Answer:
[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]
So then the value of 205 it's 0.76 deviations below the population mean on this case
Step-by-step explanation:
For this case we have the following data given:
177, 205, 210, 210, 232, 205, 185, 185, 178, 210, 206, 212, 184, 174, 185, 242, 188, 212, 215, 247, 241, 223, 220, 260, 245, 259, 278, 270, 280, 295, 275, 285, 290, 272, 273, 280, 285, 286, 200, 215, 185, 230, 250, 241, 190, 260, 250, 302, 265, 290, 276, 228, 265
And these values represent the weigths of all the team members of the Arizona Cardinals
Now if we organize the data values from the smallest to the largest we have:
174 177 178 184 185 185 185 185 188 190 200 205 205 206 210 210 210 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
For this case we can calculate the mean with the following formula:
[tex] \mu = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And if we replace we got:
[tex] \mu = 233.3396[/tex]
And for the deviation we can use the following formula:
[tex] \sigma =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And if we replace we got:
[tex] \sigma = 37.498[/tex]
And in order to calculate How many standard deviations above or below the mean was he we can use the z score formula given by:
[tex] z = \frac{x -\mu}{\sigma}[/tex]
And we assume that x=205 and if we replace we have:
[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]
So then the value of 205 it's 0.76 deviations below the population mean on this case
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]
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).
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
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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An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is a means of determining the age of certain wood and plant remains, and hence of animal or human bones or artifacts found buried at the same levels. Radiocarbon dating is based on the fact that some wood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. This isotope is accumulated during the lifetime of the plant and begins to decay at its death. Since the half-life of carbon-14 is long (approximately 5730 years),4 measurable amounts of carbon-14 remain after many thousands of years. If even a tiny fraction of the original amount of carbon-14 is still present, then by appropriate laboratory measurements the proportion of the original amount of carbon-14 that remains can be accurately determined. In other words, if Q(t) is the amount of carbon-14 at time t and Q0 is the original amount, then the ratio Q(t)/Q0 can be determined, as long as this quantity is not too small. Present measurement techniques permit the use of this method for time periods of 50,000 years or more.
(a) Assuming that Q satisfies the differential equation Q' = -rQ, determine the decay constant r for carbon-14. (b) Find an expression for Q(t) at any time t, if Q(0) = Qo. (c) Suppose that certain remains are discovered in which the current residual amount of carbon-14 is 20% of the original amount. Determine the age of these remains.
Answer:
a) r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121/year
b) Q(t) = Q₀ (e^-rt)
c) Are of the 20% remnant of Carbon-14 = 13301.14 years.
Step-by-step explanation:
Q' = -rQ
Q' = dQ/dt
dQ/dt = -rQ
dQ/Q = -rdt
Integrating the left hand side from Q₀ to Q₀/2 and the right hand side from 0 to t1/2 (half life, t1/2 = 5730 years)
In ((Q₀/2)/Q₀) = -r(t1/2)
In (1/2) = -r(t1/2)
In 2 = r(t1/2)
r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121 /year
b) Q' = -rQ
Q' = dQ/dt
dQ/dt = -rQ
dQ/Q = -rdt
Integrating the left hand side from Q₀ to Q(t) and the right hand side from 0 to t.
In (Q(t)/Q₀) = -rt
Q(t)/Q₀ = e^(-rt)
Q(t) = Q₀ (e^-rt)
c) Q(t) = Q₀ (e^-rt)
Q(t) = 0.2Q₀, t = ? and r = 0.000121/year
0.2Q₀ = Q₀ (e^-rt)
0.2 = e^-rt
In 0.2 = -rt
-1.6094 = - 0.000121 × t
t = 1.6094/0.000121 = 13301.14 years.
Hope this Helps!
Radiocarbon dating is a method used in archeological research to determine the age of artifacts based on the ratio of carbon-14 to the original amount. This technique is limited to time periods of 50,000 years or more.
Explanation:Radiocarbon dating is an important tool in archeological research that allows scientists to determine the age of wood, plant remains, and other artifacts. This method is based on the fact that some wood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. By measuring the ratio of carbon-14 to the original amount, scientists can accurately determine the age of these remains. Radiocarbon dating is limited to time periods of 50,000 years or more.
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A fire company keeps two rescue vehicles. Because of the demand on the vehicles and the chance of mechanical failure, the probability that a specific vehicle is available when needed is 90%. The availability of one vehicle is independent of the availability of the other. Find the probability that (a) both vehicles are available at a given time, (b) neither vehicle is available at a given time, and (c) at least one vehicle is available at a given time.
Answer:
(a) P (Both vehicles are available at a given time) = 0.81
(b) P (Neither vehicles are available at a given time) = 0.01
(c) P (At least one vehicle is available at a given time) = 0.99
Step-by-step explanation:
Let A = Vehicle 1 is available when needed and B = Vehicle 2 is available when needed.
Given:
The availability of one vehicle is independent of the availability of the other, i.e. P (A ∩ B) = P (A) × P (B)
P (A) = P (B) = 0.90
(a)
Compute the probability that both vehicles are available at a given time as follows:
P (Both vehicles are available) = P (Vehicle 1 is available) ×
P (Vehicle 2 is available)
[tex]P(A\cap B)=P(A)\times P(B)[/tex]
[tex]=0.90\times0.90\\=0.81[/tex]
Thus, the probability that both vehicles are available at a given time is 0.81.
(b)
Compute the probability that neither vehicles are available at a given time as follows:
P (Neither vehicles are available) = [1 - P (Vehicle 1 is available)] ×
[1 - P (Vehicle 2 is available)]
[tex]P(A^{c}\cap B^{c})=[1-P(A)]\times [1-P(B)]\\[/tex]
[tex]=(1-0.90)\times (1-0.90)\\=0.10\times0.10\\=0.01[/tex]
Thus, the probability that neither vehicles are available at a given time is 0.01.
(c)
Compute the probability that at least one vehicle is available at a given time as follows:
P (At least one vehicle is available) = 1 - P (None of the vehicles are available)
[tex]=1-[P(A^{c})\times P(B^{c})]\\=1-0.01.....(from\ part\ (b))\\ =0.99[/tex]
Thus, the probability that at least one vehicle is available at a given time is 0.99.
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.
Suppose that the probability of a defective part is 0.03. Suppose that you have a shipment of 1000 parts. What is the probability that more than 10 parts will be defective? Answer to at least five decimal places. You may find it easier to use excel than to use a calculator for this one.
Answer:
The probability that more than 10 parts will be defective is 0.99989.
Step-by-step explanation:
Let X = a part in the shipment is defective.
The probability of a defective part is, P (Defect) = p = 0.03.
The size of the sample is: n = 1000.
Thus, the random variable [tex]X\sim Bin(1000, 0.03)[/tex].
But the sample size is very large.
The binomial distribution can be approximated by the Normal distribution if the following conditions are satisfied:
np ≥ 10n (1 - p) ≥ 10Check the conditions:
[tex]np=1000\times0.03=30>10\\n(1-p)=1000\times(1-0.03)=970>10[/tex]
Thus, the binomial distribution can be approximated by the Normal distribution.
The sample proportion (p) follows a normal distribution.
Mean: [tex]\mu_{p}=0.03[/tex]
Standard deviation: [tex]\sigma_{p}=\sqrt{\frac{p(1-p)}{n} } =\sqrt{\frac{0.03(1-0.03)}{1000} } =0.0054[/tex]
Compute the probability that there will be more than 10 defective parts in this shipment as follows:
The proportion of 10 defectives in 1000 parts is: [tex]p=\frac{10}{1000}=0.01[/tex]
The probability is:
[tex]P(p>0.01)=P(\frac{p-\mu_{p}}{\sigma_{p}}> \frac{0.01-0.03}{0.0054}) =P(Z>-3.704)=P(Z<3.704)[/tex]
Use the standard normal table for the probability.
[tex]P(p>0.01)=P(Z<3.704)=0.99989[/tex]
Thus, the probability that more than 10 parts will be defective is 0.99989.
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.
Suppose two events A and B are two independent events with P(A) > P(B) and P(A ∪ B) = 0.626 and P(A ∩ B) = 0.144, determine the values of P(A) and P(B).
Answer:
P(A)= 0.606 and P(B)= 0.237
Step-by-step explanation:
Since A and B are independent
P(A ∩ B) = P(A) * P(B) → P(B) = P(A ∩ B) / P(A)
and also
P(A ∪ B)= P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = P(A) + P(A ∩ B) / P(A) - P(A ∩ B)
[P(A ∪ B) + P(A ∩ B) ]* P(A) = P(A)² + P(A ∩ B)
P(A)² - [P(A ∪ B) + P(A ∩ B) ]* P(A) + P(A ∩ B) = 0
P(A)² - [ 0.626+0.144] * P(A) + 0.144 =0
P(A)² - 0.77* P(A) + 0.144 =0
thus
P(A)₁= 0.606 or P(A)₂= 0.1647
for P(A)₁→ P(B)₁ = P(A ∩ B) / P(A)₁ = 0.144/0.606 = 0.237
thus P(A)₁ > P(B)₁ → correct
for P(A)₂→ P(B)₂ = P(A ∩ B) / P(A)₂ = 0.144/0.1647= 0.8743
thus P(A)₂ < P(B)₂ → incorrect
therefore
P(A)= 0.606 and P(B)= 0.237
Newlyweds Bryce & lauren need to rent a truck to move their belongings to their new apartment. They can rent a truck of the size they need from U-Haul for $29.95 per day plus 28 cents per mile or from Budget Truck rentals for $34.95 per day plus 25 cents per mile. After how many miles (to the nearest mile) would the budget rental be a better deal than the U-Haul one?
Answer:
After 167 miles the Budget rental deal would be better than the U-haul deal
Step-by-step explanation:
Let x be the number of miles.
Considering the U-haul plan the cost can be expressed as:
[tex]u=29.95+0.28x[/tex]
Consider the Budget rental plan the cost can be expressed as:
[tex]b=34.95+0.25x[/tex]
For the Budget rental plan to be better than the U-haul plan the relation between the two cost equation will be:
[tex]b<u[/tex]
Substitute the equations of u and b and solve for x:
[tex]b<u\\34.95+0.25x<29.95+0.28x\\34.95-29.95<0.28x-0.25x\\5<0.03x\\0.03x>5\\x>166.67\approx167[/tex]
So after 167 miles the Budget rental deal would be better than the U-haul deal.
After approximately 167 miles, Budget Truck rental becomes a cheaper option than U-Haul for Bryce and Lauren's move.
Explanation:To find the point at which Budget Truck rentals becomes cheaper than U-Haul, we must set up the cost equation for each company and solve for the number of miles (m) that makes them equal. For U-Haul, the cost (C) is given by C = 29.95 + 0.28m, and for Budget it’s C = 34.95 + 0.25m.
Setting these two equations equal, we get: 29.95 + 0.28m = 34.95 + 0.25m.
We then solve for m: 0.03m = 5. Subtracting 29.95 from both sides, then dividing by 0.03 gives: m ≈ 167 miles.
So, after 167 miles, Budget Truck becomes the cheaper option.
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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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Customers arrive at a checkout counter at times 1,4,5,10,20,22,23,28,29,35. The time it takes fortheir checkout are the amounts 4,4,4,4,3,3,5,1,5. Compute for each customer the wait time fromwhen the customer arrives until the customer begins to get served. Use the formula given in classand in Dai&Park’s text.
Answer:
wait time in order = 0,1,4,3,0,1,3,3,3,2
Step-by-step explanation:
We need to make a table for easy computation of wait time for each customer.
Customer Arrival Time taken Checkout Wait
Serial no. time for checkout time time
1 1 4 1+4= 5 0
2 4 4 4+5= 9 5-4= 1
3 5 4 9+4= 13 9-5= 4
4 10 4 13+4= 17 13-10=3
5 20 3 20+3= 23 0
6 22 3 23+3= 26 23-22=1
7 23 5 26+5= 31 26-23=3
8 28 1 31+1= 32 31-28=3
9 29 5 32+5= 37 32-29=3
10 35 37-35=2
A triangle has measures that are 45 45 90 The hypothenuse of the triangle is 10 What is the perimeter of the triangle
Answer: you need to add each side .45+45+90=180
Step-by-step explanation:
Determine whether the statement is true or false. Justify your answer.
If A and B are independent events with nonzero probabilities, then A can occur when B occurs.
Answer:Yes (A can occur when B occurs)
Step-by-step explanation: Independent events are events that are not determined by the occurrence of another. In this case EVENT A CAN OCCUR AS EVENT B OCCURS.
Justification: Landing on tails after tossing a coin and landing on a Five (5) after rolling the DIE Landing on Six (6) after rolling a DIE and landing on tails after tossing a coin.
Independent probabilities can occur together,they don't depend on each other.
7. Martin sells cars. He earns $100 per day, plus
any commission on his sales. His daily salary s in
dollars depends on the amount of commission c.
Write an equation to represent his daily salary.
Answer:the equation representing his daily salary is
s = 100 + c
Step-by-step explanation:
Let s represent the Salary that Martins earns per day.
Let c represent the commission that Martins earns on his sales on that day.
He earns $100 per day, plus
any commission on his sales. Since
his daily salary in dollars depends on the amount of commission, then an equation to represent his daily salary would be
s = 100 + c
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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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%.
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]
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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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.
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.
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.
A slice of pizza whose edges form a 32degrees angle with an outer crust edge 4 inches long was found in a gym locker. What was the diameter of the original pizza?g
Answer:
The diameter was [tex]d=\frac{45}{\pi }[/tex] in.
Step-by-step explanation:
The arc of a circle is given by
[tex]s=r\theta[/tex]
where
s = arc length
r = radius of the circle
θ = measure of the central angle in radians.
From the information given
s = 4 in
θ = 32º
To find the diameter of the original pizza, we use the formula of the diameter of a circle
[tex]d=2r[/tex]
First, we need to convert the angle to radians
[tex]\theta=32\º \cdot \frac{\pi}{180\º} =\frac{8\pi }{45}[/tex]
Next, solve for r from the arc formula
[tex]r=\frac{s}{\theta} =\frac{4}{\frac{8\pi }{45}} =\frac{45}{2\pi }[/tex]
Then, we use the diameter of a circle formula
[tex]d=2r=2(\frac{45}{2\pi })=\frac{45}{\pi }[/tex]
3.20 × 105 gallons of tar (SG = 1.20) is stored in a 20.0-ft tall storage tank. What is the total mass of the liquid in the tank?
Answer:
Mass of the liquid in the tank will be [tex]1.45\times 10^8kg[/tex]
Step-by-step explanation:
We have given volume of tar [tex]V=3.20\times 10^5gallon[/tex]
1 gallon = 3.785 liter
So [tex]3.20\times 10^5gallon=3.20\times 10^{5}\times 3.785=12.112\times 10^5liter[/tex]
Specific gravity = 1.2
Density of water [tex]=1000kg/m^3[/tex]
We know that specific gravity [tex]=\frac{density\ of\ liquid}{density\ of\ water}[/tex]
[tex]1.2=\frac{density\ of\ liquid}{density\ of\ water}[/tex]
Density of liquid [tex]=1200kg/m^3[/tex]
So mass of liquid = volume × density
= [tex]12.112\times 10^5\times 1200=1.45\times 10^8kg[/tex]
So mass of the liquid in the tank will be [tex]1.45\times 10^8kg[/tex]
POLYGONS AND CIRCLES
PLEASE HELP
Answer:
Step-by-step explanation:
The formula for finding the sum of the measure of the interior angles in a regular polygon is expressed as (n - 2) × 180.
Where
n represents the number of sides of the polygon.
5a) The polygon has 11 sides. Therefore, sum of angles is
(11 - 2) × 180 = 1620
The measure of each angle is
1620/11 = 148.3°
b) n = 24
Therefore, sum of angles is
(24 - 2) × 180 = 3960
The measure of each angle is
3960/24 = 165°
5a) n = 7
Therefore, sum of angles is
(7 - 2) × 180 = 900
The measure of each angle is
900/7 = 128.6°
5b) n = 10
Therefore, sum of angles is
(10 - 2) × 180 = 1440
The measure of each angle is
1440/10 = 144°
Determine if the following is consistent Subscript[x, 1] - 2 Subscript[x, 2] + Subscript[x, 3] = 0 Subscript[ , ] 2Subscript[x, 2] - 8Subscript[x, 3] = 8 5Subscript[x, 1] - 5Subscript[x, 3] = 10
Answer:
The systems of equation is CONSISTENT
Step-by-step explanation:
The detailed steps using crammers rule to ascertain the CONSISTENCY is as shown in the attached file
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).
Roll two dice, one white and one red. Consider these events: A : The sum is 7 B : The white die is odd C : The red die has a larger number showing than the white D : The dice match (doubles) Which pair(s) of events are disjoint (events A and B are disjoint if A ∩ B = ∅ )? Which pair(s) are independent? Which pair(s) are neither disjoint nor independent?
Final answer:
Events B and D are disjoint; events B and C, A and C, B and D are independent; events A and B, A and C, and B and D are neither disjoint nor independent.
Explanation:
The pairs of events that are disjoint (have no intersection) are:
Events B and D are disjoint because if the white die is odd (event B), then it cannot match with the red die (event D).
Events A and D are disjoint because if the sum of the dice is 7 (event A), then the dice cannot match (event D).
The pairs of events that are independent are:
Events B and C are independent because the outcome of one die does not affect the outcome of the other die.
Events A and C are independent for the same reason.
The pairs of events that are neither disjoint nor independent are:
Events A and B are neither disjoint nor independent. If the white die is odd (event B), it is still possible for the sum of the dice to be 7 (event A).
Events A and C are also neither disjoint nor independent. If the sum of the dice is 7 (event A), it is still possible for the red die to have a larger number showing than the white die (event C).
Events B and D are neither disjoint nor independent. If the white die is odd (event B), it is still possible for the dice to match (event D).
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