Answer:
Alternative hypothesis:[tex]\mu_1 -\mu_2 \neq 0[/tex]
Or in the alternative way would be:
Alternative hypothesis:[tex]\mu_1 \neq \mu_2 [/tex]
Step-by-step explanation:
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
On this case the claim that they want to test is: "The means for the two groups (American Idol and 60 Minutes) is the same". So we want to check if we have significant differences between the two means, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.
Null hypothesis:[tex]\mu_1 -\mu_2 = 0[/tex]
This null hypothesis can be expressed like this:
Null hypothesis:[tex]\mu_1 = \mu_2 [/tex]
Alternative hypothesis:[tex]\mu_1 -\mu_2 \neq 0[/tex]
Or in the alternative way would be:
Alternative hypothesis:[tex]\mu_1 \neq \mu_2 [/tex]
Final answer:
Lucy Baker's alternate hypothesis for her demographic analysis of American Idol and 60 Minutes would suggest a difference in the mean ages of the two audiences, represented as either (Ha: µ₁ ≠ µ₂), (Ha: µ₁ < µ₂), or (Ha: µ₁ > µ₂), depending on the direction of the difference she anticipates.
Explanation:
Lucy Baker's hypothesis test within her demographic analysis involves comparing the mean ages of audiences of two television programs: American Idol and 60 Minutes. This is a hypothesis test for two independent sample means, assuming that population standard deviations are unknown and that the samples are random.
Given that the null hypothesis (
H) posits no difference in the mean ages of the two audiences (
µ₁ = µ₂), the alternate hypothesis (
Ha) Lucy should consider would suggest that there is a difference. Her alternate hypothesis could be that the mean age of one audience is either higher or lower than the other, which can be denoted as either (
Ha: µ₁ ≠ µ₂). However, if she has a specific direction in mind (e.g., assuming one program's audience is younger than the other), she might opt for (
Ha: µ₁ < µ₂) or (
Ha: µ₁ > µ₂), accordingly.
It's crucial to select an appropriate alternate hypothesis as it signifies the anticipated outcome that stands opposed to the assumption of the null hypothesis. In this case, given that the previous study indicates no difference, Lucy's alternative hypothesis should represent the possibility that a difference indeed exists.
Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300. In addition, for 15% of the bills, the amount billed was greater than $1000. A sample of 900 bills is drawn. What is the probability that the average amount billed on the sample bills is greater than $500? (Round the final answer to four decimal places.)
Answer:
0.02% probability that the average amount billed on the sample bills is greater than $500.
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].
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10[/tex].
What is the probability that the average amount billed on the sample bills is greater than $500?
This probability is 1 subtracted by the pvalue of Z when [tex]X = 500[/tex]. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{500 - 465}{10}[/tex]
[tex]Z = 3.5[/tex]
[tex]Z = 3.5[/tex] has a pvalue of 0.9998.
So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.
The probability that the average amount billed in a sample of 900 bills is greater than $500 is approximately 0.0002.
Given the mean amount billed is $465
standard deviation of $300,
sample size of 900,
the probability that the average amount billed exceeds $500.
First, we need to calculate the standard error of the mean (SEM):
[tex]SEM = \(\frac{\sigma}{\sqrt{n}}\)[/tex]
Substituting the given values:
[tex]SEM = \frac{300}{\sqrt{900}} = \frac{300}{30} = 10[/tex]
Now, we can use the Z-score formula to find the probability.
The Z-score is calculated as follows:
[tex]Z = \frac{X - \mu}{SEM}[/tex]
(X = 500),
mu = 465
SEM=10
[tex]Z = \frac{500 - 465}{10} = 3.5[/tex]
Using Z-tables or a standard normal distribution calculator, we can find the probability corresponding to a Z-score of 3.5:
P(Z > 3.5) ≈ 0.0002
Therefore, the probability that the average amount billed in a sample of 900 bills is greater than $500 is approximately 0.0002 (rounded to four decimal places).
Let A = (0, 0), B = (8, 1), C = (5, −5), P = (0, 3), Q = (7, 7), and R = (1, 10). Prove that angles ABC and P QR have the same size.
Answer:
To prove
∠ABC = ∠PQR
Using euclidean distance, Length of each side can be found as
[tex]AB=\sqrt{65} ,BC=\sqrt{45} ,CA=\sqrt{50} \\PQ=\sqrt{65} ,QR=\sqrt{45} ,RP=\sqrt{50}[/tex]
As can be seen
AB ≅ PQ
BC ≅ QR
CA ≅ RP
As all the sides of ΔABC are equal and congruent to ΔPQR, this Proves that measure of all angles inside both triangles must be equal.
Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). An article reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for m = 7 normal subjects was 1.83 mm, and the sample standard deviation was 0.54; for n = 10 CTS subjects, the sample mean and sample standard deviation were 2.35 and 0.88, respectively. Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of 0.01. (Use μ1 for normal subjects and μ2 for CTS subjects.)
Answer:
[tex]t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507[/tex]
[tex]p_v =P(t_{(15)}>1.507)=0.076[/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 FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X_{CTS}=2.35[/tex] represent the mean for the sample CTS
[tex]\bar X_{N}=1.83[/tex] represent the mean for the sample Normal
[tex]s_{CTS}=0.88[/tex] represent the sample standard deviation for the sample of CTS
[tex]s_{N}=0.54[/tex] represent the sample standard deviation for the sample of Normal
[tex]n_{CTS}=10[/tex] sample size selected for the CTS
[tex]n_{N}=7[/tex] sample size selected for the Normal
[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 for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{CTS} \leq \mu_{N}[/tex]
Alternative hypothesis:[tex]\mu_{CTS} > \mu_{N}[/tex]
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{CTS}+n_{N}-2=10+7-2=15[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(15)}>1.507)=0.076[/tex]
We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,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 FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.
A family paid 12 percent of its annual after tax income on food last year. This amount was equal to 10 percent of its annual before tax income last year. Which of the following is closest to the percent of the family's annual before-tax income that was paid for tares last year? a. 8% b. 12% c. 17% d. 20%e. 25%
Answer:
Option c. 17%
Step-by-step explanation:
Data provided in the question:
Amount paid on food last year = 12% of Annual after tax income
Amount paid on food for the current year = 10% of Annual before tax income
Now,
Let the after tax income be 'x'
and tax be 'y'
Therefore,
Income before tax = x + y
Amount paid on food = 12% of x
According to the question
12% of x = 10% of (x + y)
or
0.12x = 0.10 (x + y)
or
1.2x - x = y
0.2x = y
or
x = 5y
Thus,
percent of the family's annual before-tax income that was paid for tares last year
= [Tax ÷ Income before tax] × 100%
= [ y ÷ ( x + y )] × 100%
= [ y ÷ ( 5y + y )] × 100%
or
= 0.167 × 100% ≈ 17%
Hence,
Option c. 17%
You're conducting a significance test for H0 : p = .35, Ha : p < .35. In a sample of size 40, you identify a count of 11 successes. The computed z-score and P-value are:
a. –1.06 and .1446b. –1 and .3174c. –1 and .1587d. 1 and .3174e. 1 and .8413
Answer: c. –1 and .1587
Step-by-step explanation:
As per given , we have
Null hypothesis : [tex]H_0 : p= 0.35[/tex]
Alternative hypothesis : [tex]H_1 : p< 0.35[/tex]
Since Alternative hypothesis is left-tailed ,so the test must be a left tailed test .
Z -Test statistic for proportion = [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
, where p= population proportion
[tex]\hat{p}[/tex] = sample proportions
n= Sample size.
Let x be the number of successes.
For n= 40 and x= 11
[tex]\hat{p}=\dfrac{x}{n}=\dfrac{11}{40}=0.275[/tex]
Then , [tex]z=\dfrac{0.275-0.35}{\sqrt{\dfrac{0.35(1-0.35)}{40}}}[/tex]
[tex]z=\dfrac{-0.075}{\sqrt{0.0056875}}[/tex]
[tex]z=\dfrac{-0.075}{0.075415515645}[/tex]
[tex]z=-0.99449031619\approx-1[/tex]
By using z-table ,
P-value for left-tailed test = P(z<-1)= 1-P(z<1) [∵ P(Z<-z)= 1-P(Z<z) ]
= 1-0.8413
=0.1587
Hence, the -score and P-value are –1 and 0.1587 .
So the correct option is c. –1 and .1587
In the textile industry, a manufacturer is interested in the number of blemishes or flaws occurring in each 100 feet of material.
The probability distribution that has the greatest chance of applying to this situation is the:
a. Normal distribution.
b. Binomial distribution.
c. Poisson distribution.
d. Uniform distribution.
Answer:
The probability that can be applied to the given situation is Poisson distribution
Step-by-step explanation:
The probability that can be applied to the given situation is Poisson distribution because this distribution applied when the duration of occurrence of an event is known. In the given question laws have occurred every 100 feet therefore we have the number of events that have occurred. Moreover, Poisson distribution predicts the probability of events in fixed time interval
The probability distribution that has the greatest chance of applying to the number of blemishes or flaws occurring in each 100 feet of material in the textile industry is the Poisson distribution.
The probability distribution that has the greatest chance of applying to this situation is the Poisson distribution.
The Poisson distribution is commonly used to model the number of events that occur in a fixed interval of time or space. In this case, the manufacturer is interested in the number of blemishes or flaws occurring in each 100 feet of material, which can be seen as events occurring in a fixed space.
The Poisson distribution is appropriate when the events occur randomly and independently, and the average rate of occurrence is known. It allows for both discrete and continuous distributions.
Learn more about Poisson distribution here:https://brainly.com/question/33722848
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A study was made to determine whether more Italians than Americans prefer white champagne to pink champagne at weddings. Of the 300 Italians selected at random, 74 preferred white champagne, and of the 400 Americans selected, 60 preferred white champagne. Can we conclude that a higher proportion of Italians than Americans prefer white champagne at weddings? Use a 0.05 level of significance.
Answer:
[tex]z=\frac{0.247-0.15}{\sqrt{0.191(1-0.191)(\frac{1}{300}+\frac{1}{400})}}=3.23[/tex]
[tex]p_v =P(Z>3.23)=0.000619[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at weddings it's significantly higher than the proportion of Americans.
Step-by-step explanation:
1) Data given and notation
[tex]X_{I}=74[/tex] represent the number of Italians that preferred white champagne
[tex]X_{A}=60[/tex] represent the number of Americans that preferred white champagne
[tex]n_{I}=300[/tex] sample of Italians selected
[tex]n_{A}=400[/tex] sample of Americans selected
[tex]p_{I}=\frac{74}{300}=0.247[/tex] represent the proportion of Italians that preferred white champagne
[tex]p_{A}=\frac{60}{400}=0.15[/tex] represent the proportion of Americans that preferred white champagne
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if a higher proportion of Italians than Americans prefer white champagne at weddings, the system of hypothesis would be:
Null hypothesis:[tex]p_{I} - p_{A} \leq 0[/tex]
Alternative hypothesis:[tex]p_{I} - p_{A} > 0[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{I}-p_{A}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{I}}+\frac{1}{n_{A}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{I}+X_{A}}{n_{I}+n_{A}}=\frac{74+60}{300+400}=0.191[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.247-0.15}{\sqrt{0.191(1-0.191)(\frac{1}{300}+\frac{1}{400})}}=3.23[/tex]
4) Statistical decision
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>3.23)=0.000619[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at weddings it's significantly higher than the proportion of Americans.
To determine whether a higher proportion of Italians than Americans prefer white champagne at weddings, we need to conduct a hypothesis test. First, state the null and alternative hypotheses. Then, calculate the test statistic and compare it to the critical value at a significance level of 0.05.
Explanation:To determine whether a higher proportion of Italians than Americans prefer white champagne at weddings, we need to conduct a hypothesis test. First, we need to state the null hypothesis (H0) and the alternative hypothesis (Ha). In this case, H0: p1 = p2 (the proportion of Italians who prefer white champagne is equal to the proportion of Americans who prefer white champagne) and Ha: p1 > p2 (the proportion of Italians who prefer white champagne is greater than the proportion of Americans who prefer white champagne).
Next, we calculate the test statistic using the formula: z = (p1 - p2) / √(p*(1-p)*((1/n1) + (1/n2))), where p = (x1 + x2) / (n1 + n2), x1 and x2 are the number of Italians and Americans who prefer white champagne respectively, and n1 and n2 are the sample sizes of Italians and Americans respectively.
We then compare the test statistic to the critical value at a significance level of 0.05. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that a higher proportion of Italians than Americans prefer white champagne at weddings.
Listed are 32 ages for Academy Award winning best actors in order from smallest to largest. (Round your answers to the nearest whole number.) 18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 (a) Find the percentile of 31. th percentile
Answer:
Step-by-step explanation:
The ages for Academy Award winning best actors in order from smallest to largest are
18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
The total number of terms, n is 32
31℅ of 32 = 31/100 × 32 = 0.31 × 32 = 9.92. It is approximately 10
Counting from left to right, the 10th term is 31. This means that the 31st percentile is 31
A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded.
The summary statistics are as follows.
Type A Type B
x1 = 76.3 hrs x2 = 65.1 hrs
s1 = 4.5 hrs s2 = 5.1 hrs
n1 = 11 n2 = 9
The following 98% confidence interval was obtained for μ1 - μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B:
4.90 hrs < μ1 - μ2 < 17.50 hrs
What does the confidence interval suggest about the population means?
Answer:
We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.
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".
[tex]\bar X_1 =76.3[/tex] represent the sample mean 1
[tex]\bar X_2 =65.1[/tex] represent the sample mean 2
n1=11 represent the sample 1 size
n2=9 represent the sample 2 size
[tex]s_1 =4.5[/tex] sample standard deviation for sample 1
[tex]s_2 =5.1[/tex] sample standard deviation for sample 2
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
Confidence interval
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =76.3-65.1=11.2[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n_1 +n_2 -1=11+9-2=18[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,18)".And we see that [tex]t_{\alpha/2}=\pm 2.55[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]
And replacing we have:
[tex]SE=\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=2.175[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]11.2-2.55\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=5.65[/tex]
[tex]11.2+2.55\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=16.75[/tex]
So on this case the 98% confidence interval would be given by [tex]5.65 \leq \mu_1 -\mu_2 \leq 16.75[/tex]
But let's assume that the confidence interval given is true 4.90 hrs < μ1 - μ2 < 17.50 hrs
What does the confidence interval suggest about the population means?
We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.
A student uses pens whose lifetime is an exponential random variable with mean 1 week. Use the central limit theorem to determine the minimum number of pens he should buy at the beginning of a 15-week semester, so that with probability .99 he does not run out of pens during the semester.
Answer:
Student needs pens= n = 27.04
Rounding off with upper floor function ⇒ n =28
Rounding off with lower floor function ⇒ n =27
Step-by-step explanation:
Given that lifetime of each pen is a exponential random variable with mean 1 week.
Let [tex]S_{n}[/tex] be total sum of lifetime of n pens.
So mean of [tex]S_{n}[/tex] = μ = n.1
Standard deviation of [tex]S_{n}[/tex] =[tex]\sigma=\sqrt{n}[/tex]
Probability that he doesnot run out of pens= 0.99
Considering Sn be sum of n lifetimes, using central limit theorem
[tex]\frac{S_{n}-n}{\sqrt{n}}\approx N(0,1)\\\\P(S_{n}>15)=[P(\frac{S_{n}-n}{\sqrt{n}})>\frac{15-n}{\sqrt{n}}]\\ 1-\phi(\frac{15-n}{\sqrt{n}})=\phi(-(\frac{15-n}{\sqrt{n}}))=0.99\\[/tex]
From table of standard normal distribution
[tex]\frac{15-n}{\sqrt{n}}=-2.3263\\15-n=-2.3263\sqrt{n}\\n-2.3263-15\sqrt{n}[/tex]
Solving the quadratic Equation in variable x we get
n=27.04
What are the possible rational zeros of f(x) = x4 − 4x3 + 9x2 + 5x + 14?
Answer:
±1, ±2, ±7, ±14
Step-by-step explanation:
Assume a device manufacturer tests 100 devices. The first device fails at 100 hours. The last device fails at 200 hours. What is the device MTBF:a. 10,000 Hoursb. 15,000 hours c. 20,000 hours d. 100 hours
Answer:
a. 10,000 Hours
Step-by-step explanation:
MTBF (Mean Time Between Failures) can be calculated using the formula:
[tex]MTBF={(L-F)}*{n}[/tex] where
L is the time at which last device fails (200 hours) F is the time at which first device fails (100 hours) n is the number of devices tested (100)[tex]{MTBF=(200-100)}*{100}=10000[/tex]
Final answer:
The MTBF for a set of devices tested by the manufacturer, with failures evenly distributed from 100 to 200 hours, is 15,000 hours, which is calculated using the sum of an arithmetic series formula.
Explanation:
The student is asking about the mean time between failures (MTBF) for a set of devices. MTBF is a basic measure of reliability for repairable items and represents the average time expected between failures. To calculate MTBF, you divide the total operational time by the number of failures. In this case, the device manufacturer tests 100 devices, with the first failing after 100 hours and the last after 200 hours, which suggests a linear distribution of failures over time.
Assuming a linear and even distribution of failures from 100 to 200 hours for the 100 devices, you would calculate the total operational time before each device failed and then divide by the number of devices. The calculation would be the sum of an arithmetic series:
MTBF = (100 hours + 101 hours + ... + 200 hours) / 100 devices
We can use the formula for the sum of an arithmetic series:
S = n/2 * (a1 + an)
Where:
n is the number of terms in the series (here, 100 devices),
a1 is the first term in the series (100 hours),
an is the last term in the series (200 hours),
Now apply the formula:
MTBF = 100/2 * (100 + 200)
MTBF = 50 * 300
MTBF = 15,000 hours
So, the answer is b. 15,000 hours.
The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen's School of Business in Kingston, Ontario requires a minimum LSAT score of 157 for admission. Find the 35th percentile of the LSAT scores. Give your answer accurate to one decimal place. Use the applet. (Example: 124.7) Your Answer:
To find the 35th percentile of LSAT scores, a z-score corresponding to the 35th percentile is needed, which can then be applied to the formula using the provided mean and standard deviation of LSAT scores.
Explanation:To find the 35th percentile of LSAT scores, we need to use a normal distribution with the given mean and standard deviation. We use a z-table or a percentile calculator, looking to find the z-score that corresponds with the 35th percentile. Once the z-score is identified, we can use the mean and standard deviation of the LSAT scores to calculate the actual score corresponding to that percentile.
The process involves the following calculations:
Identify the z-score that corresponds to the 35th percentile using the z-table or percentile calculator.Apply the formula: actual score = mean + (z-score * standard deviation).Unfortunately, without the z-score or the use of applets, as suggested in the question, we cannot provide the exact LSAT score that corresponds to the 35th percentile.
Because of her past convictions for mail fraud and forgery, Jody has a 30% chance each year of having her tax returns audited. What is the probability that she will escape detection for at least three years? Assume that she exaggerates, distorts, misrepresents, lies, and cheats every year. Larsen, Richard J.; Marx, Morris L.. An Introduction to Mathematical Statistics and Its Applications (Page 259). Pearson Education. Kindle Edition.
Answer:
The probability that she will escape detection for at least three years is P=0.343.
Step-by-step explanation:
If Jody has a 30% chance each year of having her tax returns audited, she also has 70% chance each year of escaping detection.
The probability of this happening 3 years in a row is:
[tex]P_n=q^n=0.7^3=0.343[/tex]
Answer:
108
Step-by-step explanation:
Let's say that one of the items the university measured in the study was reaction time. That is a typical measurement that is taken when judging concussions as well as looki the beginning of the study they recorded a baseline average reaction time for the group of 41 football players at .239 seconds. At the end of their study they retested the players and the average reaction time was 233 with a standard deviation of .021. Use this data to create a 95% confidence interval for μ and explain if the reaction time at the conclusion of the study showed a significant decrease in reaction time or not.
Answer:
The 95% confidence interval is given by: (0.226, 0.240)
On this case we can't conclude that we have a significant reduction on the reaction time since the upper bounf of the interval is higher than the value of 0.239.
Step-by-step explanation:
1) 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".
[tex]\bar X =0.233[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=0.021 represent the sample standard deviation
n=41 represent the sample size
2) Calculate the confidence interval
Since the sample size is large enough n>30 but we don't know the population deviation. The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=41-1=40[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,40)".And we see that [tex]t_{\alpha/2}=2.02[/tex]
Now we have everything in order to replace into formula (1):
[tex]0.233-2.02\frac{0.021}{\sqrt{41}}=0.226[/tex]
[tex]0.233+2.02\frac{0.021}{\sqrt{41}}=0.240[/tex]
The 95% confidence interval is given by: (0.226, 0.240)
On this case we can't conclude that we have a significant reduction on the reaction time since the upper bounf of the interval is higher than the value of 0.239.
Consider the function f left parenthesis x right parenthesis equals 4 x squared minus 3 x minus 1f(x)=4x2−3x−1 and complete parts (a) through (c).(a) Find f left parenthesis a plus h right parenthesis f(a+h);(b) Find StartFraction f left parenthesis a plus h right parenthesis minus f left parenthesis a right parenthesis Over h EndFraction f(a+h)−f(a) h;(c) Find the instantaneous rate of change of f when aequals=77.
To find f(a+h), substitute a+h into the function f(x). To find the difference quotient, subtract f(a) from f(a+h) and divide by h. To find the instantaneous rate of change of f when a = 77, substitute a = 77 into f(a).
Explanation:To find f(a+h), we substitute a+h into the function f(x).
f(a+h) = 4(a+h)^2 - 3(a+h) - 1
Expanding and simplifying:
f(a+h) = 4(a^2 + 2ah + h^2) - 3a - 3h - 1
f(a+h) = 4a^2 + 8ah + 4h^2 - 3a - 3h - 1
To find f(a), we substitute a into the function f(x).
f(a) = 4a^2 - 3a - 1
To find the difference quotient, we subtract f(a) from f(a+h) and divide by h:
(f(a+h) - f(a))/h = [(4a^2 + 8ah + 4h^2 - 3a - 3h - 1) - (4a^2 - 3a - 1)]/h
Expanding and simplifying:
(f(a+h) - f(a))/h = (8ah + 4h^2 - 3h)/h
(f(a+h) - f(a))/h = 8a + 4h - 3
To find the instantaneous rate of change of f when a = 77, we substitute a = 77 into f(a) and simplify:
f(77) = 4(77)^2 - 3(77) - 1
f(77) = 4(5929) - 231 - 1
f(77) = 23716 - 231 - 1
f(77) = 23484
We are interested in determining whether the variances of the sales at two music stores (A and B) are equal. A sample of 25 days of sales at store A has a sample standard deviation of 30, while a sample of 16 days of sales from store B has a sample standard deviation of 20. At 95% confidence, the null hypothesis _____.
a. should be rejected
b. should be revised
c. should not be rejected
d. None of these answers are correct.
Answer:
C.
Step-by-step explanation:
Hypothesis testing procedure:
Hypothesis:
The null hypothesis will be the variances of sales of two musical stores are equal and the alternative hypothesis will be the variances of sales of two musical stores are not equal
Level of significance: alpha=0.05
Test statistic: F=variance A/variance B=(30)^2/(20)^2=900/400=2.25
P-value: p=0.107
As the alternative hypothesis mentioned that the variances are not equal this leads to two tailed test. so the p-value is calculated using excel function 2*F.DIST.RT(2.25,24,15).
Conclusion:
The p-value seems to exceed the alpha=0.05 and this depicts that the null hypothesis should not be rejected.
At 95% confidence, the null hypothesis should not be rejected. The correct answer is option c. should not be rejected.
Step 1
To test whether the variances of the sales at music stores A and B are equal, we perform an F-test. The null hypothesis [tex](\(H_0\))[/tex] states that the variances are equal, while the alternative hypothesis [tex](\(H_a\))[/tex] states that they are not equal.
The F-statistic is calculated as the ratio of the larger sample variance to the smaller sample variance:
[tex]\[ F = \frac{s_1^2}{s_2^2} \][/tex]
Where [tex]\( s_1^2 \)[/tex] is the variance of store A and [tex]\( s_2^2 \)[/tex] is the variance of store B.
Step 2
In this case, the F-statistic is [tex]\( \frac{30^2}{20^2} = \frac{900}{400} = 2.25 \)[/tex].
Using a significance level of 0.05 and degrees of freedom (df) as [tex]\( n_1 - 1 \)[/tex] and [tex]\( n_2 - 1 \)[/tex], we compare this value to the critical F-value from the F-distribution table.
Since the calculated F-statistic of 2.25 is less than the critical F-value, we fail to reject the null hypothesis.
Therefore, at 95% confidence, the correct answer is c. should not be rejected.
Insert five numbers between 1/27 and 27 to form a geometric sequence. there are 2 answers
Final answer:
Two valid sequences that insert five numbers between 1/27 and 27 to form a geometric sequence are 1/27, 1/9, 1/3, 1, 3, 9, 27 using a common ratio of 3, and 1/27, -1/9, 1/3, -1, 3, -9, 27 using a common ratio of -3.
Explanation:
To insert five numbers between 1/27 and 27 to form a geometric sequence, we first identify that in a geometric sequence, each term after the first is found by multiplying the previous one by a constant called the common ratio (r). We are effectively looking for seven terms in total (including the given 1/27 and 27) which form this sequence.
The formula for the nth term of a geometric sequence is a_n = a_1 × r^(n-1), where a_n is the nth term of the sequence, a_1 is the first term, and n is the term number.
Since we are given the first term (1/27) and the 7th term (27), we can use these to find the common ratio (r), through the equation 27 = (1/27) × r^(7-1), or simplifying, 27 = (1/27) × r^6. Solving for r, we get r = 3 (considering the positive root for practical purposes in a school context).
Thus, the sequence is: 1/27, 1/9, 1/3, 1, 3, 9, 27. However, since it is mentioned there are two answers, another valid sequence can be determined by using the negative root, r = -3. Therefore, another valid sequence is: 1/27, -1/9, 1/3, -1, 3, -9, 27. These two sequences incorporate the geometric sequence properties correctly.
An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : Suppose a sample of 1418 new car buyers is drawn. Of those sampled, 354 preferred foreign over domestic cars. Using the data, construct the 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.
Final answer:
The 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars, based on a sample of 1418 buyers where 354 prefer foreign cars, is between 0.2205 and 0.2787.
Explanation:
To construct a 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars, we use the sample proportion and the Z-distribution since the sample size is large. Given that in a sample of 1418 new car buyers, 354 preferred foreign cars, we calculate the sample proportion (p-hat) as follows:
p-hat = 354 / 1418 ≈ 0.2496
The formula for a confidence interval is p-hat ± Z*(sqrt((p-hat*(1-p-hat))/n)), where Z* is the Z-score corresponding to the confidence level, and n is the sample size. For a 99% confidence interval, Z* is approximately 2.576.
Using the formula, the standard error (SE) for the proportion is calculated as:
SE = sqrt((0.2496*(1-0.2496))/1418) ≈ 0.0113
The margin of error (ME) is:
ME = Z* * SE ≈ 2.576 * 0.0113 ≈ 0.0291
Now, we can construct the 99% confidence interval:
99% CI = p-hat ± ME = 0.2496 ± 0.0291 = (0.2205, 0.2787)
Therefore, we are 99% confident that the true proportion of new car buyers who prefer foreign cars over domestic cars is between 0.2205 and 0.2787.
Your mathematics instructor claims that, over the years, 88% of his students have said that math is their favorite subject. In this year's class, however, only 21 out of 32 students named math as their favorite class. The instructor decides to construct a confidence interval for the true population proportion based on the sample value. What's the correct value for the standard error of pˆ in this case?
Answer: 0.084
Step-by-step explanation:
Formula to find standard error of [tex]\hat{p}[/tex] for finding confidence interval for p:
[tex]SE=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where [tex]\hat{p}[/tex] = sample proportion and n= sample size.
Let p be the population proportion of students named math as their favorite class.
As per given , we have
n= 32
[tex]\hat{p}=\dfrac{21}{32}=0.65625[/tex]
Substitute these values in the formula, we get
[tex]SE=\sqrt{\dfrac{0.65625(1-0.65625)}{32}}\\\\=\sqrt{0.00705}\\\\=0.0839642781187\approx0.084[/tex]
∴ The correct value for the standard error of [tex]\hat{p}[/tex] in this case = 0.084
Final answer:
The standard error of pˆ in this case is 0.0417.
Explanation:
The correct value for the standard error of pˆ in this case is 0.0417.
To calculate the standard error of pˆ, you can use the formula: SE = √((pˆ * q) / n), where pˆ is the sample proportion, q is 1 - pˆ (the proportion of non-favorites), and n is the sample size.
In this case, pˆ = 21/32 = 0.6563, q = 1 - 0.6563 = 0.3437, and n = 32. Plugging these values into the formula gives SE = √((0.6563 * 0.3437) / 32) = 0.0417.
According to the Normal model ?N(0.054?,0.015?) describing mutual fund returns in the 1st quarter of? 2013, determine what percentage of this group of funds you would expect to have the following returns. Complete parts? (a) through? (d) below. ?
a) Over? 6.8%? ?b) Between? 0% and? 7.6%? ?c) More than? 1%? ?d) Less than? 0%?
A) The expected percentage of returns that are over 6.8% is ______%
b) The expected percentage of returns that are between 0& and 7.6% is ____%
C) The expected percentage of returns that are more than 1% is ____%
D) The expected percentage of returns that are less than 0% is ____%
Type an intenger or a decimal rounded to one decimal place as needed.
Answer:
Step-by-step explanation:
Given that X, the mutual fund returns in the 1st quarter of 2013, is
N(0.054, 0.015)
a) P(X>6.8%) = [tex]1-0.8246\\=0.1754[/tex]
b) [tex]P(0<X<0.076)\\\\\\=0.9288-0.0002\\=0.9286[/tex]
c) [tex]P(X>0.01)\\=1-0.0017\\=0.9983[/tex]
d) [tex]P(X<0) = 0.0002[/tex]
A) The expected percentage of returns that are over 6.8% is _17.5_____%
b) The expected percentage of returns that are between 0& and 7.6% is __92.9__%
C) The expected percentage of returns that are more than 1% is _99.8___%
D) The expected percentage of returns that are less than 0% is _0.02___%
It is reasonable to model the number of winter storms in a season as with a Poisson random variable. Suppose that in a good year the average number of storms is 4, and that in a bad year the average is 5. If the probability that next year will be a good year is 0.6 and the probability that it will be bad is 0.4, find the expected value and variance in the number of storms that will occur.
Answer:
E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4
Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24
Step-by-step explanation:
Previous concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
Other equations useful:
Let X and Y random variables
E(X) =E[E(X|Y)] (conditional expectation)
Var(X)=E[Var(X|Y)]+Var[E(X|Y)] (Total variance)
Solution to the problem
Let A the random variable that represent the number of winter storms next year
B a binary variable, B=1 if the next year is a good year and B=0 in the other case. Then we have this:
E(A|B=1) = 4 and E(A|B=0)=5
We can use the propoerties of conditional expectation like this:
E(A)= E[E(A|B)]= E(A|B=1)P(B=1) +E(A|B=0)P(B=0)
E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4
And we can use also the properties for conditional variance we have the following values:
Var(A|B=1)=4 Var(A|B=0)=5, by the propertis of the Poisson distribution
And then the conditional variance is givne by:
[tex]Var[E(A|B)]= E(A|B=1)^2 P(B=1) +E(A|B=0)^2 P(B=0)[/tex]
And if we replace we got:
[tex]Var[E(A|B)]= 4^2 *0.6 +5^2 *0.4 =19.6[/tex]
And we have also that the expected value for the conditional variance is given by:
E[Var(A|B)]= 4*0.6 +5*0.4 =4.4
And then finally the variance for the random variable A is given by:
Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24
The expected value and variance in the number of storms are 4.4 and 4.4 respectively.
To calculate the expected number of winter storms, we can use the probability of a good or bad year along with their respective average storms. The expected value is then given by:
E(X) = (0.6 * 4) + (0.4 * 5) = 2.4 + 2 = 4.4 storms.
For the variance, since the variance of a Poisson distribution is equal to its mean, we have:
Variance in a good year = 4
Variance in a bad year = 5
The overall variance is a weighted average:
V(X) = (0.6 * 4) + (0.4 * 5) = 2.4 + 2 = 4.4
Which of the following is a required condition for a discrete probability function?
a. ∑f(x) = 0 for all values of x
b. f(x) 1 for all values of x
c. f(x) < 0 for all values of x
d. ∑f(x) = 1 for all values of x
The required condition for a discrete probability function is that the sum of the probabilities for all possible outcomes, represented by '∑f(x)', must equal 1.
Explanation:The correct answer to this question is d. ∑f(x) = 1 for all values of x. This is a required condition for a discrete probability function. In probability theory, a probability mass function (PMF), also known as a discrete probability function, provides the probabilities of discrete random variables. The function gives the probability that a discrete random variable is exactly equal to some value. The terms 'probability distribution function' and 'probability function' have also been used to denote the same concept.
As a basic rule, the sum of the probabilities for all possible outcomes (x-values) in a discrete probability function should equal 1. This is because these probabilities together represent the entire possible outcome set for your discrete random variable, which accounts for all possibilities and hence should total to 100% or, in decimal form, 1.
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Final answer:
The correct condition for a discrete probability function is that the sum of all the probabilities must equal 1, which is represented as Σf(x) = 1 for all values of x. The correct option is d.
Explanation:
The required condition for a discrete probability function is that the sum of all probabilities must equal 1, which is known as the normalization condition. This implies that when we list all possible outcomes or events that a random variable can take, the sum of their probabilities must be 1. This condition is represented as Σf(x) = 1 for all values of x. Therefore, the correct option is d. Σf(x) = 1 for all values of x.
PLEASE PLEASE HELP
Answer:68.3 degrees
Step-by-step explanation:
The diagram of the triangle ABC is shown in the attached photo. We would determine the length of side AB. It is equal to a. We would apply the cosine rule which is expressed as follows
c^2 = a^2 + b^2 - 2abCos C
Looking at the triangle,
b = 75 miles
a = 80 miles.
Angle ACB = 180 - 42 = 138 degrees. Therefore
c^2 = 80^2 + 75^2 - 2 × 80 × 75Cos 138
c^2 = 6400 + 5625 - 12000Cos 138
c^2 = 6400 + 5625 - 12000 × -0.7431
c^2 = 12025 + 8917.2
c = √20942.2 = 144.7
To determine A, we will apply sine rule
a/SinA = b/SinB = c/SinC. Therefore,
80/SinA = 144.7/Sin 138
80Sin 138 = 144.7 SinA
SinA = 53.528/144.7 = 0.3699
A = 21.7 degrees
Therefore, theta = 90 - 21.7
= 68.3 degees
Find a particular solution to y′′+25y=−40sin(5t). y″+25y=−40sin(5t).
The particular solution to the differential equation is:
[tex]y_p(t)= 16/25tcos(5t) - 8/5tsin(5t)[/tex]
Here, we have,
To find a particular solution to the given differential equation
y′′+25y=−40sin(5t),
we'll assume a particular solution of the form:
[tex]y_p(t)=Asin(5t)+Bcos(5t)[/tex]
where A and B are constants to be determined.
Now, let's find the first and second derivatives of [tex]y_p(t)[/tex] :
[tex]y_p'(t)=5Acos(5t)-5Bsin(5t)\\y_p''(t)=-25Asin(5t)-25Bcos(5t)[/tex]
Now, substitute these derivatives back into the original differential equation:
y′′+25y=(−25Asin(5t)−25Bcos(5t))+25(Asin(5t)+Bcos(5t))
Now, equate the coefficient of sin(5t) and cos(5t) on both sides of the equation:
For sin(5t):
−25A+25A=0⟹0=0
For cos(5t):
−25B+25B=−40⟹0=−40
The above equations show that there is no solution for A and B that satisfy the original equation.
This means that our initial assumption for the particular solution is not appropriate for this case.
To find a particular solution that works, we'll make another assumption. Since the right-hand side of the differential equation is −40sin(5t), we'll try a particular solution in the form:
[tex]y_p(t)=Atcos(5t)+Btsin(5t)[/tex]
where A and B are constants to be determined.
Now, let's find the first and second derivatives of this new
[tex]y_p'(t)=Acos(5t)-5Atsin(5t)+Bsin(5t)+5Btcos(5t)[/tex]
[tex]y_p''(t)=-10Asin(5t)-25Atcos(5t)+25Btsin(5t)-10Bcos(5t)[/tex]
Now, substitute these derivatives back into the original differential equation:
[tex]y''+25y=(-10Asin(5t)-25Atcos(5t)+25Btsin(5t -10Bcos(5t))+25(Atcos(5t)+Btsin(5t))[/tex]
Now, equate the coefficient of sin(5t) and cos(5t) on both sides of the equation:
For sin(5t):
25Bt=−40⟹B=− 8/5
For cos(5t):
−25At−10B=0
⟹A= 16/25
So, the particular solution to the differential equation is:
[tex]y_p(t)= 16/25tcos(5t) - 8/5tsin(5t)[/tex]
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To find a particular solution to the differential equation y″+25y=−40sin(5t), assume a particular solution in the form y(t) = Asin(5t) + Bcos(5t) and solve for the coefficients A and B.
Explanation:To find a particular solution to the given differential equation y″+25y=−40sin(5t), we can assume a particular solution in the form of y(t) = Asin(5t) + Bcos(5t). Differentiating this equation twice and substituting it back into the original differential equation, we can solve for the coefficients A and B. In this case, the particular solution is y(t) = -8sin(5t) - 0.64cos(5t).
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A new housing development offers homes with a mortgage of $222,000 for 25 years at an annual interest of 8%. Find the monthly mortgage payment.
Using the same formula from your other question:
A = P x (r/12(1+r)^t)/ (1+r/12)^t -1)
A = 222000 x (0.08/12(1+0.08/12)^300 / (1+0.08/12)^300 - 1)
A = $1,713.43 per month
What point on the line y 9x+4 is closest to the origin? Let D be the distance between the two points. What is the objective function in terms of the x-coordinate? (Type an expression.)
Answer:
The objective function in terms of the x-coordinate is [tex]d=\sqrt{82x^2+72x+16}[/tex].
The point closest to the origin is [tex](-\frac{18}{41},\frac{2\sqrt{82}}{41})[/tex].
Step-by-step explanation:
The formula for the distance from point (x, y) to the origin is
[tex]d=\sqrt{x^{2}+y^{2} }[/tex]
So, in our case, [tex]y=9x+4[/tex] and the distance is
[tex]d=\sqrt{x^{2}+(9x+4)^{2} }\\\\d=\sqrt{x^2+81x^2+72x+16} \\\\d=\sqrt{82x^2+72x+16}[/tex]
This is the objective function.
Next, we need to find the derivative of the function
[tex]d=\sqrt{82x^2+72x+16}\\\\\frac{d}{dx}d= \frac{d}{dx}\sqrt{82x^2+72x+16}\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt{u},\:\:u=82x^2+72x+16\\\\\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(82x^2+72x+16\right)\\\\\frac{1}{2\sqrt{u}}\left(164x+72\right)\\\\\mathrm{Substitute\:back}\:u=82x^2+72x+16\\\\\frac{1}{2\sqrt{82x^2+72x+16}}\left(164x+72\right)\\\\\frac{d}{dx}d=\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}[/tex]
Now, we set the derivative function equal to zero to find the critical points
[tex]\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}=0[/tex]
[tex]\frac{f\left(x\right)}{g\left(x\right)}=0\quad \Rightarrow \quad f\left(x\right)=0\\\\2\left(41x+18\right)=0\\\\\frac{2\left(41x+18\right)}{2}=\frac{0}{2}\\\\41x+18=0\\\\x=-\frac{18}{41}[/tex]
We need to check that the value that we found is a minimum point for this we analyze the intervals of increase or decrease (First derivative test).
We can use a sign chart. In a sign chart, we pick a test value at each interval that is bounded by the critical points and check the derivative's sign on that value.
This is the sign chart for our function:
[tex]\left\begin{array}{ccc}\mathrm{Interval}&\mathrm{Test \:x-value}&f'(x)\\(-\infty,-\frac{18}{41} )&-1&-9.02\\(-\frac{18}{41},\infty )&1&9.05\end{array}\right\\[/tex]
d(x) decreases before [tex]x=-\frac{18}{41}[/tex], increases after it, and is defined at [tex]x=-\frac{18}{41}[/tex]. So d(x) has a relative minimum point at [tex]x=-\frac{18}{41}[/tex].
The point closest to the origin is
[tex]d=y=\sqrt{82(-\frac{18}{41})^2+72(-\frac{18}{41})+16}=\frac{2\sqrt{82}}{41}[/tex]
[tex](-\frac{18}{41},\frac{2\sqrt{82}}{41})[/tex]
The objective function in this problem is the x-coordinate of the closest point on the line y = 9x + 4 to the origin. To find this point, we need to find the intersection of the line and the perpendicular line passing through the origin.
Explanation:The objective function in this case is the distance between the closest point on the line y = 9x + 4 and the origin. To find this point, we need to find the intersection of the line and the perpendicular line passing through the origin. The x-coordinate of this closest point will be our objective function.
The slope of the given line is 9, the negative reciprocal of which is -1/9. This means the perpendicular line will have a slope of -1/9 as well. Since the perpendicular line passes through the origin, its equation is given by y = -1/9x.
To find the intersection point, we can set the equations of the two lines equal to each other: 9x + 4 = -1/9x. Solving this equation, we find x = -4/81. Therefore, the objective function in terms of the x-coordinate is -4/81.
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Use the general slicing method to find the volume of the following solids. The solid whose base is the region bounded by the curves y=x² and y=2−x², and whose cross sections through the solid perpendicular to the x-axis are squares
Answer:
The volume is [tex]V=\frac{64}{15}[/tex]
Step-by-step explanation:
The General Slicing Method is given by
Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is
[tex]V=\int\limits^b_a {A(x)} \, dx[/tex]
Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is
The key observation is that the width is the distance between the upper bounding curve [tex]y = 2 - x^2[/tex] and the lower bounding curve [tex]y = x^2[/tex]
The width of each square is given by
[tex]w=(2-x^2)-x^2=2-2x^2[/tex]
This means that the area of the square cross section at the point x is
[tex]A(x)=(2-2x^2)^2[/tex]
The intersection points of the two bounding curves satisfy [tex]2 - x^2=x^2[/tex], which has solutions x = ±1.
[tex]2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}[/tex]
Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is
[tex]V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}[/tex]
The cash operating expenses of the regional phone companies during the first half of 1994 were distributed about a mean of $29.93 per access line per month, with a standard deviation of $2.65. Company A's operating expenses were $27.00 per access line per month. Assuming a normal distribution of operating expenses, estimate the percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean. (Round your answer to two decimal places.)
Answer:
The percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean is 73%.
Step-by-step explanation:
The Company A's operating expenses were $27.00. This is $2.93 less than the regional mean.
[tex]\Delta E=29.93-27.00=2.93[/tex]
The companies whose operating expenses are closer to the mean are the ones that have expenses $2.93 below or above the mean.
The fraction of companies that are closer to the mean is equal to the proability of having expenses between those two limits:
[tex]z_1=(M-\mu)/\sigma=-2.93/2.65=-1.105\\\\z_2=+1.105[/tex]
[tex]P(|z|\leq1.105)=P(z\leq 1.105)-P(z<-1.105)=0.86542-0.13458=0.73[/tex]
To find the percentage of companies with operating expenses closer to the mean than Company A, calculate the Z score for Company A. Then find the percentile, and double it to account for both sides of the normal distribution.
Explanation:To find out the percentage of regional phone companies that had operating expenses closer to the mean than the ones of Company A, we need to calculate the Z-score for Company A. The
Z-score
is a measure of how many standard deviations an element is from the mean. In this particular case, you can calculate the Z score using the formula:
Z = (X - μ) / σ
, where X is the value from the dataset (in this case, Company A's operating expenses), μ is the mean of the dataset, and σ is the standard deviation of the dataset.
So using the data from the question:
Z = ($27.00 - $29.93) / $2.65 = -1.109. Next, recognize that finding the percentage closer to the mean is the same as finding the percentile of company A's Z-score. You can then use a standard normal distribution table or online Z-score calculator to find the percentile associated with a Z-score of -1.109. If it's a two-tailed test (as normally implied by the word 'closer' in statistical analysis), you can multiply the cumulative probability by 2, as it would be the probability on either side of the distribution.This will give you a percentage that can be understood as the percentage of regional phone companies which had operating expenses closer to the mean than Company A's were to the mean.
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A continuous random variable may assume:
-any value in an interval or collection of intervals
-only integer values in an interval or collection of intervals
-only fractional values in an interval or collection of intervals
-only the positive integer values in an interval
Answer:
-any value in an interval or collection of intervals
Step-by-step explanation:
Discrete random variables are only integers in the interval.
Continuous random variables are all the values(integers, fractional, negative, positive) in an interval, or a collection of intervals.
The correct answer is:
-any value in an interval or collection of intervals
A continuous random variable can assume any value in an interval or collection of intervals. Unlike discrete random variables, which can only take integer values, continuous variables can take a range of values, like measurements such as height.
Explanation:A continuous random variable is a type of variable in statistics that can assume any value within an interval or collection of intervals. This is different from a discrete random variable which can only take on a finite or countable number of values (often integer values).
For example, if we measure the height of a person, it could be any value within the feasible range, say between 0 meter and 3 meters. This is a continuous variable because there are infinite possibilities between 0 and 3, including measurements like 1.73 meters or 2.45 meters etc. It's not limited to integer values (like 1 m, 2 m, etc.), nor only fractional values, nor just positive integers within an interval.
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