Answer: 95% confidence interval would be (0.449, 0.511)
Step-by-step explanation:
Since we have given that
n = 1022
p = 48% = 0.48
We need to find the 95% confidence interval first.
z = 1.96
Margin of error would be
[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.48\times 0.52}{1022}}\\\\=0.031[/tex]
95% confidence interval would be
[tex]p\pm 0.031\\\\=(0.48-0.031,0.48+0.031)\\\\=(0.449,0.511)[/tex]
It means true proportion who felt that economic growth is more important than protecting the environment is within 0.449 and 0.511 using 95% confidence.
What is an equation of the line, in point-slope form, that passes through the given point and has the given slope?
point: (11, 3); slope: 4/11
a. y - 3 = 4/11(x + 11)
b. y - 3 = 4/11(x - 11)
c. y - 3 = -4/11( x - 11)
d. y - 11 = 4/11(x - 3)
Answer:
The answer is b
Step-by-step explanation:
cause we have the equation y- y1=m(x-x1)
so y-3=4/11(x-11)
You believe that the mean BMI for college students is less than 25. The population standard deviation of BMI is unknown. You select a sample of size 30 and compute the sample mean to be 23.9 with a sample standard deviation of 3. What are the appropriate hypotheses? A. Null Hypothesis: μ< 25 Alt Hypothesis: μ= 25 B. Null Hypothesis: μ=23.9 Alt Hypothesis: μ < 23.9 C. Null Hypothesis: μ=25 Alt Hypothesis: μ>25 D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25
Answer:
D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25
Step-by-step explanation:
Previous concepts and notation
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".
[tex]\bar X= 23.9[/tex] represent the sample mean
[tex]s=3[/tex] represent the sample standard deviation
n=25 represent the sample selected
Solution
The claim that we want to test is that: "You believe that the mean BMI for college students is less than 25". So this statement needs to be on the Alternative hypothesis ([tex]\mu <25[/tex]) and the Null hypothesis would be the complement ([tex]\mu =25[/tex]) or ([tex]\mu \geq 25[/tex]).
So the best option on this case is:
D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25
There are 39 employees in a particular division of a company. Their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000. The largest number on the list is $100,000. By accident, this number is changed to $1,000,000.
a)What is the value of the mean after the change? Write your answer in units of $1000.b)What is the value of the median after the change? Write your answer in units of $1000.c)What is the value of the standard deviation after the change? Write your answer in units of $1000.
Answer:
a) New mean 78.07692308 thousands of dollars
b) The median does not vary
c) New standard deviation 150.1793799 thousands of dollars
Step-by-step explanation:
We are working in units of $1,000
a)What is the value of the mean after the change?
Let
[tex]s_1,s-2,...,s_38[/tex]
be the salaries of the employees that earn less than 100 units.
The mean of the 39 salaries is 55 units so
[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+100}{39}=55[/tex]
and
[tex]s_1+s_2+...+s_{38}+100=55*39=2145[/tex]
By accident, the 100 on the left is changed to 1,000
[tex]s_1+s_2+...+s_{38}+100+900=55*39=2145+900\Rightarrow \\\\\Rightarrow s_1+s_2+...+s_{38}+1000=3045[/tex]
Dividing by 39 both sides, we get the new mean
[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+1000}{39}=\displaystyle\frac{3045}{39}=78.07692308[/tex]
b)What is the value of the median after the change?
Since the number of data does not change and only the right end of the range of salaries is changed, the median remains the same; 55
c)What is the value of the standard deviation after the change?
The variance is 400, so
[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2}{39}=400\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2=400*39=15600[/tex]
Adding 864,000 to both sides we get
[tex](s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2+864000=15600+864000\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2=879600[/tex]
Dividing by 39 and taking the square root we get the new standard deviation
[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}=\displaystyle\frac{879600}{39}=22553.84615\Rightarrow\\\\\Rightarrow \sqrt{\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}}=150.1793799[/tex]
Final answer:
a) The new mean after the change is approximately $66,923,000. b) The new median after the change is approximately $60,961,000. c) The new standard deviation after the change is approximately $234,358,000.
Explanation:
a) To find the new value of the mean, we need to subtract the original largest number ($100,000) and add the new largest number ($1,000,000) to the sum of the salaries. The sum of the salaries can be calculated by multiplying the mean ($70,000) by the number of employees (39). So, the new sum of the salaries is $(70,000 × 39) - $100,000 + $1,000,000 = $2,610,000. Since there are still 39 employees, the new mean is $2,610,000 divided by 39, which is approximately $66,923. The value of the mean after the change is $66,923,000 (in units of $1000).
b) The median is the middle value in a list of numbers when they are ordered from smallest to largest. After the change, the values in the list would be: $55,000, $55,000, $55,000, ..., $55,000, $66,923, $66,923, ..., $1,000,000. Since there are 39 employees, the middle two values would be $55,000 and $66,923. To find the median, we take the average of these two values: ($55,000 + $66,923) / 2 = $60,961. The value of the median after the change is $60,961,000 (in units of $1000).
c) To find the new standard deviation, we need to recalculate it using the new values. First, we need to find the squared differences between each salary and the new mean. The sum of these squared differences is $(39 × ($66,923 - $66,923)^2) + $1,000,000. Then, we divide this sum by 39 and take the square root to find the new standard deviation. The value of the standard deviation after the change is approximately $234,358,000 (in units of $1000).
A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. A random sample of 49 bottles is selected, and the contents are measured. The sample yielded a mean content of 11.88 ounces with a standard deviation of 0.35 ounces. Please test if the machine fills the bottles with 12 ounces.
Answer:
yes
Step-by-step explanation:
Suppose a random sample of size 60 is selected from a population withσ = 10.Find the value of the standard error of the mean in each of the following cases. (Use the finite population correction factor if appropriate. Round your answers to two decimal places.)(a)The population size is infinite.Correct: Your answer is correct.(b)The population size isN = 60,000.Correct: Your answer is correct.(c)The population size isN = 6,000.Incorrect: Your answer is incorrect.(d)The population size isN = 600.Incorrect: Your answer is incorrect.
Answer:
a) 1.29
b) 1.29
c) 1.28
d) 1.23
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 60
Population standard Deviation = 10
a) Standard error with infinite population
[tex]\text{Standard error} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{60}} = 1.29[/tex]
For finite population with size N,
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{N-n}{N-1}}\times \displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
b) N = 60,000
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{60000-50}{60000-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.29[/tex]
c) N = 6,000
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{6000-50}{6000-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.28[/tex]
d) N = 600
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{600-50}{600-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.23[/tex]
To find the standard error of the mean, use the formula σx/√n for infinite population size and σx/√n * √((N - n)/(N - 1)) for finite population size. Use the finite population correction factor when the sample size is not negligible relative to the population size.
Explanation:To find the value of the standard error of the mean in each case, we need to use the formula for the standard error of the mean, which is σx/sqrt(n), where σ is the population standard deviation and n is the sample size.
a) Case with infinite population size:If the population size is infinite, we don't need to use the finite population correction factor. So, the standard error of the mean would be σ/√n.
b) Case with N = 60,000:In this case, the population size is finite, but the sample size is less than 5% of the population size. Therefore, we can still consider the population as effectively infinite. So, the standard error of the mean would be σ/√n.
c) Case with N = 6,000:In this case, the population size is finite and the sample size is not negligible relative to the population size. Therefore, we need to use the finite population correction factor, which is √((N - n)/(N - 1)). So, the standard error of the mean would be σ/√n * √((N - n)/(N - 1)).
d) Case with N = 600:In this case, the population size is finite and the sample size is not negligible relative to the population size. Therefore, we need to use the finite population correction factor. So, the standard error of the mean would be σ/√n * √((N - n)/(N - 1)).
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A manufacturer must decide whether to build a small or a large plant at a new location. Demand at the location can be either low or high, with probabilities estimated to be 0.4 and 0.6, respectively. If a small plant is built, and demand is high, the production manager may choose to maintain the current size or to expand. The net present value of profits is $223,000 if the firm chooses not to expand. However, if the firm chooses to expand, there is a 50% chance that the net present value of the returns will be 330,000 and a 50% chance the estimated net present value of profits will be $210,000. If a small facility is built and demand is low, there is no reason to expand and the net present value of the profits is $200,000. However, if a large facility is built and the demand turns out to be low, the choice is to do nothing with a net present value of $40,000 or to stimulate demand through local advertising. The response to advertising can be either modest with a probability of .3 or favorable with a probability of .7. If the response to advertising is modest, the net present value of the profits is $20,000. However, if the response to advertising is favorable, then the net present value of the profits is $220,000. Finally, if the large plant is built and the demand happens to be high, the net present value of the profits $800,000a) Draw a decision tree b) Determine the payoff for each decision and event node. Which alternative should the manufacturer choose?
Answer:
a then b
Step-by-step explanation:
a then b because you could mess up if you didn't draw a decision tree
Louise is collecting can tabs for charity. She already has 35 collected and intends to collect 4 every week. The equation for the number of can tabs y she has collected is y = 4x + 35, where x is the number of weeks. What do the slope and y-intercept represent?
Answer:
Step-by-step explanation:
The equation for the number of can tabs y that she has collected is
y = 4x + 35
where x is the number of weeks. To plot the graph, values for different number of weeks are inputted to get the corresponding y values.
Comparing the equation with the slope intercept form,
y = mx + c,
m represents slope.
So slope of the equation is 4. It represents the rate at which the total number of can tabs that she has collected is increasing with respect to the increase in the number of weeks. So the total number increases by 4 each week
The y intercept is the point at x = 0
It is equal to 35. It represents the number of can tabs that she had initially collected before she started collecting 4 can tabs weekly. At this point, her weekly collection is 0
In 2008, the Centers for Disease Control and Prevention reported that 34% of adults in the United States are obese. A country health service planning a new awareness campaign polls a random sample of 750 adults living there. In this sample, 228 people were found to be obese based on their answers to a health questionnaire. Do these response provide strong evidence that the 34% figure is not accurate for this region?
Answer: No, these response does not provide strong evidence that the 34% figure is not accurate for this region.
Step-by-step explanation:
Since we have given that
p = 0.34
x= 228
n = 750
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{228}{750}=0.304[/tex]
So, hypothesis would be
[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]
So, test statistic value would be
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.304-0.38}{\sqrt{\dfrac{0.38\times 0.62}{750}}}\\\\z=\dfrac{-0.076}{0.0177}\\\\z=-4.293[/tex]
At 95% confidence , z = 1.96
So, 1.96>-4.293.
So, we accept the null hypothesis.
No, these response does not provide strong evidence that the 34% figure is not accurate for this region.
The observed data doesn't provide enough evidence to suggest that the 34% figure reported by the CDC is inaccurate for this region.
let's break down the hypothesis test step by step with more detail.
1. Setting up the Hypotheses:
- Null Hypothesis (H0):The true proportion of obese adults in the region is 34%.
- Alternative Hypothesis (H1):The true proportion of obese adults in the region is not 34%.
2. Calculating the Expected Number of Obese Individuals:
To calculate the expected number of obese individuals in the sample under the assumption that the true proportion is 34%, we use the formula:
[tex]\[ \text{Expected number of obese individuals} = \text{Proportion} \times \text{Sample size} \][/tex]
So,
[tex]\[ \text{Expected number of obese individuals} = 0.34 \times 750 = 255 \][/tex]
3. Checking Conditions:
- The sample is randomly selected.
- The sample size is large enough for the Central Limit Theorem to apply (n = 750).
- Since the population size isn't provided, we'll assume it's much larger than the sample size (which is often the case with populations of adults in countries).
4. Calculating the Z-Score:
The formula for the z-score is:
[tex]\[ z = \frac{{\text{Observed proportion} - \text{Expected proportion}}}{{\sqrt{\frac{{\text{Expected proportion} \times (1 - \text{Expected proportion})}}{{\text{Sample size}}}}}} \][/tex]
So,
[tex]\[ z = \frac{{\frac{228}{750} - \frac{255}{750}}}{{\sqrt{\frac{255}{750} \times \frac{495}{750}}}} \]\[ z \approx \frac{{0.304 - 0.34}}{{\sqrt{\frac{191.25}{750}}}} \]\[ z \approx \frac{{-0.036}}{{\sqrt{0.255}}} \]\[ z \approx \frac{{-0.036}}{{0.505}} \]\[ z \approx -0.071 \][/tex]
5. Finding Critical Z-Value:
For a two-tailed test with a significance level of 0.05, the critical z-values are approximately ±1.96.
6. Making a Decision:
Since the calculated z-score (-0.071) falls within the range (-1.96, 1.96), we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the true proportion of obese adults in the region is different from 34%.
In conclusion, based on the provided sample data, we cannot confidently claim that the 34% figure reported by the CDC is inaccurate for this region.
Since an instant replay system for tennis was introduced at a majorâ tournament, men challenged 1399 refereeâ calls, with the result that 411 of the calls were overturned. Women challenged 759 refereeâ calls, and 211of the calls were overturned.
Use a 0.01 significance level to test the claim that men and women have equal success in challenging calls.H0:p1=p2H1:p1 (does not equal) p2Identify the test statistic z=___
Answer: The value of test statistic is 0.696.
Step-by-step explanation:
Since we have given that
[tex]n_1=1399\\\\x_1=411\\\\p_1=\dfrac{x_1}{n_1}=\dfrac{411}{1399}=0.294[/tex]
Similarly,
[tex]n_2=759, x_2=211\\\\p_2=\dfrac{x_2}{n_2}=\dfrac{211}{759}=0.278[/tex]
At 0.01 level of significance.
Hypothesis are :
[tex]H_0:P_1=P_2=0.5\\\\H_a:P_1\neq P_2[/tex]
So, the test statistic value would be
[tex]z=\dfrac{(p_1-p_2)-(P_1-P_2)}{\sqrt{\dfrac{P_1Q_1}{n_1}+\dfrac{P_2Q_2}{n_2}}}\\\\z=\dfrac{0.294-0.278}{\sqrt{0.5\times 0.5(\dfrac{1}{1399}+\dfrac{1}{759})}}\\\\z=\dfrac{0.016}{0.023}=0.696[/tex]
Hence, the value of test statistic is 0.696.
please solve quick!! i really need it!!
Answer:
[tex]\dfrac{1}{x^2}+\dfrac{y^2}{x^4}[/tex]
Step-by-step explanation:
Separate the fraction(s) at the plus sign and simplify.
[tex]\dfrac{x^2+y^2}{x^4}=\dfrac{x^2}{x^4}+\dfrac{y^2}{x^4}\\\\=\dfrac{1}{x^2}+\dfrac{y^2}{x^4}[/tex]
A plant in the manufacturing sector is concerned about its sulfur dioxide emissions. The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25. Recently, a "cleaner" technology has been adopted. In such a scenario, the CEO would like to investigate whether there has been a significant change in the emissions and has hired you for advice. In other words, the CEO wants to know if the mean level of emissions is different from 1000. Suppose that you are given a sample of data on weekly sulfur dioxide emissions for that plant. The sample size is 50 and x = 1006 ppm. What is the value of the test statistic?a. 2.26b. 1.70c. 4.78d. 2.59
Answer:
The correct option is b) 1.70
Step-by-step explanation:
Consider the provided information.
The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25.
Thus, μ=1000 and σ = 25
The CEO wants to know if the mean level of emissions is different from 1000.
Therefore the null and alternative hypothesis is:
[tex]H_0:\mu =1000[/tex] and [tex]H_a:\mu \neq1000[/tex]
The sample size is n = 50 and [tex]\bar x=1006[/tex] ppm.
Now calculate the test statistic by using the formula: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Substitute the respective values in the above formula.
[tex]z=\frac{1006-1000}{\frac{25}{\sqrt{50}}}[/tex]
[tex]z=\frac{6}{\frac{25}{\sqrt{50}}}=1.697\approx 1.70[/tex]
Hence, the correct option is b) 1.70
Two new mathematics learning techniques are being tested. The researchers are unsure which one, if any, will be better. Two hundred students were randomly selected from a population. Ninety of them were randomly assigned to use Technique A, and 110 of them were randomly assigned to use Technique BEach student spent 30 minutes learning the technique to which they were assigned, and then were asked to complete a task. The time to complete the task was recorded, in seconds. A shorter time indicates better mastery ofthe task. The data are below:Technique A: sample average = 26.1, sample SD = 7.2Technique B: sample average = 27.9, sample SD = 12.4(a) State hypotheses relevant to the research question.(b) Perform a test of the hypotheses from (a) using a significance level of 10%. Make sure to compute the test statistic and P-value, and make a conclusion in context.
Answer:
(a) Hypotheses relevant to the research question are
[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} < 0[/tex] (lower-tail alternative) and
[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} > 0[/tex] (upper-tail alternative).
(b) There is no evidence that one mathematic learning technique is better than the other.
Step-by-step explanation:
Let's suppose that the data related to Technique A comes from population 1 and that the data related to Technique B comes from population 2. We have large sample sizes [tex]n_{1} = 90[/tex] and [tex]n_{2} = 110[/tex]. The unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 26.1 - 27.9 = -1.8. The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}\approx[/tex] [tex]\sqrt{\frac{(7.2)^{2}}{90}+\frac{(12.4)^{2}}{110}}[/tex] = 1.4049.
(a) Hypotheses relevant to the research question are
[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} < 0[/tex] (lower-tail alternative) and
[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} > 0[/tex] (upper-tail alternative).
(b) The test statistic is [tex]Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}[/tex] and the observed value is [tex]z_{0} = \frac{-1.8}{1.4049} = -1.2812[/tex].
The p-value for the lower-tail alternative is P(Z < -1.2812) = 0.1000617 and the p-value for the upper-tail alternative is P(Z > -1.2812) = 0.8999. The p-value is greater than 0.10 in both cases, therefore, we fail to reject the null hypotheses in both cases.
The null hypothesis states that the mean completion time using Technique A equals that using Technique B. The alternative hypothesis states that they are not equal. A t-test can be used to test these hypotheses. The test statistic and the P-value are calculated and compared with the given significance level (0.10) to make a conclusion.
Explanation:Given the data provided for Techniques A and B, let’s formulate the hypothesis and carry out the test to determine the effectiveness of the techniques.
(a) State the Hypotheses
Here, the null hypothesis (H₀) is that the mean completion time using Technique A equals that using Technique B. The alternative hypothesis (H₁) is that the mean time using Technique A is not equal to that using Technique B.
(b) Hypothesis Testing
For the hypothesis test, we have to calculate the t-statistic and the P-value. Given that the data is normally distributed, we can use the two-sample t-test. The standard formula used to calculate the t-statistic is: (sample mean₁ - sample mean₂) / √[(sample variance₁/n₁) + (sample variance₂/n₂)]. After calculating these values, the resultant t-statistic can be compared to critical t values for the given level of significance (α=10%).
Next, compute the P-value using the t-statistic, which indicates the probability of observing such a difference in means under the null hypothesis. We reject the null hypothesis if the P-value is less than our significance level (α).
Without specific test statistics and P-value, we can’t make a definitive conclusion. In general, if we have a statistically significant result (P<0.10), we would reject the null hypothesis and conclude that there is a significant difference in the performance of Techniques A and B.
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The proportion of students at a college who have GPA higher than 3.5 is 19%. a. You take repeated random samples of size 25 from that college and find the proportion of student who have GPA higher than 3.5 for each sample. What is the mean and the standard error of the sampling distribution of the sample proportions?
Answer:
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=0.0785[/tex]
Step-by-step explanation:
We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-
[tex]\mu_{\hat{p}}=p[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]
, where p=population proportion and n= sample size.
Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.
i.e. p= 19%=0.19
The for sample size n= 25
The mean and the standard error of the sampling distribution of the sample proportions will be :-
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785[/tex]
Hence , the mean and the standard error of the sampling distribution of the sample proportions :
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=0.0785[/tex]
If f '(5) = 0 and f ''(5) = 0, what can you say about f ?
A. At x = 5, f has a local maximum.
B. At x = 5, f has a local minimum.
C. At x = 5, f has neither a maximum nor a minimum.
D. More information is needed to determine if f has a maximum or minimum at x = 5.
Answer:
D)
Step-by-step explanation:
Remember, if a is critical point of f then f'(a)=0. And criterion of the second derivative says that if a is a critical point of f and
1. if [tex]f''(a)<0[/tex] then f has a relative maximum in (a,f(a)),
2. if [tex]f''(a)>0[/tex] then f has a relative minimum in (a,f(a)),
3. if [tex]f''(a)=0,[/tex] Then the criterion does not decide. That is, f may have a relative maximum at a, a relative minimum at (a, f (a)) or neither.
Since [tex]f'(5)=0[/tex] then 5 is a critical point of f. Now we apply the second criterium:
since [tex]f''(5)=0[/tex] then the criterium doesn't decide, that means, more information is needed to determine if f has a maximum or minimum at x = 5.
Using the concept of critical point and the second derivative test, it is found that the correct option is:
D. More information is needed to determine if f has a maximum or minimum at x = 5.
The critical points of a function [tex]f(x)[/tex] are the values of x for which [tex]f^{\prime}(x) = 0[/tex].Applying the second derivative test, we have that:
If positive, that is, [tex]f^{\prime\prime}(x) > 0[/tex], it is a relative minimum.If negative, that is, [tex]f^{\prime\prime}(x) < 0[/tex], it is a relative maximum.If zero, that is, [tex]f^{\prime\prime}(x) = 0[/tex], we do not have sufficient information.In this problem:
[tex]f^{\prime}(5) = 0[/tex], thus, at x = 5 is a critical value.[tex]f^{\prime\prime}(5) = 0[/tex], thus, we need more information, which means that the correct option is:D. More information is needed to determine if f has a maximum or minimum at x = 5.
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If all the beads in a jar are either orange or purple and the ratio of orange to purple beads is 6 to 5, what is the ratio of the following:
a. purple to orange
b. purple to all beads
c. all beads to orange
Please help ASAP on all 3 parts!!! :(
Answer:
a. purple to orange =[tex]\frac{5}{6}[/tex]
b. purple to all beads=[tex]\frac{5){11}[/tex]
c. all beads to orange=[tex]\frac{11}{6}[/tex]
Step-by-step explanation:
Given:
Colour of beads in jar = orange or purple
The ratio of orange to purple beads = 6 : 5
Solution:
Let number of orange beads be= 6x
Number of purple beads=5x
Total no of beads= 6x+5x=11x
a. Ratio of purple to orange.
Ratio of purple to orange=[tex]\frac{\text{number of purple boads}}{\text{number of orange beads}}[/tex]
Ratio of purple to orange=[tex]\frac{5x}{6x}[/tex]
Ratio of purple to orange=[tex]\frac{5}{6}[/tex]
b. Ratio of purple to all beads
Ratio of purple to all beads=[tex]\frac{\text{number of purple beads}}{\text{Total number of beads}}[/tex]
Ratio of purple to all beads=[tex]\frac{5x}{11x}[/tex]
Ratio of purple to all beads=[tex]\frac{5}{11}[/tex]
c. Ratio of all beads to orange
Ratio of purple to all beads=[tex]\frac{\text{Total number of beads}}{\text{number of orange beads}}[/tex]
Ratio of purple to all beads=[tex]\frac{11x}{6x}[/tex]
Ratio of purple to all beads=[tex]\frac{11}{6}[/tex]
Beaker A contains 1 liter which is 30 percent oil and the rest is vinegar, thoroughly mixed up. Beaker B contains 2 liters which is 40 percent oil and the rest vinegar, completely mixed up. Half of the contents of B are poured into A, then completely mixed up. How much oil should now be added to A to produce a mixture which is 60 percent oil?
Answer:
1.25 liters of oil
Step-by-step explanation:
Volume in Beaker A = 1 L
Volume of Oil in Beaker A = 1*0.3 = 0.3 L
Volume of Vinegar in Beaker A = 1*0.7 = 0.7 L
Volume in Beaker B = 2 L
Volume of Oil in Beaker B = 2*0.4 = 0.8 L
Volume of Vinegar in Beaker B = 1*0.6 = 1.2 L
If half of the contents of B are poured into A and assuming a homogeneous mixture, the new volumes of oil (Voa) and vinegar (Vva) in beaker A are:
[tex]V_{oa} = 0.3+\frac{0.8}{2} \\V_{oa} = 0.7 \\V_{va} = 0.7+\frac{1.2}{2} \\V_{va} = 1.3[/tex]
The amount of oil needed to be added to beaker A in order to produce a mixture which is 60 percent oil (Vomix) is given by:
[tex]0.6*V_{total} = V_{oa} +V_{omix}\\0.6*(V_{va}+V_{oa} +V_{omix}) = V_{oa} +V_{omix}\\0.6*(1.3+0.7+V_{omix})=0.7+V_{omix}\\V_{omix}=\frac{0.5}{0.4} \\V_{omix}=1.25 \ L[/tex]
1.25 liters of oil are needed.
There is 1.25 litres of oil that should now be added to A to produce a mixture that is 60 per cent oil.
Given
Beaker A contains 1 litre which is 30 per cent oil and the rest is vinegar, thoroughly mixed up.
Beaker B contains 2 litres which are 40 per cent oil and the rest vinegar, completely mixed up.
Half of the contents of B are poured into A, then completely mixed up.
How much oil is in each container?Contents in beaker A implies;
15% of oil in 1 litre = 0.15 litre of oil
So that, there are 0.15 litres of oil and 0.85 litres of vinegar in beaker A.
Contents in beaker B implies:
55% of oil in 2 litres = 1.1 litres of oil
So that, thee are 1.1 litres of oil and 0.9 litres of vinegar in beaker B.
Half of the contents of B poured into A implies that beaker A now contains:
0.15 litres + 0.55 litres = 0.7 litres of oil
0.85 litres + 0.45 litres = 1.3 litres of vinegar
Then,
The percentage of oil in A is;
[tex]=\dfrac{0.7}{2} \times 100\\\\= 35[/tex]
To increase the percentage of oil to 60%, then:
0.7 litres + 1.25 litres = 1.95 litres of oil
And The new total litres of the content in beaker A = 3.25 litres
[tex]=\dfrac{1.95}{3.25}\times 100\\\\=60 \rm \ percent[/tex]
Hence, 1.25 litres of oil should now be added to A to produce a mixture that is 60 per cent oil.
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A small stock brokerage firm wants to determine the average daily sales (in dollars) of stocks to their clients.A sample of the sales for 36 days revealed average daily sales of $200,000. Assume that the standard deviation of the population is known to be $18,000.
a. Provide a 95% confidence interval estimate for the average daily sale.
b. Provide a 97% confidence interval estimate for the average daily sale.
Answer:
Step-by-step explanation:
a) For a 95% confidence level: [tex]\( \$194,120 \) to \( \$205,880 \)[/tex]
b) For a 97% confidence level: [tex]\( \$193,490 \) to \( \$206,510 \)[/tex]
To solve this problem, we'll use the information provided and apply the formula for a confidence interval estimate for the population mean when the population standard deviation is known.
Given data:
- Sample size n: 36 days
- Sample mean [tex](\( \bar{x} \))[/tex]: $200,000
- Population standard deviation [tex](\( \sigma \)): $18,000[/tex]
(a) 95% Confidence Interval
For a 95% confidence interval, the critical value [tex]\( z^* \)[/tex] from the standard normal distribution is approximately 1.96.
The formula for the confidence interval is:
[tex]\[ \text{CI} = \bar{x} \pm z^* \cdot \frac{\sigma}{\sqrt{n}} \][/tex]
Calculate the standard error:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{18,000}{\sqrt{36}} = \frac{18,000}{6} = 3,000 \][/tex]
Now, construct the confidence interval:
[tex]\[ \text{CI} = 200,000 \pm 1.96 \cdot 3,000 \][/tex]
[tex]\[ \text{CI} = 200,000 \pm 5,880 \][/tex]
Therefore, the 95% confidence interval estimate for the average daily sale is approximately [tex]\( \$194,120 \) to \( \$205,880 \)[/tex].
(b) For a 97% confidence interval, the critical value [tex]\( z^* \)[/tex] from the standard normal distribution is approximately 2.17.
Calculate the confidence interval using the same formula with the updated [tex]\( z^* \):[/tex]
[tex]\[ \text{CI} = \bar{x} \pm z^* \cdot \frac{\sigma}{\sqrt{n}} \][/tex]
[tex]\[ \text{CI} = 200,000 \pm 2.17 \cdot 3,000 \][/tex]
[tex]\[ \text{CI} = 200,000 \pm 6,510 \][/tex]
Therefore, the 97% confidence interval estimate for the average daily sale is approximately [tex]\( \$193,490 \) to \( \$206,510 \)[/tex].
Use a(t) = -32 ft/sec^2 as the acceleration due to gravity. (Neglect air resistance.) With what initial velocity must an object be thrown upward (from ground level) to reach the top of a national monument (580 feet)? (Round your answer to three decimal places.) Use a(t) = -32 ft/sec^2 as the acceleration due to gravity. (Neglect air resistance.) A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 32 feet above the ground.
(a) How many seconds after its release will the bag strike the ground? (Round your answer to two decimal places.) sec
(b) At what velocity will it strike the ground? (Round your answer to three decimal places.)
Answer:
Initial velocity 192.666 ft/sec
a) 1.41 sec
Step-by-step explanation:
Equations for
V(f) = V₀ ± at
V(f)² =( V₀)² ± 2 a*d
d = V₀*t ± a*t²/2
We are going to use a(t) = g = 32 ft/sec²
a) V₀ = ?? a = -g
Then as d = 580 ft and V(f) = 0
we have
V(f)² = ( V₀)² - 2*32*580 ⇒ ( V₀)² = 64*580 (ft²/sec²)
( V₀)² = 37120 V₀ = 192,666 ft/sec
a) t = ?? in this case V₀ = 0
d = V₀*t + gt²/2 ⇒ 32 = ( 32 t²) /2
t² = 2
t = 1.41 sec
The monthly starting salaries of students who receive an MBA degree have a population standard deviation of $120. What sample size should be selected to obtain a .95 probability of estimating the population mean monthly income within a margin of $20? [sample size]
Answer:
138
Step-by-step explanation:
Population standard deviation = [tex]\sigma = 120[/tex]
.95 probability of estimating the population mean monthly income within a margin of $20
So, Significance level = 1-0.95 = 0.05
α =0.05
Margin error = 20
[tex]ME =Z \times \frac{\sigma}{\sqrt{n}}[/tex]
Z at 0.05 = 1.96
[tex]20 =1.96 \times \frac{120}{\sqrt{n}}[/tex]
[tex]\sqrt{n} =1.96 \times \frac{120}{20}[/tex]
[tex]n =(1.96 \times \frac{120}{20})^2[/tex]
[tex]n =138.2976[/tex]
So, n = 138
Hence sample size should be 138 selected to obtain a .95 probability of estimating the population mean monthly income within a margin of $20
A school district has two high schools. The district could only afford to hire 13 guidance counselors. Determine how many counselors should be assigned to each school using Hamilton's method. School Students Enrolled Counselors to Assign Lowell 3584 Fairview 6816 The next year, a new school is opened, with 1824 students. Using the divisor from above, determine how many additional counselors should be hired for the new school: counselors.
a. Using Hamilton's method, 4 counselors should be assigned to Lowell, and 9 counselors should be assigned to Fairview.
b. The same divisor, 2 counselors should be hired for the new school.
Hamilton's method is used to allocate resources, such as counselors, among several entities based on a certain divisor. The divisor is typically calculated by dividing the total number of students by the total number of counselors available.
Let's start by determining the divisor and assigning counselors to each school based on the given information:
1. Calculate the divisor:
[tex]\[ \text{Divisor} = \frac{\text{Total Students}}{\text{Total Counselors}} \]\\\text{Divisor} = \frac{3584 + 6816}{13} = \frac{10400}{13} \approx 800 \][/tex]
2. Assign counselors to each school:
[tex]\[ \text{Counselors Assigned to Lowell} = \frac{\text{Students Enrolled in Lowell}}{\text{Divisor}} \][/tex]
[tex]\[ \text{Counselors Assigned to Fairview} = \frac{\text{Students Enrolled in Fairview}}{\text{Divisor}} \][/tex]
[tex]\[ \text{Counselors Assigned to Lowell} = \frac{3584}{800} = 4.48 \approx 4 \][/tex]
[tex]\[ \text{Counselors Assigned to Fairview} = \frac{6816}{800} = 8.52 \approx 9 \][/tex]
So, using Hamilton's method, 4 counselors should be assigned to Lowell, and 9 counselors should be assigned to Fairview.
3. New School:
If a new school with 1824 students is opened, we can use the same divisor to determine how many additional counselors should be hired for the new school:
[tex]\[ \text{Counselors for New School} = \frac{\text{Students in New School}}{\text{Divisor}} \][/tex]
[tex]\[ \text{Counselors for New School} = \frac{1824}{800} = 2.28 \approx 2 \][/tex]
So, using the same divisor, 2 counselors should be hired for the new school.
The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.
a) Find the probability of a pregnancy lasting 308 days or longer.
b) If the length of pregnancy is in the lowest 22?%, then the baby is premature.
Find the length that separates premature babies from those who are not premature.
a) the probability that a pregnancy will last 308 days or longer is?
b) babies who are born on or before ___ days are considered premature
Answer:
a) The probability that a pregnancy will last 308 days or longer is 0.0038
b) Babies who are born on or before 256 days are considered prematures.
Step-by-step explanation:
Let X be the random variable that represents the length of a pregnancy. Then, X is normally distributed with a mean of 268 days and a standard deviation of 15 days.
a) The z-score related to 308 days is z = (308-268)/15 = 2.6667, so, the probability of a pregnancy lasting 308 days or longer is P(Z > 2.6667) = 0.0038
b) We are looking for a value q such that P(X < q) = 0.22, i.e., P((X-268)/15 < (q-268)/15) = 0.22, here, Z = (X-268)/15 is a standard normal random variable and z = (q-268)/15 is the 22nd quantile of the standard normal distribution, i.e., z = -0.7722 = (q-268)/15 and (-0.7722)(15) + 268 = q, i.e., q = 256.417, so, babies who are born on or before 256 days are considered premature.
Answer:
a) P(X>308) = 0.00383
b) 256.42 days
Step-by-step explanation:
Population mean (μ) = 268 days
Standard deviation (σ) = 15 days
a) P(X>308)?
The z-score for any length of pregnancy 'X' is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
The z-score for X=308 is:
[tex]z=\frac{308-268}{15}\\z=2.6667[/tex]
A z-score of 2.6667 is equivalent to the 99.617-th percentile of a normal distribution. Thus, the probability of a pregnancy lasting 208 days or longer is:
[tex]P(X>308)=1-0.99617\\P(X>308) = 0.00383[/tex]
b) X at which P(X) < 0.22?
At the 22-nd percentile, a normal distribution has an equivalent z-score of -0.772. Therefore, the length of pregnancy, X, that separates premature babies from those who are not premature is:
[tex]-0.772=\frac{X-268}{15}\\X= 256.42[/tex]
Evaluate the expression C(190,1)
Answer:
190
Step-by-step explanation:
C(n, k) = n!/(k!(n -k)!)
C(190, 1) = 190!/(1!(189!)) = 190/1 = 190
The number of combinations of 190 things taken 1 at a time is 190.
A 2003 study of dreaming found that out of a random sample of 106 people, 80 reported dreaming in color. However, the rate of reported dreaming in color that was established in the 1940s was 0.21. Check to see whether the conditions for using a one-proportion z-test are met assuming the researcher wanted to test to see if the proportion dreaming in color had changed since the 1940s.
1. What is the normal approximation method appropriate for this test?2. compute appropriate test statistic.3. At a 0.10 level of significance, what are the critical values for the test.4. what Is the appropriate decision and conclude for the test at 0.10 level of significance (fail to reject, reject Ha)5. would your conclusion change if the test were to be conducted as an upper tailed test? why or why not supporting your answer using the p-value approach.
Answer:
Step-by-step explanation:
Hello!
You have the following experiment, a random sample of 106 people was made and they were asked if they dreamed in color. 80 persons of the sample reported dreaming in color.
The historical data from the 1940s informs that the population proportion of people that dreams in color is 0.21(usually when historical data is given unless said otherwise, is considered population information)
Your study variable is a discrete variable, you can define as:
X: Amount of people that reported dreaming in colors in a sample of 106.
Binomial criteria:
1. The number of observation of the trial is fixed (In this case n = 106)
2. Each observation in the trial is independent, this means that none of the trials will have an effect on the probability of the next trial (In this case, the fact that one person dreams in color doesn't affect or modify the probability of the next one dreaming in color)
3. The probability of success in the same from one trial to another (Or success is dreaming in color and the probability is 0.21)
So X~Bi(n;ρ)
In order to be able to run a proportion Z-test you have to apply the Central Limit Theorem to approximate the distribution of the sample proportion to normal:
^ρ ≈ N(ρ; (ρ(1-ρ))/n)
With this approximation, you can use the Z-test to run the hypothesis.
Now what the investigators want to know is if the proportion of people that dreams in color has change since the 1940s, so the hypothesis is:
H₀: ρ = 0.21
H₁: ρ ≠ 0.21
α: 0.10
The test is two-tailed,
Left critical value: [tex]Z_{\alpha/2} = Z_{0.95} = -1.64[/tex]
Right critical value: [tex]Z_{1-\alpha /2} = Z_{0.95} = 1.64[/tex]
If the calculated Z-value ≤ -1.64 or ≥ 1.64, the decision is to reject the null hypothesis.
If -1.64 < Z-value < 1.64, then you do not reject the null hypothesis.
The sample proportion is ^ρ= x/n = 80/106 = 0.75
Z= 0.75 - 0.21 = 12.83
√[(0.75*0.25)/106]
⇒Decision: Reject the null hypothesis.
p-value < 0.00001 is less than 0.10
If you were to conduct a one-tailed upper test (H₀: ρ = 0.21 vs H₁: ρ > 0.21) with the information of this sample, at the same level 10%, the critical value would be [tex]Z_{0.90}[/tex]1.28 against the 12.83 from the Z-value, the decision would be to reject the null hypothesis (meaning that the proportion of people that dreams in colors has increased.)
I hope this helps!
A local marketing company wants to estimate the proportion of consumers in the Oconee County area who would react favorably to a marketing campaign. Further, the company wants the estimate to have a margin of error of no more than 4 percent with 90 percent confidence. Of the following, which is the closest to the minimum number of consumers needed to obtain the estimate with the desired precision? A. 65 B. 93 C. 423 D. 601
Answer:
C. n=423
Step-by-step explanation:
1) Notation and important concepts
Margin of error for a proportion is defined as "percentage points your results will differ from the real population value"
Confidence=90%=0.9
[tex]\alpha=1-0.9=0.1[/tex] represent the significance level defined as "a measure of the strength of the evidence that must be present in your sample before you will reject the null hypothesis and conclude that the effect is statistically significant".
[tex]\hat p=0.5[/tex] represent the sample proportion of consumers in the Oconee County area who would react favorably to a marketing campaig. For this case since we don't have enough info we use the value of 0.5 since is equiprobable the event analyzed.
[tex]z_{\alpha/2}[/tex] represent a quantile of the normal standard distribution that accumulates [tex]{\alpha/2}[/tex] on each tail of the distribution.
2) Formulas and solution for the problem
For this case the margin of error for a proportion is given by this formula
[tex]ME=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex] (1)
For this case the confidence level is 90% or 0.9 so then the significance would be
[tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex]
With [tex]\alpha/2=0.05[/tex], we can find the value for [tex]z_{\alpha/2}[/tex] using the normal standard distribution table or excel.
The calculated value is [tex]z_{\alpha/2}=1.644854[/tex]
Now from equation (1) we need to solve for n in order to answer the question.
[tex]\frac{ME}{z_{\alpha/2}}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
Squaring both sides:
[tex](\frac{ME}{z_{\alpha/2}})^2=\frac{\hat p(1-\hat p)}{n}[/tex]
And solving for n we got:
[tex]n=\frac{\hat p(1-\hat p)}{(\frac{ME}{z_{\alpha/2}})^2}[/tex]
Now we can replpace the values
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739[/tex]
And rounded up to the nearest integer we got:
n=423
If a 25 in. by 32 in. window has a wooden frame 2 inches wide surrounding it, what is the total area of the window and frame?
1044 inches
1036 sq. inches
580 inches
1044 sq. inches
Answer:
1044 sq. inches
Step-by-step explanation:
The frame adds 2 inches on each side of the window, so the width becomes 29 in. and the length becomes 36 in.
Area = width × length
So the total area of the window and the frame is:
A = 29 × 36 = 1044 in²
Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of saturated fat for a sample of 6 brands of stick margarine (solid fat) and for a sample of 6 brands of liquid margarine and obtained the following results: Exam Image Exam Image We want to determine if there a significant difference in the average amount of saturated fat in solid and liquid fats. What is the test statistic? (assume the population data is normally distributed)
Answer:
t = 34.548
Step-by-step explanation:
Here X₁ = Solid values / n
= 26.3 +25.3+26.1+25.6+26.7+25.9/6
=155.9 / 6
=25.984
X₂ = Liquid values/n
= 16.9 +16.6+16.5+17.4+17.4+17.2 / 6
=102/6 = 17
Formula for the sample standard deviation is
[tex]s = \sqrt {Σ(x - mean )^2/n-1}[/tex]
we get
s₁ = 0.49967
s₂ = 0.3949
Construction of hypothesis
H₀ :μ₁ = μ₂
H₁ : μ₁ ≠ μ₂
Apply test statistic formula we get the value
[tex]t = X_{1} - X_{2} /\sqrt{S_{1}^2 /n + S_{2}^2 /n }[/tex]
putting all these values
t = 25.984 - 17 / √ (o.49967)²/6 + (0.39497)²/6
t = 34.548
To determine the test statistic for the difference in the average amount of saturated fat in solid and liquid fats, a two-sample t-test would be performed. Without the specific data, the precise value cannot be calculated. The formula for the test statistic in a two-sample t-test is (mean1 - mean2) / sqrt[(sd1^2/n1) + (sd2^2/n2)].
Explanation:To answer your question, we want to perform a two-sample t-test to determine if there is a significant difference in the average amount of saturated fat in solid and liquid fats. However, without the specific data of saturated fat percentage in both types of fats, we can't calculate the exact test statistic. The test statistic in a two-sample t-test is calculated using the difference between the two sample means divided by the pooled standard error of the means. The formula is as follows:
Test statistic (t) = (mean1 - mean2) / sqrt[(sd1^2/n1) + (sd2^2/n2)]
Where 'mean1' and 'mean2' are the sample means, 'sd1' and 'sd2' are the sample standard deviations, and 'n1' and 'n2' are the sample sizes of the two sets of data respectively.
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Choose the graph that represents the following system of inequalities:
y ≤ −3x + 1
y ≤ x + 3
In each graph, the area for f(x) is shaded and labeled A, the area for g(x) is shaded and labeled B, and the area where they have shading in common is labeled AB.
Correct graph is C
Step-by-step explanation:
Given two inequalities are:
1. [tex]y\leq -3x+1[/tex]
2.[tex]y\leq x+3[/tex]
Step1 : Remove the inequalities
1. [tex]y =-3x+1[/tex]
2.[tex]y = x+3[/tex]
Step2 : Finding intersection points of equations
By solving linear equation
[tex]y =-3x+1 =x+3 [/tex]
[tex]-3x+1 =x+3 [/tex]
[tex] x = -0.5[/tex]
Replacing value of x in any equations
we get,
[tex]y = x+3[/tex]
[tex]y =-0.5+3[/tex]
[tex]y = 2.5[/tex]
Therefore, Point of intersection is (-0.5,2.5)
Step3: Test of origin (0,0)
Here, If inequalities holds true for origin then, shades the graph towards the origin.
For equation 1.
[tex]y\leq -3x+1[/tex]
[tex]0\leq -3(0)+1[/tex]
[tex]0\leq +1[/tex]
True, Shade graph towards origin.
For equation 2.
[tex]y\leq x+3[/tex]
[tex]0\leq 0+3[/tex]
[tex]0\leq 3[/tex]
True, Shade graph towards origin.
Thus, Correct graph is C
Answer:
c
Step-by-step explanation:
took test
You want to estimate the mean time college students spend watching online videos each day. The estimate must be within 2 minutes of the population mean. Determine the required sample size to construct a 99% confidence interval for the population mean. Assume the population standard deviation is 2.4 minutes.
Answer: 10
Step-by-step explanation:
We know that the formula to find the sample size is given by :-
[tex]n= (\dfrac{z_{\alpha/2\cdot \sigma}}{E})^2[/tex]
, where [tex]\sigma[/tex] = population standard deviation.
[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)
E= margin of error.
Given : Confidence level : C =99%=0.99
i.e. [tex]1-\alpha=0.99[/tex]
⇒Significance level :[tex]\alpha=1-0.99=0.01[/tex]
By using z-value table ,Two -tailed z-value for [tex]\alpha=0.01 [/tex]:
[tex]z_{\alpha/2}=2.576[/tex]
E= 2 minutes
[tex]\sigma=\text{2.4 minutes}[/tex]
The required sample size will be :-
[tex]n= (\dfrac{2.576\cdot 2.4}{2})^2\\\\= (3.0912)^2\\\\=9.55551744\approx10[/tex]
Hence, the required sample size = 10
To estimate the mean time college students spend watching online videos each day within 2 minutes of the population mean at a 99% confidence level, at least 10 students should be sampled assuming a population standard deviation of 2.4 minutes.
To determine the required sample size, we use the formula for the sample size of a mean:
n = (z*σ/E)^2
Where n is the sample size, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and E is the maximum error allowed (margin of error).
For a 99% confidence interval, the z-score (from the standard normal distribution) is approximately 2.576. The population standard deviation σ is given as 2.4 minutes, and the margin of error E is 2 minutes. Using these values:
n = (2.576*2.4/2)^2
n ≈ (6.1824/2)^2
n ≈ (3.0912)^2
n ≈ 9.554
Since the sample size must be a whole number, we would round up to ensure the sample size is large enough to achieve the desired margin of error, which means at least 10 students should be surveyed to construct the 99% confidence interval for the mean time spent watching online videos.
It is important to note that the larger the sample size, the more accurate the estimate of the population mean will be.
A random sample of 200 books purchased at a local bookstore showed that 72 of the books were murder mysteries. Let 푝be the true proportion of books sold by this store that is murder mystery. Construct a confidence interval with a 95% degree of confidence
Answer:
[tex](0.2935, 0.4265)[/tex]
Step-by-step explanation:
Given that a random sample of 200 books purchased at a local bookstore showed that 72 of the books were murder mysteries.
Sample proportion p follows the following characteristics
p 0.36
q 0.64
pq/n 0.001152
Std dev 0.033941125
For 95% confidence level we have critical value as
Z critical = 1.96
Hence margin of error = 1.96*std dev
Confidence interval = (sample proportion - margin of error, sample proportion + margin of error)
=[tex](0.2935, 0.4265)[/tex]
To construct the 95% confidence interval for the proportion of murder mystery books sold at a bookstore, we use the sample proportion, Z-score for 95% confidence, and sample size to find the margin of error and then add and subtract this from the sample proportion. The 95% confidence interval is approximately (0.2935, 0.4265).
To construct a 95% confidence interval for the true proportion of books sold by the store that are murder mysteries, we can use the formula for a confidence interval for a proportion, which is [tex]p \pm Z*(\sqrt{\frac{p(1-p)}{n}})[/tex]
Sample proportion (p) = 72÷200 = 0.36
Sample size (n) = 200
For a 95% confidence interval, we use the Z-score corresponding to 95% confidence level from the standard normal distribution table, which is approximately 1.96.
The margin of error (E) is calculated as follows:
[tex]E = 1.96 * \sqrt{\frac{0.36(1-0.36)}{200}}\\E = 1.96 * \sqrt{0.001152}[/tex]
E = 1.96 * 0.033941
E ≈ 0.0665
Now we can construct the confidence interval:
The lower boundary is p - E = 0.36 - 0.0665 ≈ 0.2935
The upper boundary is p + E = 0.36 + 0.0665 ≈ 0.4265
Therefore, the 95% confidence interval is (0.2935, 0.4265).
A CAT scan of a human pancreas shows cross-sections spaced 1 cm apart. The pancreas is 12 cm long and the cross-sectional areas, in square centimeters, are 0, 7.9, 15.3, 18.1, 10.2, 10.7, 9.5, 8.5, 7.8, 5.6, 4.2, 2.7, and 0. Use the Midpoint Rule to estimate the volume of the pancreas.
Final answer:
The volume of the pancreas is estimated using the Midpoint Rule by adding the cross-sectional areas and multiplying by the distance between sections, resulting in an estimated volume of 100.5 cubic centimeters.
Explanation:
The question asks us to use the Midpoint Rule to estimate the volume of the human pancreas given its length (12 cm) and cross-sectional areas at 1 cm intervals. To do this, we add up the areas of the cross-sections (excluding the first and last since they're 0) and multiply by the distance between the sections (1 cm). This approach approximates the volume using cylindrical segments where each segment's volume is its cross-sectional area times the height (1 cm).
The given cross-sectional areas are 7.9, 15.3, 18.1, 10.2, 10.7, 9.5, 8.5, 7.8, 5.6, 4.2, and 2.7 square cm. Adding these gives a total of 100.5 square cm. Since the distance between each section is 1 cm, multiplying 100.5 by 1 cm gives an estimated pancreatic volume of 100.5 cubic centimeters.
This method, while not perfectly accurate due to it being an approximation, provides a useful estimate of the volume based on the available data. It's a practical application of the Midpoint Rule in estimating volumes from cross-sectional data.