Answer:
The 7th percentile is 34.2
Step-by-step explanation:
Problems of normally distributed samples are 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 = 46, \sigma = 8[/tex]
Find the 7 th percentile.
The value of X when Z has a pvalue of 0.07. So it is X when Z = -1.475.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.475 = \frac{X - 46}{8}[/tex]
[tex]X - 46 = -1.475*8[/tex]
[tex]X = 34.2[/tex]
The 7th percentile is 34.2
Manny bought 12 pounds of vegetables at the supermarket. 75% of the vegetables were on sale. How many pounds of vegetables were not on sale?
Answer:
3 pounds of vegetables bought were not on sale.
Step-by-step explanation:
Given:
Total pounds of vegetable bought by Manny = 12
Percentage of vegetables that are on sale = 75%
To find the pounds of vegetables that were not on sale.
Solution:
If the percentage of of vegetables that are on sale is 75%.
Then, the percentage of of vegetables that were not on sale is = [tex]100\%-75\%[/tex] = 25%
Total pounds of vegetable bought is 12 pounds, of which 25% were not on sale.
So, the pounds of vegetables that were not on sale is:
⇒ [tex]25\%\ of\ 12[/tex]
⇒ [tex]0.25\times 12[/tex]
⇒ 3
Thus, 3 pounds of vegetables bought were not on sale.
Determine if the statement is true or false, and justify your answer. If u4 is not a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly independent. True. If {u1, u2, u3, u4} is linearly dependent, then u4
Answer:The statement is true
Step-by-step explanation:
This is an implication, then if we by starting from the premise we can get to the conclusion, then the implication is true. First let us remember that a given set of n vectors, say [tex]\{u_1,\ldots, u_n\}[/tex] is l.i (linearly independent), by definition if the only solution to [tex]\alpha_1\cdot u_1 + \cdots + \alpha_n\cdot u_n = \textbf{0}, is (\alpha_1, \ldots, \alpha_n)= \textbf{0}\in \mathcal {R}^n, \mathcal {R}[/tex] the set of real numbers, that is the only linear combination equal to the null vector is the null combination, all the scalars must be zero. And since it is a definition it is if and only if.
Then if we say that the premise is true, then u4 is not a linear combination of {u1,u2,u3}, this means there do not exist [tex]\alpha_1, \alpha_2, \alpha_3 \in \mathcal{R}[/tex] not all zero, such that, [tex]\alpha_1\cdot u_1+ \alpha_2\cdot u_2+ \alpha_3\cdot u_3=u_4[/tex] which is equivalent to say that for every [tex]\alpha_1, \alpha_2, \alpha_3,\alpha_4 \in \mathcal{R}[/tex] with
[tex]\alpha_1\cdot u_1 + \alpha_2\cdot u_2+ \alpha_3\cdot u_3+ \alpha_4\cdot u_4 = \textbf{0}[/tex] implies [tex]\alpha_1= \alpha_2= \alpha_3= \alpha_4=0[/tex], then the given set is linearly independent as by definition and just as it is stated in the conclusion, therefore the affirmation is true.
what are the coordinates of point A
Answer:
-5,7 those are the coordinates on the graph
Answer:
-5,7 this should be the answer
A closed box with a square bottom is three times high as it is wide. a) Express the surface area of the box in terms of its width. b) Express the volume of the box in terms of its width. c) Express the surface area in terms of the volume. d) If the box has a volume of 24 m³, what is its surface area?
Answer:
a) [tex]S(s) = 14s^2[/tex]
b) [tex]V(s) = 3s^3[/tex]
c) [tex]S(s) = \dfrac{14V(s)}{3s}[/tex]
d) 56 square meter
Step-by-step explanation:
We are given the following in the question:
A closed box with a square bottom is three times high as it is wide.
Let s be the side of square and h be the height.
[tex]h = 3s[/tex]
a) Surface area of box
[tex]2(lb + bh + hl)[/tex]
where l is the length, b is the breadth and h is the height.
Putting values:
[tex]S = 2(s^2 + sh +sh)\\S = 2(s^2 + 3s^2 + 3s^2)\\S(s) = 14s^2[/tex]
b) Volume of box
[tex]l\times b \times h[/tex]
where l is the length, b is the breadth and h is the height.
Putting values:
[tex]V = s\times s\times h\\V= s\times s\times 3s\\V(s) = 3s^3[/tex]
c) Surface area in terms of volume
[tex]S(s) = 14s^2 = \dfrac{14V(s)}{3s}[/tex]
d) Surface area
Volume = 24 m³
[tex]V(s) = 24\\3s^3 = 24\\s^3 = 3\\s = 2[/tex]
[tex]S(2) = 14(2)^2 = 56\text{ square meter}[/tex]
Following is the data of new car color preferences for U.S. buyers. New Car Color Preferences for U.S. Buyers Color Percent Blue 12 Green 7 Natural 12 Red 13 Silver/Grey 24 White 16 Black 13 Other 3 Total 100
Answer:
The chart A is correct
Pareto Chart
Step-by-step explanation:
Given chart is missing (Attached)
Find:
- Which chart represents the correct data.
- What other chart can be used to express the given data
Solution:
- Use the given values for each color and compare with the three charts A,B and C given.
For Blue = A (12) , B(12) , C(11)
For Green = A(7) , B(13) , C(12)
- Hence, The chart A is correct.
- Any other chart which can correctly express the information given should be a chart that uses bars or frequency to expresses the percentages. Pareto Chart expresses both bars and line chart(curve) to express the frequency of the data.
You draw one card at random from a standard deck of 52 playing cards. Find the probability that (a) the card is an even-numbered card, (b) the card is a heart or a diamond, and (c) the card is a nine or a face card.?
Answer:
[tex]P(even )=\frac{20}{52} =\frac{5}{13}[/tex]
[tex]P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}[/tex]
[tex]P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}[/tex]
Step-by-step explanation:
You draw one card at random from a standard deck of 52 playing cards
Total cards = 52
5 even numbers in each suit. 4 times 5 = 20 even number cards
[tex]P(even )=\frac{20}{52} =\frac{5}{13}[/tex]
(b) the card is a heart or a diamond
there are 13 heart and 13 diamonds
[tex]P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}[/tex]
(c) the card is a nine or a face card
There are 4 nine and 12 face cards
[tex]P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}[/tex]
The random variable is the number of nonconforming solder connections on a printed circuit board with 970 connections. Determine the range (possible values) of the random variable.
Answer: X = {0,1,2,3,4,5,....970}
Explanation: Let X → random nonconforming solder connections on a printed circuit board. The number has to be a whole number to indicate the connections. This gives all possible range of X = {0,1,2,3,4,5,....970}
The number of solder connections would always be discrete, hence, a positive integer value. The range of possible values which the variable could have are {1, 2, 3,..., 970}
Number of solder connections would always be a whole number, as it cannot be negative or be a fractional value, hence, the possible range of values will be discrete.
Hence, the possible range can be expressed in the form :
X = {1, 2, 3,..., 970}Learn more : https://brainly.com/question/17225268
Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year.
Answer:
[tex]n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689[/tex]
So the answer for this case would be n=689 rounded up to the nearest integer
Step-by-step explanation:
Assuming this complete question: "Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.
Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year."
Solution for the problem
First we need ti find the estimation for the standard deviation using the Rule of thumb, with the following formula:
[tex] s \approx \frac{R}{4}[/tex]
Where R is the range defined as :
[tex] R = Max - Min = 8-0 = 8[/tex]
So then the deviation would be approximately:
[tex] s \approx 4[/tex]
Important 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 (ME) 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".
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
And on this case we have that [tex]ME =\pm 0.25[/tex] and we are interested in order to find the value of n, if we solve n from equation (1) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (2)
We can assume that the estimator for the population deviation from the rule of thumb is [tex]\hat \sigma = s= 4[/tex]
The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.64[/tex], replacing into formula (2) we got:
[tex]n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689[/tex]
So the answer for this case would be n=689 rounded up to the nearest integer
To estimate textbook ages, an assumed range of 1 to 20 years might give an estimated standard deviation of 4.75 years. Using the formula for sample size, about 275 textbooks would need to be sampled to estimate the mean age to within 0.25 years with 90% confidence.
Explanation:To estimate the minimum and maximum ages for typical textbooks currently used in college courses, we need more specific information. However, if we assume a range of 1 to 20 years, this would give us a range of 19 years. We can use the range rule of thumb for estimating the standard deviation, which is the range divided by 4. So the estimated standard deviation of the textbook ages is 19 / 4 = 4.75 years.
To find the required sample size to estimate the mean age of textbooks, we use the equation n = (Z*σ/E)^2 where Z is the z-score corresponding to the desired confidence level, σ is the standard deviation, and E is the maximum tolerable error. For a 90% confidence level, the z-score is 1.645. Thus with σ = 4.75 and E = 0.25. Filling these values into our equation we get: n = (1.645*4.75/0.25)^2 = 274.68. Rounded up, we need a sample of size 275 textbooks to estimate the mean age to within 0.25 years with 90% confidence.
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A system of identifying call letters uses call letters that begin either with E or W. Some have a total of 2 letters, and others have 3 letters. How many different call letters are possible?
Answer:
If repetition is allowed then the different call letters possible are 1404.
If repetition is allowed then the different call letters possible are 1250.
Step-by-step explanation:
The call letter are of either 2 letter or 3 letters.
And for them to be identified they should start with either E or W.
If Repetition is allowed:Consider the call letters with 2 letter:
Possible number of call letters starting with E : E _
The second place can be filled any of the 26 letters.
Possible number of call letters starting with W : W _
The second place can be filled any of the 26 letters.
Consider the call letters with 3 letter:
Possible number of call letters starting with E : E _ _
The second place and third place can be filled with any of the 26 letters.
Possible number of call letters starting with W : W _ _
The second place and third place can be filled with any of the 26 letters.
Total number of possible letters: 26 + 26 + (26)² + (26)² = 1404
Thus, if repetition is allowed then the different call letters possible are 1404.
If Repetition is not allowed:Consider the call letters with 2 letter:
Possible number of call letters starting with E : E _
The second place can be filled any of the 25 letters.
Possible number of call letters starting with W : W _
The second place can be filled any of the 25 letters.
Consider the call letters with 3 letter:
Possible number of call letters starting with E : E _ _
The second place can be filled with any of the remaining 25 letters and the third can be filled with the remaining 24 letters.
Possible number of call letters starting with W : W _ _
The second place can be filled with any of the remaining 25 letters and the third can be filled with the remaining 24 letters.
Total number of possible letters: 25 + 25+ (25 × 24) + (26 × 24) = 1250
Thus, if repetition is allowed then the different call letters possible are 1250.
Marine biologists have been studying the effects of acidification of the oceans on weights of male baluga whales in the Arctic Ocean. One of the studies involves a random sample of 16 baluga whales. The researchers want to create a 95% confidence interval to estimate the true mean weight of male baluga whales. Their data follow a normal distribution. The population standard deviation of weights of male baluga whales is LaTeX: \sigma = 125????=125 kg, and the researchers feel comfortable using this standard deviation for their confidence interval.
Use this information to answer Questions.
. Assuming the relevant requirements are met, calculate the margin of error in estimating the true mean weight of male baluga whales in the Artic Ocean.
15.31 kg
51.40 kg
61.25 kg
80.49 kg
Assuming the relevant requirements are met, what sample size would be required if the researchers wanted the margin of error to be 45 kg?
25
30
35
40
Are the requirements for the use of a confidence interval met? Explain.
Yes. The distribution of sample means is normal because the data are normal.
Yes. The distribution of sample means is normal because the sample size is large.
No. The distribution of sample means is not normal because the sample size is small.
No. The fact that the data are normal does not imply that the distribution of sample means is normal.
Answer: a) margin of error = 61.25, b) sample size when margin of error is 45 = 30
Step-by-step explanation:
The formulae to get the margin of error of a confidence interval is given as
Margin of error = critical value * (σ/√n)
Where σ = population standard deviation = 125
n = sample size = 16
Critical value =Zα/2 = 1.96 ( this is so because we are performing a 95% confidence level test then level of significance (α) will be 5% and since our test is of two values, it will be 2 tailed).
Margin of error = 1.96 * (125/√16)
Margin of error = 1.96 * 125/4
Margin of error = 1.96 * 31.25
Margin of error = 61.25
Question b)
Margin of error = 45
Critical value =Zα/2 = 1.96
Population standard deviation = σ = 125
Sample size =n =??
By recalling the formulae
Margin of error = critical value * (σ/√n)
45 = 1.96 * (125/√n)
45 = (1.96 * 125)/√n
45 = 245/√n
45 * √n = 245
√n = 245/ 45
√n = 5.444
n = (5.444)²
n= 29.64 which is approximately 30.
Final answer:
The margin of error calculation for estimating the mean weight of male baluga whales, determining the necessary sample size, and confirming the requirements for a confidence interval.
Explanation:
The margin of error in estimating the true mean weight of male baluga whales in the Arctic Ocean can be calculated using the formula:
Margin of Error = Z * (Standard Deviation / √sample size)
Given Z value for 95% confidence interval is 1.96,
Margin of Error = 1.96 * (125 / √16) = 61.25 kg.
To find the required sample size for a margin of error of 45 kg:
Set up the formula: Margin of Error = Z * (Standard Deviation / √sample size)Plug in the values: 45 = 1.96 * (125 / √sample size)Solve for the sample size: sample size ≈ 35The requirements for the use of a confidence interval are considered met when the distribution of sample means is normal.
In this case, the distribution of sample means is normal because the sample size is large, ensuring the validity of utilizing a confidence interval for estimation.
A statistician selected a sample of 16 accounts receivable and determined the mean of the sample to be $5,000 with a standard deviation of $400. He reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. He neglected to report what confidence level he had used. Based on the above information, determine the confidence level that was used. Assume the population has a normal distribution.
Answer:
The confidence level that was used is 0.25% .
Step-by-step explanation:
We are given that mean of the selected sample of 16 accounts is $5,000 and a standard deviation of $400.
It has also been reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. This represents the Confidence Interval for population mean.
But we have to find that at what confidence level this information about range of population men has been stated.
Since we know that Confidence Interval for population mean is given by :
C.I. for population mean = Sample mean(xbar) [tex]\pm[/tex] z value * [tex]\frac{Standard deviation}{\sqrt{n} }[/tex]
i.e.,if we have 95% C.I. = xbar [tex]\pm[/tex] 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] .
So, our Confidence Interval for population is written as :
[$4,739.80 , $5,260.20] = $5000 [tex]\pm[/tex] z value * [tex]\frac{400}{\sqrt{16} }[/tex]
$5000 - z value * 100 = $4739.80 { Solving these we get Z value = 2.602}
$5000 + z value * 100 = $5260.20
In z table we find that at value of 2.60 the probability is 0.99534 so subtracting this from 1 we get confidence level for one tail i.e.0.5%(approx).
Therefore, for two tail Confidence level will be 0.5%/2 = 0.25% .
Using the t-distribution, it is found that a confidence level of 98% was used.
We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.
The information given is:
Sample mean of [tex]\overline{x} = 5000[/tex]. Sample standard deviation of [tex]s = 400[/tex]. Sample size of [tex]n = 16[/tex].The margin of error is of:
[tex]M = t\frac{s}{\sqrt{n}}[/tex]
In which t is the critical value.
For this problem, the margin of error is:
[tex]M = \frac{5260.2 - 4739.8}{2} = 260.2[/tex]
Hence, the critical value is found solving the equation of the margin of error for t.
[tex]M = t\frac{s}{\sqrt{n}}[/tex]
[tex]260.2 = t\frac{400}{\sqrt{16}}[/tex]
[tex]100t = 260.2[/tex]
[tex]t = \frac{260.2}{100}[/tex]
[tex]t = 2.602[/tex]
Looking at the t-table, with 16 - 1 = 15 df, t = 2.602 is associated with a confidence level of 98%.
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The following sample data are from a normal population: 10, 9, 12, 14, 13, 11, 6, 5.
a. What is the point estimate of the population mean?
b. What is the point estimate of the population standard deviation (to 2 decimals)?
c. With 95% confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)?
d. What is the 95% confidence interval for the population mean (to 1 decimal)?
The point estimate of the population mean is 10, the point estimate for the population standard deviation is ~3.03. The margin of error for 95% confidence of the estimation of the mean is ~2.4. Therefore, the 95% confidence interval for the population mean is (7.6, 12.4).
Explanation:To answer your questions, we begin by calculating basic measures of central tendency and variability.
a. The point estimate of the population mean is the average value of the data. It's calculated by adding up all the values and dividing by the number of values. In this case, (10+9+12+14+13+11+6+5)/8 = 10.
b. The point estimate of the population standard deviation (sd) is the root mean square deviation of the values from the mean. It's calculated by determining the square root of the variance, but let's use a calculator to get sd ≈ 3.03.
c. The margin of error for the estimation of the population mean at a 95% confidence level is calculated using the standard deviation and the size of the sample (for a t-distribution). This gives: MoE = t * (sd/sqrt(n)) ~ 2.3 * (3.03/sqrt(8)) = 2.44 (1 decimal place).
d. Finally, the 95% confidence interval for the population mean is determined by adding and subtracting the margin of error from the point estimate of the mean. Therefore, the 95% confidence interval for the mean is (10 - 2.4 , 10 + 2.4) = (7.6, 12.4).
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a. Point Estimate of Population Mean
Firstly, a point estimate for the population mean is the average of the sample data. For the given data (10, 9, 12, 14, 13, 11, 6, 5), the point estimate is calculated by adding all the values and dividing by the number of observations:
(10 + 9 + 12 + 14 + 13 + 11 + 6 + 5) / 8 = 10
b. Point Estimate of Population Standard Deviation
To find the standard deviation, we use the formula for the sample standard deviation:
s =√[tex](\sum (x_i - \bar{x})^2[/tex] / (n - 1))
After calculating using the given data, we find that the point estimate of the population standard deviation, rounded to two decimal places, is s = 3.32.
c. Margin of Error for Mean Estimation
With the margin of error, we create the confidence interval by subtracting and adding the margin of error from the point estimate of the mean. This gives us the range of values in which we are 95% confident that the true population mean lies. The standard error of the sample mean is 3.24 / √8 ≈ 1.146,
so the margin of error at 95% confidence is 1.96 * 1.146 ≈ 2.24 (rounded to 1 decimal).
d. 95% Confidence Interval for Population Mean
The 95% confidence interval for the population mean is 10.375 ± 2.24, which gives us a range of (8.13, 12.62) (rounded to 1 decimal).
In the 1939 movie The Wizard of Oz, upon being presented with a Th.D. (Doctor of Thinkology), the Scarecrow proudly exclaims, "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side." Did the Scarecrow get the Pythagorean Theorem right? In particular, describe four errors in the Scarecrow's statement.
Answer:
The Scarecrow got it wrong.
Step-by-step explanation:
1. The Pythagoras theorem does not apply to every isosceles triangle. It only applies to a triangle that has one of its sides as a right angle i.e. 90 degrees.
2. The Pythagoras theorem deals with squares of sides not square roots of sides.
3. Pythagoras theorem does not state that the square of any side is equal to the sum of the remaining two. Instead, it states that the square of the hypotenuse of a right angled triangle is equal to the sum of the square of the remaining two sides.
4. The Pythagoras theorem does not mention anything about an isosceles triangle. It deals only with Right angled triangles.
The Scarecrow's statement has errors in terms of the type of triangle it refers to, the mathematical operation used, which sides of the triangle he's considering, and the side with which the sum is equated.
Explanation:The Scarecrow in the 1939 film The Wizard of Oz did not correctly state the Pythagorean Theorem. There are a few errors in his statement:
Firstly, the Pythagorean theorem applies to right triangles, not isosceles triangles.Secondly, the theorem talks about the squares of the sides of the triangle, not the square roots.Thirdly, it's about the sum of the squares of the two shorter sides, not any two sides.Lastly, this sum equals the square of the longest side (hypotenuse), not its square root.The correct statement of the Pythagorean Theorem is: 'In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.'
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A patio was to be laid in a design with one tile in the
first row, two tiles in the second row, three
tiles in the third row, and so on. Mr. Tong had 60
tiles to use. How many tiles should be placed
in the bottom row to use the most tiles?
Answer:
10
Step-by-step explanation:
The number of tiles in the design is 1 + 2 + 3 + ...
We can model this as an arithmetic series, where the first term is 1 and the common difference is 1. The sum of the first n terms of an arithmetic series is:
S = n/2 (2a₁ + d (n − 1))
Given that a₁ = 1 and d = 1:
S = n/2 (2(1) + n − 1)
S = n/2 (n + 1)
Since S ≤ 60:
n/2 (n + 1) ≤ 60
n (n + 1) ≤ 120
n must be an integer, so from trial and error:
n ≤ 10
Mr. Tong should use 10 tiles in the final row to use the most tiles possible.
In each figure a || b. Determine whether the third figure is a translation image of the first figure. Write yes or no. Explain your answer.
Answer:
No
Step-by-step explanation:
The second and third figures are translations of each other. The first figure appears to be a reflection of the second, not a translation. Hence the first and third figures are not translations of each other.
__
If an image is translated, all of the pre-image line segments are parallel to the image line segments. That is not the case for the first and third figures, in which the top line segments go in different directions.
Answer:
b. No; it is a reflection followed by a translation.
Step-by-step explanation:
Nine cars (3 Pontiacs [labeled 1-3], 4 Fords [labeled 4-7], and 2 Chevrolet's [labeled 8-9]) are divided into 3 groups for car racing. each group consistes of 3 cars, and they are allocated to 3 tracks (1-3), respectively.
1. How many different ways can you arrange the 9 cars?
Answer:
The number of different ways to arrange the 9 cars is 362,880.
Step-by-step explanation:
There are a total of 9 cars.
These 9 cars are to divided among 3 racing groups.
The condition applied is that there should be 3 cars in each group.
Use permutation to determine the total number of arrangements of the cars.
For group 1:There are 9 cars and 3 to be allotted to group 1.
This can happen in [tex]^9P_{3}[/tex] ways.
That is, [tex]^9P_{3}=\frac{9!}{(9-3)!} =504[/tex] ways.
For group 2:There are remaining 6 cars and 3 to be allotted to group 2.
This can happen in [tex]^6P_{3}[/tex] ways.
That is, [tex]^6P_{3}=\frac{6!}{(6-3)!} =120[/tex] ways.
For group 3:There are remaining 3 cars and 3 to be allotted to group 3.
This can happen in [tex]^3P_{3}[/tex] ways.
That is, [tex]^3P_{3}=\frac{3!}{(3-3)!} =6[/tex] ways.
The total number of ways to arrange the 9 cars is: [tex]^9P_{3}\times ^6P_{3}\times ^3P_{3}=504\times120\times6=362880[/tex]
Thus, the number of different ways to arrange the 9 cars is 362,880.
A baker blends 600 raisins and 400 chocolate chips into a dough mix and, from this, makes 500 cookies. (a) Find the probability that a randomly picked cookie will have no raisins. (b) Find the probability that a randomly picked cookie will have exactly two chocolate chips. (c) Find the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
Answer:
a) P(X = 0) for raisins = 0.33287
b) P(X = 2) for chocolate chips = 0.14379
c) P(X ≥ 2) for both bits = 0.59399
Step-by-step explanation:
The average amount of raisin per cookie is 600/500 = 1.2
The average amount of chocolate chips per cookie = 400/500 = 0.8
a) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For raisin, Mean = λ = 1.1
x = 0
P(X = 0) = (e⁻¹•¹)(1.1⁰)/(0!) = 0.33287
b) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For chocolate chips, Mean = λ = 0.8
x = 2
P(X = 2) = (e⁻⁰•⁸)(0.8²)/(2!) = 0.14379
c) the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
For this, the average number of bit in a raisin = (600+400)/500 = 2 bits per raisin.
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
For P(X = 0)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 0
P(X = 0) = (e⁻²)(2⁰)/(0!) = 0.13534
For P(X = 1)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 1
P(X = 1) = (e⁻²)(2¹)/(1!) = 0.27067
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
P(X ≥ 2) = 1 - (0.13534 + 0.27067) = 1 - 0.40601 = 0.59399
Using the Poisson distribution, it is found that there is a:
a) 0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.
b) 0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.
c) 0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
x is the number of successes e = 2.71828 is the Euler number [tex]\mu[/tex] is the mean in the given interval.Item a:
600 raisins in 500 cookies, hence, the mean is:
[tex]\mu = \frac{600}{500} = 1.2[/tex]
The probability is P(X = 0), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.2}1.2^{0}}{(0)!} = 0.3012[/tex]
0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.
Item b:
400 chips in 500 cookies, hence, the mean is:
[tex]\mu = \frac{400}{500} = 0.8[/tex]
The probability is P(X = 2), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-0.8}0.8^{2}}{(2)!} = 0.1438[/tex]
0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.
Item c:
1000 bits, as 600 + 400 = 1000, in 500 cookies, hence, the mean is:
[tex]\mu = \frac{1000}{500} = 2[/tex]
The probability is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which:
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}2^{0}}{(0)!} = 0.1353[/tex]
[tex]P(X = 1) = \frac{e^{-2}2^{1}}{(1)!} = 0.2707[/tex]
Hence:
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1353 + 0.2707 = 0.4060[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.406 = 0.594[/tex]
0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
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Suppose George loses 66% of all staring contests. (a) What is the probability that George loses two staring contests in a row? (b) What is the probability that George loses four staring contests in a row? Assume that each game played is independent of the rest.
Answer:
a) 43.56% probability that George loses two staring contests in a row.
b) 18.97% probability that George loses four staring contests in a row.
Step-by-step explanation:
In each staring contest, George has a 66% probability of losing.
(a) What is the probability that George loses two staring contests in a row?
[tex]P = (0.66)^{2} = 0.4356[/tex]
43.56% probability that George loses two staring contests in a row.
(b) What is the probability that George loses four staring contests in a row? Assume that each game played is independent of the rest.
[tex]P = (0.66)^{4} = 0.1897[/tex]
18.97% probability that George loses four staring contests in a row.
The probability of George losing two staring contests in a row is approximately 43.56%, while the probability of him losing four staring contests in a row is approximately 19.7%.
Explanation:To calculate the probability that George loses two staring contests in a row, we need to multiply the probability of losing one staring contest by itself. Since George loses 66% of all staring contests, the probability of losing one contest is 0.66. Therefore, the probability of losing two contests in a row is 0.66 multiplied by 0.66, which is approximately 0.4356 or 43.56%.
To calculate the probability that George loses four staring contests in a row, we again need to multiply the probability of losing one contest by itself four times. So, the probability is 0.66 raised to the power of 4, which is approximately 0.197 or 19.7%.
Suppose you find $20. If you choose to use the $20 to go to the football game, your opportunity cost of going to the game is a. nothing, because you found the money. b. $20 (because you could have used the $20 to buy other things). c. $20 (because you could have used the $20 to buy other things) plus the value of your time spent at the game. d. $20 (because you could have used the $20 to buy other things) plus the value of your time spent at the game, plus the cost of the dinner you purchased at the game.
the property of society getting the most from its scarce recources ... Fairness might require that everyone get an equal share because they were ... If you choose to use the $20 to go to the football game, your opportunity cost of going to the game is. $20 because you could have used that money on other things plus the value
The opportunity cost of going to the football game is $20 because you could have used that money to buy other things.
Explanation:The correct answer to this question is b. $20 (because you could have used the $20 to buy other things).
Opportunity cost is the value of the next best alternative that you give up when making a decision. In this case, if you choose to use the $20 to go to the football game, you are giving up the opportunity to buy other things with that money.
There are no additional costs like the value of your time spent at the game or the cost of dinner because they are not explicitly mentioned in the question.
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"Suppose the scenario that the standard devation of semiannual changes in the price of wheat is $0.79, the standdard devation of changes in the futures contract on wheat over the same tim eperiod is $0.93, and the correlation coeffieient relating the assest and futures contract is $0.86. What is the optimal headge ratio for the six month contract on wheat
Answer:
The optimal Hedge Ratio is 0.7305.
Step-by-step explanation:
Optimal Hedge ratio is given as
[tex]HR_{optimal}=\epsilon_{correlation} \times \frac{\sigma_{current}}{\sigma_{future}}[/tex]
Here
HR_optimal is the Hedge Ratio for the next 6 months which is to be calculated.ε_correlation is the correlation coefficient relating the assets and futures contract whose value is give as $0.86.σ_current is the standard deviation of the semiannual changes of the wheat which is given as $0.79σ_future is the standard deviation of the changes in the future over the same time period which is given as $0.93So the Hedge Ratio is given as
[tex]HR_{optimal}=\epsilon_{correlation} \times \frac{\sigma_{current}}{\sigma_{future}}\\HR_{optimal}=0.86 \times \frac{0.79}{0.93}\\HR_{optimal}= \$0.7305[/tex]
So the optimal Hedge Ratio is 0.7305.
In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is infected with the virus is 0.4, independently of other computers. A technician tests the computers, one after another, to see if they are infected.1. What is the probability that she has to test at least 5 computers to find the first (if any) defective one?2. Find the probability that on this day at least 5 computers are infected.
Answer:
(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.
(2) The probability at least 5 computers are infected is 0.949.
Step-by-step explanation:
The probability that a computer is defective is, p = 0.40.
(1)
Let X = number of computers to be tested before the 1st defect is found.
Then the random variable [tex]X\sim Geo(p)[/tex].
The probability function of a Geometric distribution for k failures before the 1st success is:
[tex]P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...[/tex]
Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:
P (X ≥ 5) = 1 - P (X < 5)
= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]
[tex]=1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078[/tex]
Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.
(2)
Let Y = number of computers infected.
The number of computers in the company is, n = 20.
Then the random variable [tex]Y\sim Bin(20,0.40)[/tex].
The probability function of a binomial distribution is:
[tex]P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...[/tex]
Compute the probability at least 5 computers are infected as follows:
P (Y ≥ 5) = 1 - P (Y < 5)
= 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)] [tex]=1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904[/tex]
Thus, the probability at least 5 computers are infected is 0.949.
Suppose that W1 is a random variable with mean μ and variance σ21 and W2 is a random variable with mean μ and variance σ2. From Example 5.4.3, we know that cW1 + (1 − c)W2 is an unbiased estimator of μ for any constant c > 0. If W1 and W2 are independent, for what value of c is the estimator cW1 + (1 − c)W2 most efficient?
Answer:
Step-by-step explanation:
The concept of variance in random variable is applied in solving for the value of c for the estimator cW1 + (1 − c)W2 to be most efficient. Appropriate differentiation of the estimator with respect to c will give the value of c when the result is at minimum.
The detailed analysis and step by step approach is as shown in the attachment.
Construct a 90% confidence interval for (P1-P2) in each of the following situations. a. n1-400, p1-0.67; n2-400, p2 = 0.56. b. n1 = 180; p1 = 0.31; nz" 250, p2 = 0.25. c. n1 = 100; p1 = 0.46; n2 = 120, pz" 0.61.
Answer:
a) [tex](0.67-0.56) - 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.054[/tex]
[tex](0.67-0.56) + 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.166[/tex]
And the 90% confidence interval would be given (0.054;0.166).
b) [tex](0.31-0.25) - 1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=-0.0122[/tex]
[tex](0.31-0.25) +1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=0.132[/tex]
And the 90% confidence interval would be given (-0.0122;0.132).
c) [tex](0.46-0.61) - 1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.260[/tex]
[tex](0.46-0.61) +1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.0404[/tex]
And the 90% confidence interval would be given (-0.260;-0.0404).
Step-by-step explanation:
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".
Part a
[tex]p_1[/tex] represent the real population proportion for sample 1
[tex]\hat p_1 =0.67[/tex] represent the estimated proportion for sample 1
[tex]n_A=400[/tex] is the sample size required for sample 1
[tex]p_2[/tex] represent the real population proportion for sample 2
[tex]\hat p_2 =0.56[/tex] represent the estimated proportion for sample 2
[tex]n_2=400[/tex] is the sample size required for sample 2
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]
For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.64[/tex]
And replacing into the confidence interval formula we got:
[tex](0.67-0.56) - 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.054[/tex]
[tex](0.67-0.56) + 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.166[/tex]
And the 90% confidence interval would be given (0.054;0.166).
Part b
[tex](0.31-0.25) - 1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=-0.0122[/tex]
[tex](0.31-0.25) +1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=0.132[/tex]
And the 90% confidence interval would be given (-0.0122;0.132).
Part c
[tex](0.46-0.61) - 1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.260[/tex]
[tex](0.46-0.61) +1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.0404[/tex]
And the 90% confidence interval would be given (-0.260;-0.0404).
Find the volume of a cube with side length of 7 in.
A.) 147 in3
B.) 343 in3
C.) 49 in3
D.) 215 in3
Answer:
343 in³ - B
Step-by-step explanation:
Volume of cube = l³ = 7 in x 7 in x 7 in = 343 in³
Find the mean, median, and mode of the following data. If necessary, round to one more decimal place than the largest number of decimal places given in the data.MLB Batting Averages0.3270.2950.3180.3100.2850.2800.3140.3230.3100.2950.2830.2790.2770.3130.3270.3060.3270.3170.2920.275
Answer:
Mean = 0.3026
Median = 0.308
Mode = 0.327
Step-by-step explanation:
We are given the following data set:
0.327, 0.295, 0.318, 0.310, 0.285, 0.280, 0.314, 0.323, 0.310, 0.295, 0.283, 0.279, 0.277, 0.313, 0.327, 0.306, 0.327, 0.317, 0.292, 0.275
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{6.053}{20} = 0.3026[/tex]
Sorted data: 0.275, 0.277, 0.279, 0.280, 0.283, 0.285, 0.292, 0.295, 0.295, 0.306, 0.310, 0.310, 0.313, 0.314, 0.317, 0.318, 0.323, 0.327, 0.327, 0.327
[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]
[tex]\text{Median} = \dfrac{10^{th} + 11^{th}}{2} = \dfrac{0.306 + 0.310}{2} = 0.308[/tex]
Mode is the mos frequent observation in the data.
Mode = 0.327
It appeared 3 times.
The mean of the MLB Batting Averages is around 0.306, the median is 0.308, and the mode is 0.327.
Explanation:In the field of mathematics, especially in statistics, mean, median and mode are measures of central tendency that provide an overview of the data set. Given the MLB Batting Averages, we start by arranging the data in ascending order. After that, we calculate the mean by adding up all the values and dividing by the number of values. The median is the middle value in an ordered dataset, and the mode is the value that appears most often in a dataset.
Mean: (0.327+0.295+0.318+0.310+0.285+0.280+0.314+0.323+0.310+0.295+0.283+0.279+0.277+0.313+0.327+0.306+0.327+0.317+0.292+0.275)/20 ≈ 0.306 Median: The middle numbers are 0.310 and 0.306, so the median is (0.310 + 0.306) / 2 = 0.308 Mode: The number 0.327 appears three times, more than any other number, so the mode is 0.327. Learn more about Central Tendency here:
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if the equation of m is given by y= ax + c, which of the following represents the perpendicular slope to m?
A. - 1/a
B. -a
C. 1/a
D. c
Answer: A. - 1/a
Step-by-step explanation:
Two lines are said to be perpendicular if the product of their slope equals -1 . That is , if the slope of the first line is [tex]m_{1}[/tex] and the slope of the second line is [tex]m_{2}[/tex] , if they are perpendicular , then :
[tex]m_{1}[/tex] x [tex]m_{2}[/tex] = -1.
Therefore : the perpendicular slope to m is [tex]\frac{-1}{a}[/tex]
A large tank holds 400 gallons of salt water. A salt water solution with concentration 2 lb/gal is being pumped into the tank at a rate of 3 gal/min. The tank is simultaneously being emptied at a rate of 1 gal/min. If 100 pounds of salt was dissolved in the tank initially, how much salt is in the tank after 250 minutes?
Final answer:
The question involves calculating how much salt remains in a tank after 250 minutes when saltwater of a certain concentration is added and the solution is simultaneously drained at a different rate. A basic calculation suggests an addition of 1500 pounds to the initial 100 pounds, but the real answer must consider dilution and the tank's capacity, which complicates the calculation.
Explanation:
The question involves calculating the amount of salt in a tank after 250 minutes when saltwater is being pumped in and out at different rates. Initially, the tank has 100 pounds of salt and is then filled with a solution at a rate of 3 gallons per minute, with a concentration of 2 pounds of salt per gallon. Meanwhile, the tank is emptied at a rate of 1 gallon per minute.
To find the amount of salt in the tank after 250 minutes, we need to consider both the inflow and outflow of the tank. The net flow rate is 2 gallons per minute (3 gallons in - 1 gallon out). This leads to an increase in the volume of the tank by 2 gallons for every minute.
Every minute, 6 pounds of salt (3 gallons x 2 pounds/gallon) are added to the tank. However, since the tank's volume increases by 2 gallons per minute but 1 gallon is removed, we should adjust for the concentrate reduction due to dilution. Over 250 minutes, the total added salt is 250 minutes x 6 pounds/minute = 1500 pounds. However, the calculation must be adjusted to consider the dilution effect which is beyond the scope of this immediate calculation.
Initially, the tank contains 100 pounds of salt. Without accounting for dilution or maximum capacity effects, a simple addition suggests that after 250 minutes, 1600 pounds of salt would be in the tank. Yet, this simplistic approach does not account for dilution effect properly or the fact the tank has a maximum capacity, indicating that a more complex calculation is required for precise results.
1. How do you multiply powers with the same base?
2. How do you divide powers with the same base?
3. How do you find the power of a power?
4. Simplify. (x^2*x^4)^3
____________
X^8
Answer:
[tex]\frac{(x^2\times x^4)^3}{x^8}[/tex] [tex]=x^{10}[/tex]
Step-by-step explanation:
1.
Power of the same base multiply by adding their exponent.
Example: [tex]a^m\times a^n = a^{(m+n)}[/tex]
2.
Power of the same base divide by subtracting their exponent.
Example:[tex]a^m \div a^n = a^{(m-n)}[/tex]
3.
Find power of a power we may multiple the exponent.
Example:[tex](a^m)^n = a^{mn}[/tex]
4.
[tex]\frac{(x^2\times x^4)^3}{x^8}[/tex]
[tex]=\frac{(x^6)^3}{x^8}[/tex] [ using multiplication rule]
[tex]=\frac{(x^{18})}{x^8}[/tex] [ using power of power rule]
[tex]=x^{18-8}[/tex] [ Using division rule]
[tex]=x^{10}[/tex]
You are considering buying a piece of industrial equipment to automate a part of your production process. This automation will save labor costs by as much as $35,000 per year over 10 years. The equipment costs $200,000. Should you purchase the equipment if your interest rate is 12
Answer:
No.
Step-by-step explanation:
The equipment wll save $35,000 per over for 10 years, which totals to $350,000.
If the equipment is bought on a simple interest rate of 12% annually for ten years, it will cost:
[tex]A = P (1 + rt),\\where\ A =\ Final\ amount,\ r=rate\of\interest\annually,\ t=time,\ P= Principal\ value\\\\P = 200,000r = 0.12t = 10 \\\\A = 200,000 ( 1 + 0.12\times 10)\\A = 200,000 (1 + 1.2)\\A = 200,000 (2.2)\\A = 440,000[/tex]
We will need to pay $440,000 in total for the machine in over ten years.
If we compare both values, it can be deduced that industrial equipment is more expensive than labor cost.
The decision to purchase the equipment depends on the calculation of Net Present Value, which factors in the cost of the equipment, annual saving, and interest rate. If the calculated NPV is greater than zero, it's worth investing, otherwise, it's not.
Explanation:The subject of this question falls under the category of Business Finance, specifically the concept of Net Present Value (NPV). Given the costs of the equipment, the annual saving, and the interest rate (assumed to be 12%), we can calculate NPV. The net savings account for an annuity, which we assume to be $35,000 annually, and we end up with an NPV of cells (= 35000*(1-(1+0.12)^-10)/0.12 -200000). This value helps determine whether the investment is worthwhile. If the NPV is greater than 0, you should purchase the equipment as it indicates that the project's return exceeds the cost. However, if the NPV is less than 0, you should not purchase the equipment as it indicates that the project's cost exceeds the expected return.
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A technology committee of wants to perform a test to see if the mean amount of time students are spending in a lab has increased from 54 minutes. Here are the data from a random sample of 12 students.
52 74
57 54
55 137
76 74
63 7
53 63
Are there any? outliers? Select the correct choice below and fill in any answer boxes within your choice.?Yes, 2
b) What do you conclude about the? claim? If there are? outliers, perform the test with and without the outliers present
Write appropriate hypotheses for the test.
H0: u=54
HA: u=>54
The Test Statistic is t= _________
The P- Value is ___________
If there are outliers , calculate the test statistic for the data set without outliers.
The test statistic is t=______________
The P-Value is ____________
To identify outliers from the data, the interquartile range (IQR) is used. Hypotheses are established to test if the mean time spent in the lab has increased. The test statistic and P-value are calculated with and without outliers to reach a conclusion.
Explanation:To determine if there are any outliers in the provided data, we can use the interquartile range (IQR) method. Calculating IQR and then finding the lower and upper bounds will help us identify outliers. An outlier is a value that falls below Q1 - 1.5 * IQR or above Q3 + 1.5 * IQR. The hypotheses for our test are as follows:
H0: \\( \\mu = 54 \\) minutesHA: \\( \\mu > 54 \\) minutesThe test statistic and P-value are calculated using the sample mean, standard deviation, and sample size, and comparing them with the presumed population mean of 54 minutes.
If outliers are identified and removed, we would calculate a new test statistic and P-value for the data set without the outliers to see if there is a significant change.