]T(t)=20\int e^{-0.4 t}[/tex]
[tex]T(t)=\frac{20}{-0.4}e^{-0.4t}+C[/tex]
Using the formula
[tex]\int e^{ax}dx=\frac{e^{ax}}{a}+C[/tex]
Substitute t=0 and T(0)=-8
[tex]-8=-50+C[/tex]
[tex]C=-8+50=42[/tex]
Substitute the value of C
[tex]T(t)=-50e^{-0.4t}+42[/tex]
A pine tree is 64 feet 3 inches tall. How many inches tall is the pine tree?
Answer:
3 inches you are right
Step-by-step explanation:
A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by
cin()-2+sin(t/4) lb/gal. 40-18000, sin(t)- cos(t)-46324e( )
The complete question is;
A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine the amount of salt in the tank at any time
t, if the concentration in the inflow is variable and given by
c(t) = 2 + sin(t/4) lb/gal.
Answer:
dA/dt = (8 + 4sin(t/4)) - (A_t/100)
Step-by-step explanation:
Rate is given as;
dA/dt = R_in - R_out
R_in = (concentration of salt inflow) x (input rate of brine)
So, R_in = (2 + sin(t/4)) x 4 = (8 + 4sin(t/4))
The solution is being pumped out at the same rate, thus it is accumulating at the same rate.
After t minutes, there will be 400 + (0 x t) gallons left = 400 gallons left
Thus,
R_out =(concentration of salt outflow) x (output rate of brine)
R_out = (A_t/400) x 4 = A_t/100
Thus,
A_t = (100/t)(8t - 16cos(t/4) - 60)
Since we want to find the amount of salt, A(t), let's integrate;
Thus, A = 8t - 16cos(t/4) - (A_t/100)t
A = 60 from the question. Thus,
60 = 8t - 16cos(t/4) - (A_t/100)t
Let's try to make A_t the subject;
(A_t/100)t = 8t - 16cos(t/4) - 60
A_t = (100/t)(8t - 16cos(t/4) - 60)
Moira said, "If a right triangle has one side with a length of 3 units and one side with a length
of 4 units, then the third side has a length of 5 units."
Moira's claim is not true. Give an example of a right triangle with sides of length 3, 4, and a
number other than 5. Explain your reasoning.
If the two sides with length 3 and 4 are the two legs, then the missing side, i.e. the hypotenuse, is indeed 5.
But it could also be the case that 4 is the hypotenuse and 3 is one of the legs. In this case, the missing side is the other leg, so we calculate it using
[tex]\sqrt{4^2-3^2}=\sqrt{16-9}=\sqrt{7}[/tex]
So, a right triangle with legs [tex]\sqrt{7}[/tex] and 3 has an hypotenuse of 4.
Moira's statement correctly applies the Pythagorean theorem, implying it is impossible to have a right triangle with sides of 3, 4, and a value other than 5 and still conform to the requirements of a right-angled triangle.
The Pythagorean theorem, where a2 + b2 = c2 applies, and c represents the length of the hypotenuse, while a and b represent the lengths of the other two sides. However, the question asked to provide an example that opposes Moira's claim by asking for a right triangle with sides of length 3, 4, but with a hypotenuse different from 5. This is a trick question because, according to the Pythagorean theorem, if a triangle has sides of 3 and 4 units, the hypotenuse must indeed be 5 units to satisfy the equation 32 + 42 = 52 (9 + 16 = 25). Therefore, there cannot be a right-angle triangle with the sides 3, 4, and a number other than 5, as it would violate the Pythagorean theorem.
There is another way to rank the difficulty of test questions using Networks: Each student gives scores for each problem, then we construct a network whose nodes/vertices are the problems of the exam. There is an arc from problem A to B for student k that did better in problem B than in problem A, (i.e., if sA, sB are the scores of those problems, sB > sA). The weight of the arc yk associated to student k equals (approximately) the difference of score the student received in those two problems sB −sA = yk. Explain why the Massey’s method we saw in class can be used for rating exam problems by difficulty.
Answer:
Step-by-step explanation:
Truly, on a single student level, the Massey method can be applied i.e. problems are ranked according to their level of difficulty for a single student.
Let us say there are 4 problems P1, P2, P3, P4 set in an exam where a student has to solve any two.
Assuming student 1 solves and scores more in P1 than in P2, r1>r2
student 2 more in P1 than P4 r1>r4
student 3 more in P2 than P4 r2>r4
student 4 more in P4 than P3 r4>r3
student 5 more in P2 than P3 r2>r3
and such many more cases.
then the Massey's model can be written as
P1 P2 P3 P4
\begin{bmatrix} 1 &-1 &0 &0\\ 1 &-1 &0 &0 \\ 0 &1 & 0 &-1 \\ 0 & 0 &-1 & 1 \\ 0 & 1 & -1 & 0 \end{bmatrix}.
\times \begin{bmatrix}r1 \\r2 \\r3 \\r4 \end{bmatrix} =
\begin{bmatrix}y_{1} \\ y_{2} \\ y_{3} \\ y_{4}\\ y_{5}\end{bmatrix}.
If Pi gives more score than Pj, the entry below Pi shall be 1 and below Pj shall be -1.
The rest of the entries shall be 0.
r1,r2,r3,r4,r5 are ranks .
There may be more rows depending upon more combination of 2 problems.
Should in case more than one student solve the same combination and get scores in the same order, for instance, 12 students solve P1 and P2, for 7 of them r1>r2, then their scores can be averaged.
Normally, such systems are overdetermined, i.e., more rows than the number of unknowns, then it is solved by the Least square method ., as there shall be no a solution with least error is found out.
The overall ranking, in reverse order, shall give the ranking according to increasing difficulty.
Which of the following expressions are greater than 1? Choose all that apply.
A.
1675×5
B.
27×4
C.
33100×3
D.
1233×3
E.
1021×2
F.
319×5
Answer:
I think the answers are A, C, D, and E
Step-by-step explanation:
[For Questions 1 & 2]
There was once a crooked but witty man Douglas charged for the crime of
felony. He was kept in a prison cell which was guarded by a hefty officer. The
cell was situated at the beginning of a long straight corridor partitioned by five
doors. The doors operated on different time switches so that the first, which
separated the cell from the corridor, opened every 1 minute 45 seconds, the
second every 1 minute 10 seconds, the third every 2 minutes 55 seconds, the
fourth every 2 minutes 20 seconds, and the fifth, which was at the end of the
corridor, every 35 seconds. Every once in a while, the five doors opened
simultaneously. When this happened, the guard arrived, looked down the
corridor to check the cell, and then left. Douglas calculated that in making his
escape it would take 20 seconds to cover the distance between consecutive
doors, which was longer than the amount of time a door stayed open. He also
knew that if he stayed in the corridor for longer than two and a half minutes, at a
stretch, an alarm would sound. So he had to escape in the shortest possible time.
Given that Douglas was smart enough to keep the track of all time.
Question 1:How much time had already passed when Douglas started
moving?
A. 18m 40sec B. 19m 15sec C 19m 50sec D. Prisoner cannot escape
Question 2:How long before the guard returned does Douglas cleared the
last door?
12m 50sec
B 13m 25sc
D. Douglas
Answer:
B. 19 min 15 sec
B. 13 min 25 sec
Step-by-step explanation:
Door 1 opens every 1 min 45 sec, or 105 sec.
Door 2 opens every 1 min 10 sec, or 70 sec.
Door 3 opens every 2 min 55 sec, or 175 sec.
Door 4 opens every 2 min 20 sec, or 140 sec.
Door 5 opens every 35 sec.
The greatest common factor is 35 seconds, so we can measure the time in units of 35 seconds.
Door 1 opens every 3 units.
Door 2 opens every 2 units.
Door 3 opens every 5 units.
Door 4 opens every 4 units.
Door 5 opens every 1 unit.
The least common multiple of 3, 2, 5, 4, and 1, is 60. So every 60 units, all five doors will open, and the guard will look down the corridor to check on the prisoner. Douglas must escape before this time.
In order to escape in the shortest time possible, Douglas should time his escape so that each door opens 1 unit after the door before it. It takes Douglas 20 seconds to move from one door to another, so he will have enough time to get to the next door before it opens.
Let's say Douglas starts moving when Door 1 opens for the nth time. In other words, 3n units have passed before he starts moving. That means Door 2 should open after 3n + 1 units. Door 3 should open after 3n + 2 units. Door 4 should open after 3n + 3 units. And Door 5 should open after 3n + 4 units.
Since Door 2 opens every 2 units, 3n + 1 should be a multiple of 2.
Since Door 3 opens every 5 units, 3n + 2 should be a multiple of 5.
Since Door 4 opens every 4 units, 3n + 3 should be a multiple of 4.
Since Door 5 opens every 1 unit, 3n + 4 should be a multiple of 1.
By trial and error, n = 11.
So Douglas starts moving after 33 units, or 1155 seconds, or 19 min 15 sec.
Douglas clears the fifth door after 37 units, which leaves 23 units to spare, or 805 seconds, or 13 min 25 sec.
A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 80% confidence interval for the mean and are assuming that the population standard deviation for the number of energy drinks consumed each week is 1. The study found that for a sample of 256 teenagers the mean number of energy drinks consumed per week is 4.7. Construct the desired confidence interval. Round your answers to one decimal place.
Answer:
The 80% confidence interval for the mean
(4.6199 , 4.7801)
Step-by-step explanation:
Explanation:-
Assuming that the population standard deviation for the number of energy drinks consumed each week is 1
Given the Population standard deviation 'σ' = 1
The study found that for a sample of 256 teenagers the mean number of energy drinks consumed per week is 4.7
given sample size 'n' = 256
mean of the sample 'x⁻' = 4.7
confidence interval for the mean
The 80% confidence interval for the mean is determined by
[tex](x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} + Z_{\alpha }\frac{S.D}{\sqrt{n} } )[/tex]
the z-score of 80% level of significance = 1.282
[tex](4.7 - 1.282\frac{1}{\sqrt{256} } , 4.7 + 1.282\frac{1}{\sqrt{256} } )[/tex]
(4.7 - 0.0801 , 4.7 +0.0801)
(4.6199 , 4.7801)
Conclusion:-
The 80% confidence interval for the mean
(4.6199 , 4.7801)
How to find the perimeter of a quadrilateral
It's the same as any other shape. Find the lengths of the sides and add them up.
Answer:
To find the perimeter of a quadrilateral you would add the lengths of each side together.
Example:
The perimeter is the sum or all the lengths.
If there was a rectangle with a length of 5 cm and a width of 3 cm.
To work this out you would first multiply 5 by 2, which is 10. Then you would multiplying 3 by 2, which is 6. Then you would add 6 and 10, which is 16.
1) Multiply 5 By 2.
[tex]5*2=10[/tex]
2) Multiply 3 by 2.
[tex]3*2=6[/tex]
3) Add 10 and 6.
[tex]10+6=16[/tex]
A builder makes drainpipes that drop 1 cm over a horizontal distance of 30cm to prevent clogs a certain drainpipe needs to cover a horizontal distance of 700cm whats is the length of the drainpipe? round your answer to the nearest tenth of a centimeter
Step-by-step explanation:
Given a builder makes drainpipes that drop 1 cm over a horizontal distance of 30cm to prevent clogs. The ratio of the vertical drop to the horizontal distance covered is [tex]\frac{1}{30}[/tex] cm.
The angle of inclination,
tan(θ) = [tex]\frac{1}{30}[/tex]
θ = 1.9º
By trignometric ratio,
cos(θ) = [tex]\frac{horizontal distance}{length of the drainpipe}[/tex]
length of the drainpipe = [tex]\frac{700}{cos(1.91)}[/tex] = [tex]\frac{700}{0.999}[/tex]
length of the drainpipe = 700.7 cm
Answer:
The correct Answer for Khan Academy is 700.4
Step-by-step explanation:
30 cm
30 cm
26 cm
find the area
Answer:
23400 cm
Step-by-step explanation:
30 cm x 30 cm x 26 cm = 23,400 cm
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is believed that the machine is underfilling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289. Assume the population is normally distributed. A level of significance of 0.01 will be used. State the null and alternative hypotheses.
Answer:
Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram {means that the machine is not under filling the bags}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram {means that the machine is under filling the bags}
Step-by-step explanation:
We are given that a manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting.
It is believed that the machine is under filling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289.
Let [tex]\mu[/tex] = mean weight bag filling capacity of machine.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram {means that the machine is not under filling the bags}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram {means that the machine is under filling the bags}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean = 429 grams
s = sample standard deviation = [tex]\sqrt{289}[/tex] = 17 grams
n = sample of bags = 21
So, test statistics = [tex]\frac{429-430}{\frac{17}{\sqrt{21} } }[/tex] ~ [tex]t_2_0[/tex]
= -0.269
Now at 0.01 significance level, the t table gives critical value of -2.528 at 20 degree of freedom for left-tailed test. Since our test statistics is higher than the critical value of t as -0.269 > -2.528, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that the machine is not under filling the bags.
Final answer:
The student's question involves setting up null and alternative hypotheses for a hypothesis test related to a bag filling machine's weight accuracy. The null hypothesis posits no difference from the 430-gram setting, while the alternative hypothesis suggests underfilling. The test uses a 0.01 significance level to determine whether to reject the null hypothesis.
Explanation:
The student is asking about hypothesis testing in the context of quality control for a bag filling machine. The null hypothesis (H0) assumes that there's no difference between the sample's mean weight and the population mean weight that the machine is set to, which is 430 grams; hence H0:
μ = 430 grams. The alternative hypothesis (Ha) suggests that the actual mean weight is less than the setting, given the concern about underfilling; thus Ha: μ < 430 grams.
For the given scenario, the bag sample's mean is 429 grams with a variance of 289 grams. Since a 0.01 level of significance will be used, if the test statistic calculated from the sample data falls into the critical region determined by this significance level, the null hypothesis would be rejected in favor of the alternative hypothesis.
You have prepared 10 types of treats for your 5 cats. You don’t know which treat each of your cats will go for, so you have bought for each type enough treats for all your cats. Assume that each cat is equally likely to choose any type of treats, and let X be the number of pairs of cats that will choose the same type of treats. Compute E(X) and Var(X).
Answer:
E(X) = 1.5
Var(X) = 2.325
Step-by-step explanation:
X - the number of pairs of cats (out of the 5 cats) to choose the same type of treat, can take values of:
X = (0, 1, 2, 3, 4, 5)
Also probability of choosing a treat, since they are all equally likely is: f(x) = 1/10
E(X) - expectation of x, is given by:
E(X) = Summation [X*f(x)]
E(X) = 0x(1/10) + 1x(1/10) + 2x(1/10) + 3x(1/10) + 4x(1/10) + 5x(1/10)
= (1/10) x (1+2+3+4+5)
E(X) = 3/2 = 1.5
Also, variance is:
Var(X) = Summation f(x)*[X - E(X)]^2
= (1/10)x(0-1.5)^2 + (1/10)x(1-1.5)^2 +(1/10)x(2-1.5)^2 +(1/10)x(3-1.5)^2 +(1/10)x(4-1.5)^2 +(1/10)x(5-1.5)^2
= (1/10)x[2.25 + 0.25 + 2.25 + 6.25 + 12.25]
Var(X) = 2.325
Statistics students at a community college wonder whether the cars belonging to students are, on average, older than the cars belonging to faculty. They select a random sample of 58 cars in the student parking lot and find the average age to be 8.5 years with a standard deviation of 6.2 years. A random sample of 41 cars in the faculty parking lot have an average age of 5.1 years with a standard deviation of 3.5 years. Note: The degrees of freedom for this problem is df = 93.033791. Round all results to 4 decimal places
Answer:
There is evidence to support the claim that the cars belonging to students are, on average, older than the cars belonging to faculty.
Step-by-step explanation:
We have to perform a hypothesis test on the difference between means.
The claim is that the cars belonging to students are, on average, older than the cars belonging to faculty.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]
being μ1: population mean age of students cars, μ2: population mean age of faculty cars.
The significance level is assumed to be 0.05.
The information about the students cars sample is:
Mean M1: 8.5 years.
Standard deviation s1: 6.2 years.
Sample size n1: 58 cars.
The information about the faculty cars sample is:
Mean M2: 5.1 years.
Standard deviation s2: 3.5 years.
Sample size n2: 41 cars.
The difference between means is:
[tex]M_d=M_1-M_2=8.5-5.1=3.4[/tex]
The standard error of the difference between means is:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}} =\sqrt{\dfrac{6.2^2}{58}+\dfrac{3.5^2}{58}}=\sqrt{ 0.6628 +0.2988 }=\sqrt{0.9615}\\\\\\s_{M_d}=0.9806[/tex]
Now we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{3.4-0}{0.9806}=3.4673[/tex]
The degrees of freedom are 93.033791, so the P-value for this right tail test is:
[tex]P-value=P(t>3.4673)=0.0004[/tex]
As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected.
There is evidence to support the claim that the cars belonging to students are, on average, older than the cars belonging to faculty.
For 100 births, P(exactly 5555 girls)=0.0485 and P(5555 or more girls)=0.184. Is 5555 girls in 100 births a significantly high number of girls? Which probability is relevant to answering that question? Consider a number of girls to be significantly high if the appropriate probability is 0.05 or less.
Answer:
[tex]P(X \geq 55) = 0.184 \geq 0.05[/tex], so 55 girls in 100 births is not a significantly high number of girls
The appropriate probability to determine if a value x is significantly high is the probability of being equal or higher than x. It this probability is 0.05 or less, x is significantly high.
Step-by-step explanation:
A number x is said to be significantly high if:
[tex]P(X \geq x) \leq 0.05[/tex]
The appropriate probability to determine if a value x is significantly high is the probability of being equal or higher than x. It this probability is 0.05 or less, x is significantly high.
Is 55 girls in 100 births a significantly high number of girls?
We have that:
[tex]P(X \geq 55) = 0.184 \geq 0.05[/tex], so 55 girls in 100 births is not a significantly high number of girls
To help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom 15% of the students in the district, based on the scores on a reading achievement exam given in the child's dominant language. The reading-score for the pooled students in the district is approximately normally distributed with a mean of 122 points, and standard deviation of 18 points. a. Find the probability of students score above 140 points? b. What is the 15th percentile of the students eligible for the program?
Answer:
a) [tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel and we got:
[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]
b) [tex]z=-1.036<\frac{a-122}{18}[/tex]
And if we solve for a we got
[tex]a=122 -1.036*18=103.35[/tex]
So the value of height that separates the bottom 15% of data from the top 85% is 103.35.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(122,18)[/tex]
Where [tex]\mu=122[/tex] and [tex]\sigma=18[/tex]
We are interested on this probability
[tex]P(X>140)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel and we got:
[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]
Part b
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X<a)=0.15[/tex] (a)
[tex]P(X>a)=0.85[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.15 of the area on the left and 0.85 of the area on the right it's z=-1.036. On this case P(Z<-1.036)=0.15 and P(z>-1.036)=0.85
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.15[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.15[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.036<\frac{a-122}{18}[/tex]
And if we solve for a we got
[tex]a=122 -1.036*18=103.35[/tex]
So the value of height that separates the bottom 15% of data from the top 85% is 103.35.
Stop&Shop is having a sale on paper towels: 5 rolls for $3. How much will 1 roll(s) cost?
Answer:
I believe it would be 0.60 cents
Step-by-step explanation:
roll 5
The cost 3
Divide the 5 by 5 to get 1 and divde 3 by 6 to get 0.60
Answer:
$0.60
Step-by-step explanation:
To work this out you would divide $3 by 5, which is 0.6. This is because we know that $3 is for 5 and so to work out 1 you would divide by 5.
1) Divide 3 by 5.
[tex]3/5=0.6[/tex]
1 roll of paper towels is $0.6.
Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.
Let W1 be the set of all polynomials of the form p(t)=at2, where a is in ℝ.
Let W2 be the set of all polynomials of the form p(t)=t2+a, where a is in ℝ.
Let W3 be the set of all polynomials of the form p(t)=at2+at, where a is in ℝ.
Answer:
1. Yes.
2. No.
3. Yes.
Step-by-step explanation:
Consider the following subsets of Pn given by
1.Let W1 be the set of all polynomials of the form [tex]p(t)=at^2[/tex], where a is in ℝ.
2.Let W2 be the set of all polynomials of the form [tex]p(t)=t^2+a[/tex], where a is in ℝ.
3. Let W3 be the set of all polynomials of the form [tex]p(t)=at^2+at[/tex], where a is in ℝ.
Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:
- The 0 vector of V is in W.
- Given v,w in W then v+w is in W.
- Given v in W and a a real number, then av is in W.
So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.
- First property:
Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.
For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.
- Second property:
W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials
[tex]p_1 (t) = at^2, p_2(t)=bt^2[/tex].
We must check that p_1+p_2(t) is in W1.
Note that
[tex]p_1(t)+p_2(t) = at^2+bt^2 = (a+b)t^2[/tex]
Since a+b is another real number, we have that p1(t)+p2(t) is in W1.
W3. Consider two elements in W3. Say [tex]p_1(t) = a(t^2+t), p_2(t)= b(t^2+t)[/tex]. Then
[tex]p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)[/tex]
So, again, p1(t)+p2(t) is in W3.
- Third property.
W1. Consider an element in W1 [tex] p(t) = at^2[/tex]and a real scalar b. Then
[tex] bp(t) = b(at^2) = (ba)t^2)[/tex].
Since (ba) is another real scalar, we have that bp(t) is in W1.
W3. Consider an element in W3 [tex] p(t) = a(t^2+t)[/tex]and a real scalar b. Then
[tex] bp(t) = b(a(t^2+t)) = (ba)(t^2+t)[/tex].
Since (ba) is another real scalar, we have that bp(t) is in W3.
After all,
W1 and W3 are subspaces of Pn for n= 2
and W2 is not a subspace of Pn.
W1 is a subspace of ℙn, W2 is not a subspace of ℙn, and W3 is a subspace of ℙn.
Explanation:For a set to be a subspace of ℙn, it must satisfy three conditions: the zero vector must be in the set, the set must be closed under addition, and the set must be closed under scalar multiplication.
W1:To determine if W1 is a subspace of ℙn, we check if it satisfies the three conditions:
The zero vector is a polynomial of the form p(t) = 0t^2, which is in W1.If p(t) and q(t) are polynomials of the form p(t) = at^2 and q(t) = bt^2, then their sum is (a+b)t^2, which is also in W1.If p(t) is a polynomial of the form p(t) = at^2 and c is a scalar, then c * p(t) = (ca)t^2, which is in W1.Therefore, W1 is a subspace of ℙn.
W2:The zero vector is a polynomial of the form p(t) = t^2 + 0, which is in W2.If p(t) and q(t) are polynomials of the form p(t) = t^2 + a and q(t) = t^2 + b, then their sum is (t^2 + a) + (t^2 + b) = 2t^2 + (a + b), which is not in the form required by W2. Therefore, W2 is not closed under addition and is not a subspace of ℙn.W3:The zero vector is a polynomial of the form p(t) = 0t^2 + 0t, which is in W3.If p(t) and q(t) are polynomials of the form p(t) = at^2 + at and q(t) = bt^2 + bt, then their sum is (a + b)t^2 + (a + b)t, which is also in W3.If p(t) is a polynomial of the form p(t) = at^2 + at and c is a scalar, then c * p(t) = (ca)t^2 + (ca)t, which is in W3.Therefore, W3 is a subspace of ℙn.
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The titanium content in an aircraft-grade alloy is an important determinant of strength. A sample of 20 test coupons reveals the following titanium content (in percent). 8.32 8.05 8.93 8.65 8.25 8.46 8.52 8.35 8.36 8.41 8.42 8.30 8.71 8.75 8.60 8.83 8.50 8.38 8.29 8.46 Use the normal approximation for the sign test to test the claim that the median titanium content is equal to 8.5%. Use α=0.05. Calculate the observed number of plus differences. r+= Enter your answer; r+ Calculate to 2 decimal places the test statistic. z0= Enter your answer; z0 Calculate to 3 decimal places the P-value. P-value = Enter your answer; P-value H0 : μ˜=8.5 versus H1: μ˜≠8.5. Conclusion: Choose your answer; Conclusion
Answer:
attached
Step-by-step explanation:
attached
To test the claim that the median titanium content is 8.5%, we use the normal approximation for the sign test. We calculate the observed number of plus differences, the test statistic, and the P-value. We fail to reject the null hypothesis because the P-value is larger than the significance level.
Explanation:To test the claim that the median titanium content is equal to 8.5%, we can use the normal approximation for the sign test. The sign test is used when the data is not normally distributed. In this case, we have 20 test coupons with titanium content in percent. To calculate the observed number of plus differences (r+), we count the number of values greater than 8.5% and the number of values less than 8.5% in the sample. In this case, there are 7 values greater than 8.5% and 13 values less than 8.5%. Therefore, r+ = 7.
To calculate the test statistic (z0), we use the formula z0 = (r+ - n/2) / sqrt(n/4), where n is the sample size. In this case, n = 20. Plugging in the values, we get z0 = (7 - 20/2) / sqrt(20/4) = -1.5.
To calculate the P-value, we need to find the probability of observing a test statistic as extreme as -1.5 or more extreme in a standard normal distribution. We can use a statistical table or software to find the P-value. In this case, the P-value is 0.133, calculated using a standard normal distribution table.
Since the P-value (0.133) is larger than the significance level α (0.05), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the median titanium content is different from 8.5%.
Peter is a revenue manager of a 300-room hotel. Over this past weekend (including both Saturday and Friday evenings) he sold 450 of his guestrooms with an average daily rate (ADR) of $300.00. What as his hotel's RevPAR for thepast weekend?
Answer:
$225
Step-by-step explanation:
-RevPar is defined as the room revenue divided by the number of rooms available:
[tex]RevPar=\frac{Room \ Revenue}{Rooms \ Available}[/tex]
-The average number of rooms sold per day in the two days is:
[tex]mean =\frac{450}{2}\\\\=225\ rooms[/tex]
The PevPar can then be calculated as:
[tex]RevPar=\frac{Room \ Revenue}{Rooms \ Available}\\\\=\frac{ADR\times mean occupancy}{Rooms \ Available}\\\\=\frac{300\times 225}{300}\\\\=225[/tex]
Hence, the PevPar is $225
1. A sample of n = 9 scores has ex = 108. What is the sample mean?
The value of the sample mean is 12.
Given
A sample with a mean of n = 9 has [tex]\rm \sum x=108[/tex].
What is formula is used to calculate the sample mean?The formula is used to calculate sample mean is;
[tex]\rm Sample \ mean = \dfrac{\sum x}{n}\\\\[/tex]
Substitute all the values in the formula
Then,
The sample mean is;
[tex]\rm Sample \ mean = \dfrac{\sum x}{n}\\\\\rm Sample \ mean = \dfrac{108}{9}\\\\ \rm Sample \ mean = 12[/tex]
Hence, the value of the sample mean is 12.
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what is the expression 5 square root of -4 + 4 square root of -25 in standard form
Answer:
30i
Step-by-step explanation:
[tex]5 \sqrt{ - 4} + 4 \sqrt{ - 25} \\ = 5 \sqrt{4 \times - 1} + 4 \sqrt{25 \times - 1} \\ = 5 \sqrt{4} \times \sqrt{ - 1} + 4 \sqrt{25} \times \sqrt{ - 1} \\ = 5 \times 2 \times i + 4 \times 5 \times i...( \because \sqrt{ - 1} = i) \\ = 10i + 20i \\ = 30i[/tex]
Find the number that makes the ratio equivalent to 1:9. 3:
The ratio 3:x equivalent to 1:9 results in x being equal to 27.
To find the number that makes the given ratio equivalent to 1:9 we can set up a proportion using the information given which is a ratio of 3 to an unknown number (let's call it x).
First we write down the two ratios as fractions and set them equal to each other to find x:
1/9 = 3/x
We then cross-multiply to solve for x:
1 × x = 9 × 3
This gives us:
x = 27
So, the number that makes the ratio equivalent to 1:9 when compared to 3 is 27.
A Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. A salesman suggests that he try a statistics software package that gets students more involved with computers, predicting that it will increase students' scores. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly. The professor will have to pay for the software only if he chooses to continue using it. In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.a. What is the test statistic?b. What is the P-value?c. What is the appropriate conclusion?i. Fail to reject H_0. The change is not statistically significant. The software does not appear to improve exam scores.ii. Fail to reject H_0. The change is statistically significant. The software does appear to improve exam scores.iii. Reject H_0. The change is not statistically significant. The software does not appear to improve exam scores.iv. Reject H_0. The change is statistically significant. The software does appear to improve exam scores.
Answer:
Reject [tex]H_0[/tex] . The change is statistically significant. The software does appear to improve exam scores.
Step-by-step explanation:
We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.
In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.
Let [tex]\mu[/tex] = mean scores on the final exam.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 114 points {means that the mean scores on the final exam does not increases after using software}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 114 points {means that the mean scores on the final exam increase significantly after using software}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average points = 117 points
s = sample standard deviation = 8.7 points
n = sample of students = 218
So, test statistics = [tex]\frac{117-114}{\frac{8.7}{\sqrt{218} } }[/tex] ~ [tex]t_2_1_7[/tex]
= 5.091
Now, P-value of the test statistics is given by the following formula;
P-value = P( [tex]t_2_1_7[/tex] > 5.091) = Less than 0.05%
Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the change is statistically significant. The software does appear to improve exam scores.
The test statistic is 3.448 and the P-value is 0.0014. The appropriate conclusion is that the software does appear to improve exam scores.
Explanation:The test statistic can be calculated using the formula:
test_statistic = (sample_mean - population_mean) / (sample_standard_deviation / sqrt(sample_size))
Substituting in the given values:
test_statistic = (117 - 114) / (8.7 / sqrt(218)) = 3.448
The P-value can be calculated using a statistical software or calculator. For this question, the P-value is 0.0014
Since the P-value is less than 0.05 (significance level), we can reject the null hypothesis. The appropriate conclusion is: iv. Reject H_0. The change is statistically significant. The software does appear to improve exam scores.
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Twenty eight is 40% of
Answer:
11.20 is maybe the answer
Answer:
70
Step-by-step explanation:
Assume the unknown value is 'Y'
28 = 40% x Y
28 = 40/100 x Y
Multiplying both sides by 100 and dividing both sides of the equation by 40 we will arrive at:
Y = 3 x 100/40
Y = 70%
MATH 1325 – EXAM 4 NAME: ______________________________ SHOW ALL WORK. ANSWERS WITHOUT WORK WILL RECEIVE NO CREDIT. YOU MUST USE A PENCIL. READ ALL DIRECTIONS. POINTS WILL BE DEDUCTED FOR FAILURE TO FOLLOW DIRECTIONS. TRUE/FALSE – WRITE THE WORD THAT BEST DESCRIBES THE GIVEN STATEMENT BY WRITING EITHER "TRUE" OR "FALSE" IN THE SPACE PROVIDED TO THE LEFT OF THE PROBLEM. __________ 1. THE ABSOLUTE MAXIMUM OF A FUNCTION ALWAYS OCCURS WHERE THE DERIVATIVE HAS A CRITICAL FUNCTION. __________ 2. IMPLICIT DIFFERENTIATION CAN BE USED TO FIND dy dx WHEN x IS DEFINED IN TERMS OF y . __________ 3. IN A RELATED RATES PROBLEM, THERE CAN BE MORE THAN TWO QUANTITIES THAT VARY WITH TIME. __________ 4. A CONTINUOUS FUNCTION ON AN OPEN INTERVAL DOES NOT HAVE AN ABSOLUTE MAXIMUM OR MINIMUM. __________ 5. IN A RELATED RATES PROBLEM, ALL DERIVATIVES ARE WITH RESPECT TO TIME. MULTIPLE CHOICE – CHOOSE THE ONE ALTERNATIVE THAT BEST COMPLETES THE STATEMENT OR ANSWERS THE QUESTION BY CIRCLING THE CORRECT LETTER. 6. FIND THE MAXIMUM ABSOLUTE EXTREMUM AS WELL AS ALL VALUES OF x WHERE IT OCCURS ON THE SPECIFIED DOMAIN
True, True, True, False, True. Without the multiple-choice question options, I can't provide a specific response.
Let's address each question and provide the appropriate response:
1. True/False: "The absolute maximum of a function always occurs where the derivative has a critical function."
- Answer: False. The absolute maximum of a function can occur at critical points, but it can also occur at endpoints of the domain or other non-critical points.
2. True/False: "Implicit differentiation can be used to find \( \frac{dy}{dx} \) when \( x \) is defined in terms of \( y \)."
- Answer: True. Implicit differentiation is a technique used to find derivatives of functions that are not explicitly defined in terms of one variable.
3. True/False: "In a related rates problem, there can be more than two quantities that vary with time."
- Answer: True. Related rates problems involve the rates of change of multiple quantities that are related to each other through an equation or situation.
4. True/False: "A continuous function on an open interval does not have an absolute maximum or minimum."
- Answer: False. A continuous function on a closed interval always has an absolute maximum and minimum by the Extreme Value Theorem.
5. True/False: "In a related rates problem, all derivatives are with respect to time."
- Answer: True. Related rates problems involve finding rates of change with respect to time.
Multiple Choice:
6. Since the multiple-choice question is not provided, I can't offer a response. If you have the options, feel free to share them, and I can help you choose the correct one.
Consider the triangular pyramids shown.
Pyramid 1
Pyramid 2
Pyramid 1 is a triangular pyramid. The base has a base of 12 and height of 2.5. A side has a base of 12 and height of 13.1. 2 sides have a base of 6.5 and height of 14.
[Not drawn to scale]
Pyramid 2 is a triangular pyramid. The base has a base of 12 and height of 4.5. A side has a base of 12 and height of 6.5. 2 sides have a base of 7.5 and height of 8.
[Not drawn to scale]
The base of each pyramid is an isosceles triangle. Which has the greater surface area?
Pyramid 1 is greater by 58.6 square units.
Pyramid 1 is greater by 70.6 square units.
Pyramid 2 is greater by 12 square units.
Pyramid 2 is greater by 24 square units.
9514 1404 393
Answer:
(a) Pyramid 1 by 58.6 square units
Step-by-step explanation:
The area of each triangle is given by the formula ...
A = 1/2bh
The total area is the area of all four triangles that make up the outer surface of the pyramid.
Pyramid 1
A = 1/2(12×2.5 +12×13.1 +2(6.5×14)) = 184.6 . . . square units
Pyramid 2
A = 1/2(12×4.5 +12×6.5 +2(7.5×8)) = 126 . . . square units
Pyramid 1 has the larger area by ...
184.6 -126 = 58.6 . . . square units
If the coordinates of the endpoints of a diameter of the circle are known, the equation of a circle can be found. First, find the midpoint of the diameter, which is the center of the circle. Then find the radius, which is the distance from the center to either endpoint of the diameter. Finally use the center-radius form to find the equation.
Answer:
The equation of the circle is [tex](x+3)^2+(y-5)^2 = 17[/tex]
Step-by-step explanation:
The complete question is
If the coordinates of the endpoints of a diameter of the circle are known, the equation of a circle can be found. First, find the midpoint of the diameter, which is the center of the circle. Then find the radius, which is the distance from the center to either endpoint of the diameter. Finally use the center-radius form to find the equation.
Find the center-radius form of the circle having the points (1,4) and (-7,6) as the endpoints of a diameter.
Consider that, if both points are the endpoints of a diameter, the center of the circle is the point that is exactly in the middle of the two points (that is, the point whose distance to each point is equal). Given points (a,b) and (c,d), by using the distance formula, you can check that the middle point is the average of the coordinates. Hence, the center of the circle is given by
[tex](\frac{1-7}{2}, \frac{4+6}{2}) = (-3,5)[/tex].
We will find the radius. Recall that the radius of the circle is the distance from one point of the circle to the center. Recall that the distance between points (a,b) and (c,d) is given by [tex]\sqrt[]{(a-c)^2+(b-d)^2}[/tex]. So, let us use (1,4) to calculate the radius.
[tex] r = \sqrt[]{(1-(-3))^2+(4-5)^2} = \sqrt[]{17}[/tex].
REcall that given a point [tex](x_0,y_0)[/tex]. The equation of a circle centered at the point [tex](x_0,y_0)[/tex] is
[tex](x-x_0)^2+(y-y_0)^2 = r^2[/tex]
In our case, [tex](x_0,y_0)=(-3,5)[/tex] and [tex] r=\sqrt[]{17}[/tex]. Then, the equation is
[tex](x-(-3))^2+(y-5)^2 = (x+3)^2+(y-5)^2 = 17[/tex]
In a certain Algebra 2 class of 22 students, 5 of them play basketball and 11 of them play baseball. There are 3 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball
Final answer:
The probability that a randomly chosen student from the Algebra 2 class plays basketball or baseball is 13/22, which is approximately 0.59 or 59% when calculated using the principle of inclusion and exclusion in probability.
Explanation:
To calculate the probability that a student chosen randomly from the Algebra 2 class plays basketball or baseball, you need to use the principle of inclusion and exclusion in probability theory. The total number of students in the class is 22. There are 5 students who play basketball and 11 who play baseball, but since 3 students play both sports, these students have been counted twice in the total of 5+11=16. Therefore, the correct number of students who play at least one of the sports is 5+11-3=13.
The probability that a randomly chosen student plays either basketball or baseball is hence the number of students who play either one or both sports divided by the total number of students in the class. This gives us 13/22. To convert this fraction to a decimal, we can divide 13 by 22, which gives us approximately 0.59. Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is approximately 0.59, or 59%.