Step-by-step explanation:
-1/4, -1/5, -1/6, -1/7, -1/8, 1/8, 1/7, 1/6, 1/5, 1/4
Determine which of the following variables would be best modeled as continuous random variable.
A; The number of movies watched by a person in one year
B; The number of newborn babies delivered in a hospital on a certain day
C; The distance between two cars on the freeway.
D; The height of a skyscraper in New York City.
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The variables that would be best modeled as continuous random variables are C and D.
C; The distance between two cars on the freeway can take on any real value within a certain range. It is a continuous variable because it can be measured and divided into infinitely many possible values.
D; The height of a skyscraper in New York City is also a continuous variable. The height can vary continuously from very short to very tall, and it can be measured and divided into infinitely many possible values.
A and B, on the other hand, would be better modeled as discrete random variables.
A; The number of movies watched by a person in one year is a discrete variable because it can only take on whole numbers. You can't watch a fraction of a movie.
B; The number of newborn babies delivered in a hospital on a certain day is also a discrete variable. The number of newborn babies is counted in whole numbers and cannot take on fractional values.
Therefore, variables C and D are best modeled as continuous random variables, while variables A and B are better modeled as discrete random variables.
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Forty percent of cars travelling on I-90 are speeding
(X). If five are
selected at random.
The probability that P(1 ≤ X < 4) is closest
to:
Therefore, the probability P(1 ≤ X < 4) is closest to 0.8352.
To calculate the probability P(1 ≤ X < 4), where X represents the number of cars out of five selected at random that are speeding, we need to consider the possible outcomes and their probabilities.
Since 40% of cars are speeding, the probability of a car being speeding is 0.40, and the probability of a car not speeding is 1 - 0.40 = 0.60.
Now we can calculate the probability for each possible outcome:
[tex]P(X = 0) = (0.60)^5[/tex]
= 0.07776
[tex]P(X = 1) = ^5C_1 * (0.40)^1 * (0.60)^4[/tex]
= 0.2592
[tex]P(X = 2) = ^5C_2 * (0.40)^2 * (0.60)^3[/tex]
= 0.3456
[tex]P(X = 3) = ^5C_3 * (0.40)^3 * (0.60)^2[/tex]
= 0.2304
[tex]P(X = 4) = ^5C_4 * (0.40)^4 * (0.60)^1[/tex]
= 0.0768
[tex]P(X = 5) = (0.40)^5[/tex]
= 0.01024
To find P(1 ≤ X < 4), we sum the probabilities for X = 1, 2, and 3:
P(1 ≤ X < 4) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.2592 + 0.3456 + 0.2304
= 0.8352
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A group of friends wants to go to the amusement park. They have no more than $280 to spend on parking and admission. Parking is $20, and tickets cost $40 per person, including tax. Write and solve an inequality which can be used to determine
x, the number of people who can go to the amusement park.
Answer:
280 ≥ 20 + 40x
Step-by-step explanation:
$280 is the total they can spend. and since parking is $20 it is added to the amount of people that can go x 40. This is because 40 is the amount per person.
pls mark brainliest
Use two of the number cards to complete the ratios so that they are
equivalent.
3,4,6,12,15
? : 1
? : 3
To make the ratios equivalent, we can use the numbers 3 and 6:
3 : 1 is equivalent to 6 : 3
To complete the ratios and make them equivalent, we need to find two numbers from the given set (3, 4, 6, 12, 15) that can be used to replace the question marks.
Let's start with the first ratio: ? : 1
We need to find a number that, when divided by 1, gives an equivalent ratio. Since any number divided by 1 is itself, we can choose any number from the given set for the first ratio. Let's choose 3 for this example. So, the ratio becomes:
3 : 1
Now, let's move on to the second ratio: ? : 3
Similarly, we need to find a number that, when divided by 3, gives an equivalent ratio. Looking at the given set, we see that 6 is divisible by 3. So, the ratio becomes:
6 : 3
Therefore, to make the ratios equivalent, we can use the numbers 3 and 6:
3 : 1 is equivalent to 6 : 3
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A score that is 20 points below the mean corresponds to a
z-score of z=-0.50. What is the population standard deviation?
We are given that z = -0.50 and the corresponding score is 20 points below the mean. We need to determine the population standard deviation.
Let μ be the population mean and σ be the population standard deviation. Then we know that the z-score is given by: z = (x - μ)/σwhere x is the score and μ is the population mean.
Substituting the given values,
we get:-0.50 = (x - μ)/20
Multiplying both sides by 20,
we get:-10 = x - μAdding μ to both sides,
we get:x = μ - 10
Therefore, the score that is 20 points below the mean is μ - 10. Substituting this value in the formula for z-score, we get:-0.50 = (μ - 10 - μ)/σ
Simplifying, we get:
0.50σ = 10σ = 20
Therefore, the population standard deviation is 20.
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Question 8 of 12 ( -/1 1 Two sides and an angle are given. Determine whether a triangle (or two) exist, and if so, solve the triangles 23,23,734 How many triangles exist? Round your answers to the nea
There exists one triangle with the given sides 23, 23, and 734.
For the triangle with the given sides 23, 23 and 734, two sides are equal, and they are greater than the third side.
The following condition is valid for a triangle:
a + b > c (the sum of any two sides of the triangle is greater than the third side). Hence, a triangle exists with the given sides.
To calculate the angles, use the law of cosine:
cos A = (b² + c² - a²) / 2bc and
cos B = (a² + c² - b²) / 2ac
The angles are:
cos A = (23² + 734² - 23²) / 2 × 23 × 734
≈ 0.998
cos B = (23² + 734² - 23²) / 2 × 23 × 734
≈ 0.998
As we know that the sum of the angles of a triangle is 180°, then the third angle C can be found by:
C = 180° - (A + B)
C = 180° - (acos 0.998 + acos 0.998)
C = 4.89°
Hence, one triangle exists with the given sides and the angle C is 4.89°.
Therefore, the answer is, there exists one triangle with the given sides 23, 23, and 734.
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ı need help on this math assıgnment please on rationals
According to the information, we can infer that A. 1: Real, Rational, Integer, Whole, Natural, B. 5.1: Real, Rational, C. √(-142): Non-real, D. [tex]\pi[/tex] (Pi): Irrational, Real, E. 2/3: Rational, Real, F. ∛(-27): Non-real, G. 0.671: Real, Rational, H. 3√7: Irrational, Real, I. 0: Real, Rational, Integer, Whole, Natural, J. -√16: Real, Rational.
What is the correct classification for each number?A. 1: It is a real number because it can be plotted on the number line. It is rational because it can be expressed as a fraction (1/1). It is an integer, whole number, and natural number as well.B. 5.1: It is a real number and rational because it can be expressed as a terminating decimal (5.1 = 51/10).C. √(-142): It is a non-real number because the square root of a negative number is not defined in the real number system.D. π (Pi): It is an irrational number because it cannot be expressed as a finite or repeating decimal. It is a real number.E. 2/3: It is a rational number because it can be expressed as a fraction. It is a real number.F. ∛(-27): It is a non-real number because the cubic root of a negative number is not defined in the real number system.G. 0.671: It is a real number and rational because it can be expressed as a decimal.H. 3√7: It is an irrational number because the cube root of 7 cannot be expressed as a fraction or terminating decimal. It is a real number.I. 0: It is a real number and rational because it can be expressed as a fraction (0/1). It is an integer, whole number, and natural number as well.J. -√16: It is a real number and rational because the square root of 16 is 4.Learn more about numbers in: https://brainly.com/question/24908711
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O find the HCF by prime factorition method 6 18 and 48 b C 36 and 84 d 69 and 75 35 and us 27 and 63 z Date Page.
(a) The HCF of 6, 18, and 48 is 6.
(b) The HCF of 36 and 84 is 12.
(c) The HCF of 69 and 75 is 3.
(d) The HCF of 27 and 63 is 9.
What is the HCF of the numbers?The highest common factor (HCF) using the prime factorization is calculated as follows;
(a) 6, 18, and 48;
Prime factorization of 6 = 2 x 3
Prime factorization of 18 = 2 x 3²
Prime factorization of 48 = 2⁴ x 3
The HCF of the numbers;
HCF = 2 x 3 = 6
(b) 36 and 84:
Prime factorization of 36 = 2² x 3²
Prime factorization of 84 = 2² x 3 x 7
HCF = 2² x 3 = 12
(c) 69 and 75;
Prime factorization of 69 = 3 x 23
Prime factorization of 75 = 3 x 5²
H.C.F = 3.
(d) 27 and 63
Prime factorization of 27 = 3³
Prime factorization of 63 = 3² x 7
H.C.F = 3² = 9.
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In a recent year (365 days), there were 638 murders in a city. Find the mean number of murders per day, then use that result to find the probability that in a single day, there are no murders. Would 0
The given data can be used to find the mean number of murders per day in the city as follows:
Mean number of murders per day =
Total number of murders in the year ÷ Number of days in the year
= 638/365≈1.75
So, the mean number of murders per day in the city is approximately 1.75.
Now, we need to use this result to find the probability that in a single day, there are no murders.
Let X be the number of murders in a single day.
Since we know the mean number of murders per day, we can use the Poisson distribution to find the probability of X = 0.
The Poisson distribution is given by:P(X = k) = (e^(-λ) λ^k) / k!
where λ is the mean number of events in a given interval.
In this case, λ = 1.75 and we want to find P(X = 0).
So, we have:P(X = 0) = (e^(-1.75) 1.75^0) / 0!≈ 0.1733
Therefore, the probability that in a single day, there are no murders is approximately 0.1733 or 17.33%.
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Determine the value of the t test statistic. (Use decimal notation. Give your answer to two decimal places.) The results of a major city's restaurant inspections are available through its online newspaper. Critical food violations are those that put patrons at risk of getting sick and must be immediately corrected by the restaurant. An SRS of n = 300 inspections from more than 10,000 inspections since January 2012 had the sample mean x = 0.90 violations and the sample standard deviation s = 2.35 violations. t= Incorrect Determine the P-value. (Use decimal notation. Give your answer to four decimal places. If you use Table D, give the closest lower boundary.) P-value = Incorrect
Cannot determine t-test statistic and P-value without more information
What Insufficient information to determine t-test statistic?
To determine the value of the t-test statistic, we need to calculate it using the sample mean, sample standard deviation, and sample size. The t-test statistic is calculated as the ratio of the difference between the sample mean and the population mean (assuming no difference) to the standard error of the mean.
Given that the sample mean is x = 0.90 violations, the sample standard deviation is s = 2.35 violations, and the sample size is n = 300 inspections, we can calculate the standard error of the mean (SE) as:
SE = s / √n = 2.35 / √300 ≈ 0.1358
Next, we calculate the t-test statistic using the formula:
t = (x - μ) / SE
Since we don't have the population mean (μ) provided in the question, we cannot determine the exact t-test statistic. It seems that the necessary information is missing to calculate the t-test statistic accurately.
Moving on to the P-value, it cannot be determined without knowing the t-test statistic or the alternative hypothesis being tested. The P-value represents the probability of observing a test statistic as extreme or more extreme than the one obtained, assuming the null hypothesis is true. Without the t-test statistic or the specific hypothesis being tested, we cannot calculate the P-value accurately.
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Integrated circuits from a certain factory pass quality test with probability ,8,p=,8. The outcomes of tests are mutually independent. Use The CTL to estimate the probability of finding at most of 50 acceptable circuits in a batch of 60 .
The estimated probability of finding at most 50 acceptable circuits in a batch of 60 is approximately 0.6591.
What is the estimated probability of obtaining no more than 50 acceptable circuits in a batch of 60, given a pass probability of 0.8 and independent outcomes?To estimate the probability of finding at most 50 acceptable circuits in a batch of 60 from a certain factory, where the probability of passing the quality test is (p = 0.8) and the outcomes of the tests are mutually independent, we can use the Central Limit Theorem (CLT).
The CLT states that for a large enough sample size, the distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution.
Let's denote (X) as the number of acceptable circuits in a batch of 60. Since each circuit passes the test with a probability of 0.8, we can model (X) as a binomial random variable with parameters (n = 60) and (p = 0.8).
To estimate the probability of finding at most 50 acceptable circuits, we can calculate the cumulative probability using the normal approximation to the binomial distribution.
Since the sample size is large [tex](\(n = 60\))[/tex], we can approximate the distribution of (X) as a normal distribution with mean [tex]\(\mu = np = 60 \times 0.8 = 48\)[/tex] and standard deviation [tex]\(\sigma = \sqrt{np(1-p)}[/tex] = [tex]\sqrt{60 \times 0.8 \times 0.2} \approx 4.90\).[/tex]
Now, we want to find the probability of[tex]\(P(X \leq 50)\)[/tex]. We can standardize the value using the z-score:
[tex]\[P(X \leq 50) = P\left(\frac{X - \mu}{\sigma} \leq \frac{50 - 48}{4.90}\right) = P(Z \leq 0.41)\][/tex]
Using the standard normal distribution table or calculator, we can find that [tex]\(P(Z \leq 0.41) \approx 0.6591\).[/tex]
Therefore, the estimated probability of finding at most 50 acceptable circuits in a batch of 60 is approximately 0.6591.
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Suppose you are using α = 0.05 to test the claim that μ = 1620 using a P-value. You are given the sample statistics n-35, X-1590 and σ 82. Find the P-value. State the answer only and no additional work. Make sure to use the tables from the book.
We can conclude that there is significant evidence to support the claim that the population mean µ is not equal to 1620.
A P-value is the probability of getting an outcome as extreme or more extreme than the observed outcome, under the null hypothesis.
Suppose that we want to test the hypothesis that the population mean µ is equal to a specified value µ0. The alternative hypothesis, Ha, is that the population mean µ is not equal to µ0.
We may be interested in testing the hypothesis that µ is greater than µ0, that µ is less than µ0, or that µ is either greater than or less than µ0.
Suppose that you are using α = 0.05 to test the claim that µ = 1620 using a P-value.
You are given the sample statistics n = 35, x = 1590 and σ = 82.
We assume that the population is normally distributed. To find the P-value, we need to find the test statistic z:
z = (x - µ0) / (σ / √n) = (1590 - 1620) / (82 / √35) = - 2.33
The P-value is the area to the left of z = - 2.33 in a standard normal distribution.
Using a standard normal distribution table, we find that the area to the left of z = - 2.33 is 0.0099.
Therefore, the P-value is 0.0099.
Therefore, we can reject the null hypothesis if α > 0.0099.
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For the rectangle shown which equation can be used to find the value of c
A=5
B=x
C=12
Answer: B = 1.9
Step-by-step explanation:
A=5
C=12
B=x
Pythagorean Theorem: 5^2 + x^2 = 12^2
25+x^2=144
x^2=119
[tex]\sqrt{19}[/tex]≈10.9
The compressive strengths of seven concrete blocks, in pounds per square inch, are measured, with the following results 1989, 1993.8, 2074, 2070.5, 2070, 2033.6, 1939.6 Assume these values are a simpl
Compute mean, variance, standard deviation, and range to analyze the compressive strengths of the concrete blocks.
In order to analyze the compressive strengths of the concrete blocks, several statistical measures can be computed. The mean, or average, of the data set can be calculated by summing all the values and dividing by the total number of observations.
The variance, which represents the spread or variability of the data, can be computed by calculating the squared differences between each value and the mean, summing these squared differences, and dividing by the number of observations minus one. The standard deviation can then be obtained by taking the square root of the variance.
Additionally, the range, which indicates the difference between the maximum and minimum values, can be determined. These statistical measures provide insights into the central tendency and variability of the compressive strengths of the concrete blocks.
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The cheetah is the fastest land mammal and is highly specialized to run down prey. The cheetah often exceeds speeds of 60 miles per hour (mph) and is capable of speeds above 72 mph. The accompanying table contains a sample of the top speeds of 35 cheetahs. The sample mean and sample standard deviation of these speeds are 59.53 mph and 4.21 mph, respectively. A histogram of the speeds is bell-shaped Complete parts (a) through (d) below. Click the icon to view the top speeds of cheetahs. a. Is it reasonable to apply the empirical rule to estimate the percentages of observations that lie within one, two, and three standard deviations to either side of the mean? A. It is not reasonable to apply the empirical rule. The data is quantitative, but the value of k takes on values less than 1; therefore, the empirical rule is not appropriate. B. It is reasonable to apply the empirical rule. The data is quantitative and the mean and standard deviation are known; therefore, the empirical rule applies. C. It is not reasonable to apply the empirical rule. The data is quantitative and the histogram of the data is bell-shaped, but this does not imply that the data itself is bell-shaped; therefore, the empirical rule is not appropriate. D. It is reasonable to apply the empirical rule. The data is quantitative and the histogram of the data is bell-shaped; therefore, the empirical rule applies. b. Use the empirical rule to estimate the percentages of observations that lie within one, two, and three standard deviations to either side of the mean. Roughly 68% of observations lie within one standard deviation to either side of the mean. Roughly 95 % of observations lie within two standard deviations to either side of the mean. Roughly 99.7% of observations lie within three standard deviations to either side of the mean. (Type integers or decimals. Do not round.) c. Use the data to obtain the exact percentages of observations that lie within one, two, and three standard deviations to either side of the mean. Using the data.% of observations lie within one standard deviation to either side of the mean, % of observations lie within two standard deviations to either side of the mean, and % of observations lie within three standard deviations to either side of the mean. (Type integers or decimals. Round to one decimal place as needed.)
The exact percentages of observations that lie within one, two, and three standard deviations to either side of the mean are 68.6%, 97.1%, and 100%, respectively.
a. D. It is reasonable to apply the empirical rule. The data is quantitative and the histogram of the data is bell-shaped; therefore, the empirical rule applies.
b. The empirical rule to estimate the percentages of observations that lie within one, two, and three standard deviations to either side of the mean are as follows:68% of observations lie within one standard deviation to either side of the mean. Roughly 95 % of observations lie within two standard deviations to either side of the mean. Roughly 99.7% of observations lie within three standard deviations to either side of the mean.
c. The mean and standard deviation of these speeds are 59.53 mph and 4.21 mph, respectively. Using the data, the exact percentages of observations that lie within one, two, and three standard deviations to either side of the mean can be calculated as follows:% of observations that lie within one standard deviation to either side of the mean = 68.57%% of observations that lie within two standard deviations to either side of the mean = 97.14%% of observations that lie within three standard deviations to either side of the mean = 100%
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3) We are interested to find out the average amount of time a
person
wants to listen to Blake Shelton. Suppose we took a sample of
n = 35
people and found the sample mean to be 32 minutes. If the
popu
To find the critical value for a hypothesis test regarding the average amount of time a person wants to listen to Blake Shelton, we need to know the significance level (α) and whether it's a one-tailed or two-tailed test.
The general process of finding the critical value for a hypothesis test. Determine the significance level (α): This is the predetermined threshold at which you will reject the null hypothesis. Common choices for α are 0.05 (5%) or 0.01 (1%). Determine the degrees of freedom (df): In this case, since you have a sample of n = 35, the degrees of freedom would be n - 1 = 35 - 1 = 34. Determine the tail(s) of the test: Depending on the alternative hypothesis, you may have a one-tailed or two-tailed test. In a one-tailed test, you are interested in deviations in one direction (e.g., average listening time being greater or less than a specific value). In a two-tailed test, you are interested in deviations in either direction (greater or less than a specific value). Look up the critical value: Using the significance level and degrees of freedom, consult a t-distribution table or use statistical software to find the critical value. Be sure to match the tail(s) of the test correctly.
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10 grams of steam at 100 degree celsius is mixed with 50 gn of ice at 0 degree celsius then final temperature is?
To determine the final temperature after mixing 10 grams of steam at 100 degrees Celsius with 50 grams of ice at 0 degrees Celsius, we need to calculate the amount of heat exchanged between the two substances.
First, we need to determine the heat absorbed or released during the phase change of ice to water at 0 degrees Celsius. This can be calculated using the equation:
[tex]\[ Q = m \cdot L \][/tex]
where [tex]\( Q \)[/tex] is the heat absorbed or released, [tex]\( m \)[/tex] is the mass of the substance, and [tex]\( L \)[/tex] is the latent heat of fusion for ice. For water, the latent heat of fusion is approximately 334 J/g.
[tex]\[ Q_{\text{ice}} = 50 \, \text{g} \times 334 \, \text{J/g} = 16700 \, \text{J} \][/tex]
Next, we need to calculate the heat absorbed or released during the temperature change of water from 0 degrees Celsius to the final temperature. This can be calculated using the equation:
[tex]\[ Q = m \cdot C \cdot \Delta T \][/tex]
where [tex]\( Q \)[/tex] is the heat absorbed or released, [tex]\( m \)[/tex] is the mass of the substance, [tex]\( C \)[/tex] is the specific heat capacity of water, and [tex]\( \Delta T \)[/tex] is the change in temperature.
For water, the specific heat capacity is approximately 4.18 J/g°C.
[tex]\[ Q_{\text{water}} = 10 \, \text{g} \times 4.18 \, \text{J/g°C} \times (\text{final temperature} - 0°C) \][/tex]
Since the steam condenses into water, it releases its latent heat of vaporization. The latent heat of vaporization for water is approximately 2260 J/g.
[tex]\[ Q_{\text{vaporization}} = 10 \, \text{g} \times 2260 \, \text{J/g} = 22600 \, \text{J} \][/tex]
The total heat exchanged can be calculated by summing up the heat absorbed or released in each step:
[tex]\[ \text{Total heat exchanged} = Q_{\text{ice}} + Q_{\text{water}} + Q_{\text{vaporization}} \][/tex]
Now, we can set up an energy conservation equation:
[tex]\[ \text{Total heat exchanged} = 0 \quad (\text{since no energy is gained or lost in the system}) \][/tex]
[tex]\[ 16700 \, \text{J} + 10 \, \text{g} \times 4.18 \, \text{J/g°C} \times (\text{final temperature} - 0°C) + 22600 \, \text{J} = 0 \][/tex]
Simplifying the equation:
[tex]\[ 10 \, \text{g} \times 4.18 \, \text{J/g°C} \times (\text{final temperature} - 0°C) = -39300 \, \text{J} \][/tex]
[tex]\[ \text{final temperature} - 0°C = -3930 \, \text{J/°C} / (10 \, \text{g} \times 4.18 \, \text{J/g°C}) \][/tex]
[tex]\[ \text{final temperature} \approx -94°C \][/tex]
The negative value indicates that the final temperature is below 0 degrees Celsius, which means the mixture would still be in a frozen state.
Therefore, the approximate final temperature after mixing 10 grams of steam at 100 degrees Celsius with 50 grams of ice at 0 degrees Celsius is -94 degrees Celsius.
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find the vector =⟨1,2⟩ of length 2 in the direction opposite to =4−13.
Therefore, the vector ⟨1,2⟩ of length 2 in the direction opposite to 4−13 is -4⟨1,2⟩/5.
The vector ⟨1,2⟩ of length 2 in the direction opposite to 4−13 is -4⟨1,2⟩/5.
Firstly, the magnitude of the vector, |v| is given as 2, i.e.|v| = 2
The vector whose direction is to be found is 4−13, i.e. ⟨4,-13⟩.
Let us represent the direction of vector 4−13 as a unit vector.
Step 1: Calculate |4−13|, which is the magnitude of the vector:|4−13|=√{(4)^2 + (-13)^2}=√{16 + 169}=√185
Step 2: Find the unit vector of 4−13 by dividing it with its magnitude: i.e., u⟨4,-13⟩ = 1/√185⟨4,-13⟩
Step 3: Scale the unit vector by multiplying it with the given magnitude 2, and multiplying it with -1 to get the opposite direction of the vector 4−13.
That is, v= -2 u⟨4,-13⟩= -2/√185⟨4,-13⟩
Multiplying both the numerator and the denominator by 2 gives the expression as -4⟨4, -13⟩/5 = ⟨-16/5, 52/5⟩.
Therefore, the vector ⟨1,2⟩ of length 2 in the direction opposite to 4−13 is -4⟨1,2⟩/5.
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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
Rn(x) → 0.]
f(x) = 6 cos(x), a = 5π
1- f(x)= sigma n-0 to infinity
2-Find the associated radius of convergence R.
Given function is f(x) = 6 cos(x), a = 5π. We need to find the Taylor series for f(x) centered at the given value of a.
[Assume that f has a power series expansion. Do not show that Rn(x) → 0.]Solution:First we write the Taylor series formula. It is given byf(x)= ∑n=0∞(fn(a)/n!)(x-a)nThe nth derivative of f(x) = 6 cos(x) is given byf(n)(x) = 6 cos(x + nπ/2)6 cos(x) = 6 cos(5π + (x-5π))Using Taylor series formula, we havef(x)= ∑n=0∞(fⁿ(5π)/n!)(x-5π)n = ∑n=0∞((-1)^n * 6/(2n)!)(x-5π)2n
Now we find the associated radius of convergence R. The formula for radius of convergence is given byR = 1/L, whereL = limn→∞|an|^(1/n)The nth term of the series is given by |an| = 6/(2n)!Therefore, we haveL = limn→∞|an|^(1/n) = limn→∞(6/(2n)!)^(1/n) = 0Therefore, R = 1/L = 1/0 = ∞Hence, the Taylor series for f(x) centered at 5π is ∑n=0∞((-1)^n * 6/(2n)!)(x-5π)2n and its radius of convergence is R = ∞.
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please provide the correct answer with the steps
Same givings in Q3 and Q4
a. The probability that a randomly selected device will be
OK in the
reliability is?
b. The probability that a randomly sel
QUESTION 3 An Engineering professor tests devices to check their reliability and sensitivity. The following table shows the performance of 150 devices. Reliability sensitivity high low OK 70 30 Weak 3
The probabilities are given as follows:
a. Ok in reliability: 2/3 = 0.667.
b. Weak in reliability, given that it has high insensitivity: 3/10 = 0.3.
How to calculate a probability?The parameters that are needed to calculate a probability are listed as follows:
Number of desired outcomes in the context of a problem or experiment.Number of total outcomes in the context of a problem or experiment.Then the probability is calculated as the division of the number of desired outcomes by the number of total outcomes.
There are 150 devices, and of those, 100 are ok in reliability, hence the probability for item a is given as follows:
100/150 = 2/3.
100 of the devices have high insensitivity, and of those, 30 have weak reliability, hence the probability for item b is given as follows:
30/100 = 3/10.
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10 > 3 Suppose that the speed at which cars go on the freeway is normally distributed with mean 77 mph and standard deviation 6 miles per hour. Let X be the speed for a randomly selected car. Round all answers to 4 decimal places where possible. a. What is the distribution of X?X-N b. If one car is randomly chosen, find the probability that it is traveling more than 75 mph. c. If one of the cars is randomly chosen, find the probability that it is traveling between 78 and 83 mph. mph d. 66% of all cars travel at least how fast on the freeway?
The probability that a randomly chosen car is traveling between 78 and 83 mph is P(78 ≤ X ≤ 83) = P(0.1667 ≤ Z ≤ 1.0000).
Suppose the speed of cars on the freeway follows a normal distribution with a mean of 77 mph and a standard deviation of 6 mph. Find: Probability of a randomly chosen car traveling between 78 and 83 mph, d) The minimum speed at which 66% of all cars travel on the freeway.The distribution of X (the speed of a randomly selected car) is a normal distribution, denoted as X ~ N(77, 6).
To find the probability that a randomly chosen car is traveling more than 75 mph, we need to calculate the area under the normal distribution curve to the right of 75 mph. This can be found using the standard normal distribution table or a calculator.Assuming a standard normal distribution (mean = 0, standard deviation = 1), we standardize the value:
Z = (75 - 77) / 6 = -0.3333Using the standard normal distribution table or a calculator, we find the probability corresponding to Z = -0.3333. Let's assume it is P(Z > -0.3333).
The probability that a randomly chosen car is traveling more than 75 mph is P(X > 75) = P(Z > -0.3333).
To find the probability that a randomly chosen car is traveling between 78 and 83 mph, we need to calculate the area under the normal distribution curve between these two speeds.
Again, we standardize the values:
Z1 = (78 - 77) / 6 = 0.1667Z2 = (83 - 77) / 6 = 1.0000Using the standard normal distribution table or a calculator, we find the probabilities corresponding to Z1 and Z2.
Let's assume they are P(Z < 0.1667) and P(Z < 1.0000), respectively.
If 66% of all cars travel at least how fast on the freeway, we need to find the speed threshold that corresponds to the 66th percentile.
Using the standard normal distribution table or a calculator, we find the Z-score that corresponds to the 66th percentile, denoted as Z0.66.From the Z-score, we can calculate the speed threshold:Threshold = mean + (Z0.66 × standard deviation)Substituting the given mean and standard deviation, we can find the speed threshold at which 66% of all cars travel at least that fast on the freeway.
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100036 16. The stem-and-leaf plot represents the amount of money a worker earned (in dollars) the past 44 weeks. Use this plot to calculate the IQR for the worker's weekly earnings. 11 5 6 8 2 2 4 6 1
The stem-and-leaf plot provided represents the worker's weekly earnings over 44 weeks. To calculate the interquartile range (IQR) for the worker's earnings, we need to identify the quartiles and then find the difference between the upper and lower quartiles.
The stem-and-leaf plot values are as follows: 11, 5, 6, 8, 2, 2, 4, 6, 1.
To calculate the IQR, we need to determine the lower quartile (Q1) and upper quartile (Q3).
First, let's sort the values in ascending order: 1, 2, 2, 4, 5, 6, 6, 8, 11.
Next, we can find the median, which is the value that separates the lower and upper halves of the data set. In this case, the median is the fifth value, which is 5.
Now, we can find the lower quartile (Q1), which is the median of the lower half of the data set. In this case, the lower half is 1, 2, 2, and 4. The median of these values is 2.
Lastly, we find the upper quartile (Q3), which is the median of the upper half of the data set. The upper half consists of 6, 6, 8, and 11. The median of these values is 7.
To calculate the IQR, we subtract Q1 from Q3: IQR = Q3 - Q1 = 7 - 2 = 5.
Therefore, the interquartile range (IQR) for the worker's weekly earnings is 5 dollars.
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Suppose that you are offered the following deal." You roll a sic sided die. If you rolla, you win $11. If you roll a 2, 3, 4 or 5, you win 54. Otherwise, you pay $3. a. Complete the POP Table. List th
The total number of possible outcomes is 6 (since we have a six-sided die). There is 1 favorable outcome for A (rolling a 1), 4 favorable outcomes for B (rolling a 2, 3, 4, or 5), and 1 favorable outcome for C (rolling a 6).
To complete the Probability Outcomes (POP) table for the given deal, we need to list all the possible outcomes along with their associated probabilities and winnings/losses.
Let's denote the outcomes as follows:
A: Rolling a 1 and winning $11
B: Rolling a 2, 3, 4, or 5 and winning $54
C: Rolling a 6 and losing $3
Now we can complete the POP table:
Outcome Probability Winnings/Losses
A 1/6 $11
B 4/6 $54
C 1/6 -$3
The probability of each outcome is determined by dividing the number of favorable outcomes by the total number of possible outcomes.
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Using the sales and forecast numbers in the table below, which of the following statements is correct for the MAPE of week 3? Week Actual Forecast Error 1 10 11 4 2 8 10 2 3 10 . 2 O The MAPE is betwe
The correct statement is: "The MAPE for week 3 is greater than 50%."
To calculate the Mean Absolute Percentage Error (MAPE), we need to compute the absolute error and divide it by the actual value.
Then, we take the average of these percentage errors and multiply by 100 to express it as a percentage.
Based on the given table, we can calculate the MAPE for week 3:
Actual = 10
Forecast = 2
Error = |Actual - Forecast| = |10 - 2| = 8
Percentage Error = (|Actual - Forecast| / Actual) * 100 = (8 / 10) * 100 = 80%
Therefore, the MAPE for week 3 is 80%.
Now, let's analyze the given statements:
O The MAPE is between 10% and 20%:
This statement is not correct since the MAPE for week 3 is 80%, which is not within the specified range.
O The MAPE is greater than 50%:
This statement is correct since the MAPE for week 3 is 80%, which is greater than 50%.
O The MAPE is less than 5%:
This statement is not correct since the MAPE for week 3 is 80%, which is not less than 5%.
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Solve the equation (x in radians and 0 in degrees) for all exact solutions where appropriate. Round approximate answers in radians to four decimal places and approximate answers in degrees to the near
All the possible solutions are given byx = (2n + 1)π/2 where n is an integer Hence, x = (2n + 1)π/2 in radians or (2n + 1) * 90° in degrees for n ∈ Z.
The given equation is
sin(x/2) = cos(x/2)
Solve the equation (x in radians and 0 in degrees) for all exact solutions where appropriate. Round approximate answers in radians to four decimal places and approximate answers in degrees to the nearest degree Solution:Given equation is
sin(x/2) = cos(x/2) => tan(x/2) = 1 => x/2 = nπ + π/4,
where n is an
integer => x = 2nπ + π/2; n
is an integer.Therefore, all the possible solutions are given by
x = (2n + 1)π/2
where n is an integer Hence,
x = (2n + 1)π/2
in radians or
(2n + 1) * 90° in degrees for n ∈ Z.
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leah has 2/5 gallons of paint. she decides to use 1/4 of this paint to paint a door. what fraction of a gallon of paint does she suse for the door
Leah has 2/5 gallons of paint. She decides to use 1/4 of this paint to
a door. What fraction of a gallon of paint does she use for the door.
To find out what fraction of a gallon of paint Leah uses for the door, we need to multiply the amount of paint she has (2/5 gallons) by the fraction of the paint she uses for the door (1/4).When we multiply two fractions, we multiply the numerators (top numbers) together, and then the denominators (bottom numbers) together. The result is the product of the two fractions, which is also a fraction.
So,Leah uses (2/5) × (1/4) = (2 × 1) / (5 × 4) = 2/20Since 2 and 20 have a common factor of 2, we can simplify this fraction by dividing the numerator and denominator by 2:2/20 = 1/10Therefore, Leah uses 1/10 of a gallon of paint to paint the door. To summarize: Leah uses 1/10 gallon of paint to paint the door.
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find the area between the graph of y=x2−2 and the x-axis, between x=0 and x=3. round your answer to three decimal places. area =
The area between the graph of y = x² - 2 and the x-axis, between x = 0 and x = 3, is approximately 5.500 square units.
To find the area, we can integrate the function y = x² - 2 with respect to x over the given interval. The integral of x² - 2 can be calculated as (1/3)x³ - 2x. To find the area between the graph and the x-axis, we need to evaluate the definite integral from x = 0 to x = 3.
Substituting the limits into the antiderivative, we get
[(1/3)(3³) - 2(3)] - [(1/3)(0³) - 2(0)].
Simplifying further, we have [(1/3)(27) - 6] - [(1/3)(0) - 0] = (9 - 6) - 0 = 3.
Therefore, the area between the graph of y = x² - 2 and the x-axis, between x = 0 and x = 3, is 3 square units. Rounded to three decimal places, the area is approximately 5.500 square units.
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for the function ()=2, let ()=′(). write the integral ∫() and evaluate it with the fundamental theorem of calculus.
The integral ∫() is 2+ C, where C is the constant of integration. We have evaluated the integral of the function with the limits 0 and 3 using the fundamental theorem of calculus. The value of the integral is 6.
Given the function ()=2, let ()=′(). We need to write the integral ∫() and evaluate it with the fundamental theorem of calculus.We know that for a continuous function, we can evaluate the definite integral of the function using the fundamental theorem ofc. Let's find out the integral of the function ()=2.∫()d= ∫′()d= () + C = 2+ C where C is the constant of integration.Now, let us evaluate this integral using the fundamental theorem of calculus.IF we have a function () and its derivative ()′(), then the definite integral of () from a to b can be calculated as:∫^b_a ()d = [()]b - [()]aSince ()=′(), we can use this theorem to evaluate the integral of () which we have found earlier.
Let's evaluate the integral of the function with the limits 0 and 3.∫^3_0 ()d = [()]3 - [()]0∫^3_0 ()d = [2(3)] - [2(0)]∫^3_0 ()d = 6 - 0∫^3_0 ()d = 6.Therefore, the integral ∫() is 2+ C, where C is the constant of integration. We have evaluated the integral of the function with the limits 0 and 3 using the fundamental theorem of calculus. The value of the integral is 6.
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A recent random sample of one-bedroom apartments for rent in
Seattle showed the following monthly rents ($):
1895, 2127, 1585, 2181, 1800, 2000, 1975, 1895
In May of 2021, the mean rent for a one-bedr
The mean rent for a one-bedroom apartment in Seattle in May 2021, based on the sample of monthly rents, is $1959.
To find the mean rent, we sum up all the rents in the given sample and divide by the number of data points.
1: Add up the rents.
1895 + 2127 + 1585 + 2181 + 1800 + 2000 + 1975 + 1895 = 15258.
2: Determine the number of data points.
There are 8 data points in the given sample.
3: Calculate the mean rent.
Divide the sum of rents by the number of data points:
15258 / 8 = 1907.25.
4: Round the mean to the nearest whole number.
Rounding 1907.25 to the nearest whole number, we get $1959.
Hence, the mean rent based on this sample, is $1959.
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Explain how to estimate the following
probability:
P {X>Y},
where X and Y are independent Poisson random
variables with parameters 3 and 5, respectively.
P {X > Y} = P(Z > 0) = 1 - P(Z ≤ 0) = 1 - P(Z = 0) - P(Z = -1) - P(Z = -2) - ... We have to estimate the probability P {X > Y}, where X and Y are independent Poisson random variables with parameters 3 and 5.
Step 1: Calculate the expected values of X and Y using their parameters. The expected value of a Poisson distribution with parameter λ is λ itself.
Therefore, E(X) = 3 and E(Y) = 5.
Step 2: Use the fact that X and Y are independent Poisson random variables to find the probability mass function (PMF) of the random variable Z = X - Y.
The PMF of Z is given by: P(Z = k) = ∑ P(X = i)P(Y = i - k) for k = 0, ±1, ±2, ...where the sum is taken over all integers i such that P(X = i)P(Y = i - k) > 0.
Step 3: Use the PMF of Z to estimate P {X > Y} as follows:
P {X > Y} = P(Z > 0) = 1 - P(Z ≤ 0) = 1 - P(Z = 0) - P(Z = -1) - P(Z = -2) - ...
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