Answer:
0.8 is the probability that a single detection system will detect a missile attack.
Step-by-step explanation:
We are given the following information:
We treat detection a missile attack as a success.
P(detecting a missile attack) = 80% = 0.8
Then the number of missile attack follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 1
We have to evaluate:
[tex]P(x = 1)\\= \binom{1}{1}(0.8)^1(1-0.8)^0\\= 0.8[/tex]
0.8 is the probability that a single detection system will detect a missile attack.
Answer:
(a) Probability that a single detection system will detect an attack is 0.80
Step-by-step explanation:
We are given that a reliability question is whether a detection system will be able to identify an attack and issue a warning. Assuming that a particular detection system has a 0.80 probability of detecting a missile attack.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials(samples) taken = 1 detection system
r = number of success
p = probability of success which in our question is probability of
detecting a missile attack, i.e., 80%
LET X = a particular detection system
Also, it is given that a single detection system is taken,
So, it means X ~ [tex]Binom(n=1,p= 0.80)[/tex]
Now, Probability that a single detection system will detect an attack is given by = P(X = 1)
P(X = 1) = [tex]\binom{1}{1}0.8^{1} (1-0.8)^{1-1}[/tex]
= [tex]1 \times 0.8 \times 1[/tex] = 0.80 .
Your DVD membership costs $16 per month for 10 DVD rentals. Each additional DVD rental is $2. a. Write an equation in two variables that represents the monthly cost of your DVD rentals. b. Identify the independent and dependent variables. c. How much does it cost to rent 15 DVDs in one month?
C(15) = $26
Step-by-step explanation:
1 point) Consider the following game of chance based on the spinner below: Each spin costs $2. If the spinner lands on A the player wins a quarter, if the spinner stops on D the player wins $9 otherwise the player wins nothing. Calculate the players expected winnings. Express your answer to at least three decimal places in dollar form. .
The game of chance discussed is a question about probability and expected value in mathematics. To calculate the expected winnings of the game, we use given game information and probabilities. If the probabilities are not given, the question usually assumes a fair spinner, i.e., all outcomes are equally likely.
Explanation:The subject at hand deals with probability and expected value, which are mathematical concepts typically covered in a high school math curriculum. The game described illustrates these concepts. Each possible outcome of the game (A or D, otherwise lose) corresponds to an event that has a certain probability. These probabilities are all added together to determine the expected value of the game in dollars.
Suppose the probabilities of landing on A and D are p(A) and p(D), and the probability of not landing on either A or D is 1 - p(A) - p(D), then the expected value of the game is: Expected Value = $2 * [p(A)*0.25 + p(D)*9 + (1 - p(A) - p(D))*0] .
To find the expected value, we would need to know the probabilities of landing on each of these segments on the spinner. If these probabilities are not given in the problem, it can be assumed that the spinner is fair (i.e., all outcomes are equally likely). If there are n total segments on the spinner, then p(A) = p(D) = 1/n, and the probability of not landing on A or D would be (n-2)/n. Substitute these probabilities into the expected value equation can give the answer.
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In preparation for the upcoming school year, a teacher looks at raw test scores on the statewide standardized test for the students in her class. Instead of looking at the scores relative to the norms in the state, the teacher wants to understand the scores relative to the students who will be in the class. To do so, she decides to convert the test scores into z-scores relative to the mean and standard deviation of the students in the class. The mean test score in her upcoming class is 49, and the standard deviation is 20.7. The teacher wants to identify those students who may need extra challenges. As a first cut, she decides to look at students who have z-scores above z = 2.00 Identify the test score corresponding to a z-score of above z=2.00. Round to the nearest whole number.
Answer:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And we can solve for the value of X like this:
[tex] X = \mu + z*\sigma[/tex]
And since we know that z=2 we can replace and we got:
[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]
Step-by-step explanation:
Previous concepts
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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we can assume the distribution for X is given by:
[tex]X \sim N(49,20.7)[/tex]
Where [tex]\mu=49[/tex] and [tex]\sigma=20.7[/tex]
And for this case the z score is given by:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And we can solve for the value of X like this:
[tex] X = \mu + z*\sigma[/tex]
And since we know that z=2 we can replace and we got:
[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]
To find the test score corresponding to a z-score above z=2.00, use the formula x = (z * standard deviation) + mean. Plugging in the values, the test score is approximately 90 when rounded to the nearest whole number.
Explanation:The teacher wants to identify students who have z-scores above z=2.00. To find the corresponding test score,
we can use the formula for z-score:
z = (x - mean) / standard deviation
Rearranging the formula, we get:
x = (z * standard deviation) + mean
Substituting z=2.00, standard deviation=20.7, and mean=49, we have:
x = (2.00 * 20.7) + 49
Simplifying the equation, we get:
x = 41.4 + 49 = 90.4
Therefore, the test score corresponding to a z-score above z=2.00 is approximately 90 when rounded to the nearest whole number.
Almost all medical schools require applicants to take the Medical College Admission Test (MCAT). To estimate the mean score of those who took the MCAT at WSSU, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 6.5. Suppose that (unknown to you) the mean score of those taking the MCAT at WSU is 25.0. You sampled 25 students. What is the probability that the mean score of your sample is between 22 and 28
Answer:
97.92% probability that the mean score of your sample is between 22 and 28
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 25, \sigma = 6.5, n = 25, s = \frac{6.5}{\sqrt{25}} = 1.3[/tex]
What is the probability that the mean score of your sample is between 22 and 28
This is the pvalue of Z when X = 28 subtracted by the pvalue of Z when X = 22. So
X = 28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{28 - 25}{1.3}[/tex]
[tex]Z = 2.31[/tex]
[tex]Z = 2.31[/tex] has a pvalue of 0.9896
X = 22
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{22 - 25}{1.3}[/tex]
[tex]Z = -2.31[/tex]
[tex]Z = -2.31[/tex] has a pvalue of 0.0104
0.9896 - 0.0104 = 0.9792
97.92% probability that the mean score of your sample is between 22 and 28
A group of students bakes 100 cookies to sell at the school bake sale. The students want to ensure that the price of each cookie offsets the cost of the ingredients. If all the cookies are sold for $0.10 each, the net result will be a loss of $4. If all the cookies are sold for $0.50 each. The students will make a $36 profit. First, write the linear function p(x) that represents the net profit from selling all the cookies, where x is the price of each cookie. Then, determine how much profit the students will make if they sell the cookies for $0.60 each. Explain. Tell how your answer is reasonable.
Answer:
46
Step-by-step explanation:
-Let b be the constant in the linear equation.
#The linear equation can be expressed as:
[tex]p(x)=100x+b[/tex]
Substitute the values in the equation to find b:
[tex]p(x)=100x+b\\\\-4=100(0.1)+b\\\\b=-14\\\\\#or\\\\36=100(0.5)+b\\\\b=-14[/tex]
We know have the constant value b=-14, substitute the values of b and x in the p(x) function:
[tex]p(x)=100x+b\\\\p(x)=100(0.6)-14\\\\p(x)=60-14\\\\p(x)=46[/tex]
Hence, the profit when selling price is $0.60 is $46
#From our calculations, it's evident that the cookies production has a very high fixed cost which can only be offset by raisng the selling price or the number of units sold at any given time.
If the students sell the cookies for $0.60 each, they will make a profit of $46.
To solve this problem, let's first define the variables and set up the linear function p(x) that represents the net profit based on the selling price x per cookie.
Given information:
- Selling each cookie for $0.10 results in a net loss of $4.
- Selling each cookie for $0.50 results in a net profit of $36.
From this information, we can set up two equations based on the net profit:
1. When selling each cookie for $0.10:
[tex]\[ R = 100 \cdot 0.10 = 10 \] \[ P(0.10) = R - C = 10 - C = -4 \] \[ C = 10 + 4 = 14 \][/tex]
(Total cost of ingredients)
2. When selling each cookie for $0.50:
[tex]\[ R = 100 \cdot 0.50 = 50 \] \[ P(0.50) = R - C = 50 - C = 36 \] \[ C = 50 - 36 = 14 \][/tex]
Total cost of ingredients)
So, the total cost of ingredients C is $14 regardless of the selling price, since it's consistent in both scenarios.
Now, let's define the linear function P(x) :
[tex]\[ P(x) = R - C \][/tex]
Where ( R = 100x ) (total revenue from selling 100 cookies at x dollars each), and ( C = 14 ) (total cost of ingredients).
Therefore,
[tex]\[ P(x) = 100x - 14 \][/tex]
This function P(x) gives us the net profit when each cookie is sold for x dollars.
Now, to find out how much profit the students will make if they sell the cookies for $0.60 each:
[tex]\[ x = 0.60 \]\[ P(0.60) = 100 \cdot 0.60 - 14 \]\[ P(0.60) = 60 - 14 \]\[ P(0.60) = 46 \][/tex]
So, if the students sell each cookie for $0.60, they will make a profit of $46.
Explanation of Reasonableness:
The function [tex]\( P(x) = 100x - 14 \)[/tex] is a linear function that accurately represents the relationship between the selling price x and the net profit ( P(x) ). The function is derived from the given conditions where selling at $0.10 results in a loss and selling at $0.50 results in a profit, confirming the slope and intercept of the function.
A population has a mean muequals71 and a standard deviation sigmaequals24. Find the mean and standard deviation of a sampling distribution of sample means with sample size nequals64.
Answer:
Mean 71
Standard deviation 3
Step-by-step explanation:
We use the Central Limit Theorem to solve this question.
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution with a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 71, \sigma = 24, n = 64[/tex]. So
Mean 71
Standard deviation [tex]s = \frac{24}{\sqrt{64}} = 3[/tex]
A recent study¹ examined several variables on collegiate football players, including the variable Years, which is number of years playing football, and the variable Percentile, which gives percentile on a cognitive reaction test. The regression line for predicting Percentile from Years is:
Percentile = 102 - 3.34 Years.
¹ Singh R, et al., "Relationship of Collegiate Football Experience and Concussion with Hippocampal Volume and Cognitive Outcomes", JAMA, 311(18), 2014. Data values are estimated from information in the paper.
Predict the cognitive percentile for someone who has played football for 7 years and for someone who has played football for 16 years. Enter the exact answers.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The cognitive percentile for someone who has played for 7 years is 78.62
The cognitive percentile for someone who has played for 16 years is 48.56
Step-by-step explanation:
From the question the regression line for predicting percentile from years is given as
percentile = [tex]102 - 3.34 \ Years[/tex]
For someone who has played for 7 years his cognitive percentile would be
[tex]Percentile = 102 - 3.34(7)[/tex]
[tex]=78.62[/tex]
For someone who has played for 16 years his cognitive percentile would
be [tex]Percentile = 102 - 3.34(16)[/tex]
[tex]=48.56[/tex]
Final answer:
The predicted cognitive percentile for someone who has played football for 7 years is 78.62, while for someone who has played for 16 years, it is 48.56, using the regression equation Percentile = 102 - 3.34 Years.
Explanation:
To predict the cognitive percentile for a football player who has played for 7 years, we use the given regression equation Percentile = 102 - 3.34 Years. By plugging 7 into the equation for Years, we get:
Percentile = 102 - 3.34 × 7
Percentile = 102 - 23.38
Percentile = 78.62
So, for someone who has played football for 7 years, the predicted cognitive percentile would be 78.62.
Similarly, for someone who has played football for 16 years:
Percentile = 102 - 3.34 × 16
Percentile = 102 - 53.44
Percentile = 48.56
Therefore, the predicted cognitive percentile for someone who has played for 16 years is 48.56.
The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). (1 − 2x − x2)y'' + 2(1 + x)y' − 2y = 0; y1 = x + 1
Answer and Step-by-step explanation:
The answer is attached below
To identify a second solution y2 from a given differential equation and its solution y1, it is necessary to extract P(x) from the differential equation, compute integral -∫P(x) dx, and multiply the result by the function y1.
Explanation:First, we need to find the function P(x) in the equation y2 = y1(x) e−∫P(x) dx (5). Looking at the given differential equation (1 − 2x − x2)y'' + 2(1 + x)y' − 2y = 0; y1 = x + 1, we can rearrange terms and find that P(x) is equal to -2(1+x)/(1-2x-x2). Then, we can calculate the integral -∫P(x) dx, and multiply this by our given solution y1 to find the second solution y2. It's important to remember that when carrying out these steps, accuracy is crucial since each step builds on the last.
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A study reports that college students work, on average, between 4.63 and 12.63 hours a week, with confidence coefficient .95. Which of the following statements are correct? MARK ALL THAT ARE TRUE. There are four correct answers. You must mark them all to get credit. Group of answer choices The interval was produced by a technique that captures mu 95% of the time. 95% of all college students work between 4.63 and 12.63 hours a week. 95% of all samples will have x-bar between 4.63 and 12.63. The probability that mu is between 4.63 and 12.63 is .95. 95% of samples will produce intervals that contain mu. The probability that mu is included in a 95% CI is 0.95. We are 95% confident that the population mean time that college students work is between 4.63 and 12.63 hours a week.
The correct statements are that the interval was produced by a technique that captures mu 95% of the time, 95% of all college students work between 4.63 and 12.63 hours a week, 95% of all samples will have x-bar between 4.63 and 12.63, and the probability that mu is between 4.63 and 12.63 is .95.
Explanation:The correct statements are:
The interval was produced by a technique that captures mu 95% of the time.95% of all college students work between 4.63 and 12.63 hours a week.95% of all samples will have x-bar between 4.63 and 12.63.The probability that mu is between 4.63 and 12.63 is .95.These statements are correct because a confidence interval is a range of values that is likely to contain the true population mean. With a confidence coefficient of .95, we can say that there is a 95% confidence level that the population mean falls within the interval.
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The daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.
If R is a random number between 0 and 1, then which of the following correctly models daily demand for gasoline?
a) 1200 + 350 R
b) 1200 + 350*NORMSDIST(R)
c) NORM.INV(R, 1200, 350)
d) Both b) and c) are correct.
Answer:
c) NORM.INV(R, 1200, 350)
Step-by-step explanation:
Given that the daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.
X = demand for gasolene at a local gas station is N(1200, 350)
R is any random number between 0 and 1.
Daily demand for gasolene would be
X = Mean + std deviation * z value, where Z = normal inverse of a value between 0 and 1.
The norm inv (R, 1200, 350) for R between 0 and 1 gives all the values of X
Hence correct choice would be
Option c) NORM.INV(R, 1200, 350)
The circumference of a sphere was measured to be 74 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm2 What is the relative error?
Using differentials, the estimated maximum error in the calculated surface area of a sphere with a measured circumference of 74 cm and a possible error of 0.5 cm is 24 cm². The relative error is approximately 5%.
Explanation:The subject concerns the application of differentials in estimating the maximum error in the calculated surface area of a sphere. Given the circumference C = 74 cm with a possible error δC = 0.5 cm, we can calculate the radius r = C / (2π). With the surface area formula of a sphere A = 4πr², differentiating this equation gives dA = 8πr dr. By substituting the values, the maximum error in calculated surface area δA = dA = 8πr δr = 8π(C/2π) (δC/2π) = 2C δC / π. Plugging the values of C = 74 cm and δC = 0.5 cm, we get δA ≈ 24 cm² which is the maximum error in the calculated surface area. For the relative error, it is the absolute error divided by the actual measurement, hence, the relative error is δA/A = δA / 4πr² = (2C δC / π) / 4π(C/2π)² ≈ 0.05 or 5%.
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To find a formula for the moose population, calculate the rate of change and use it in the formula P = 190t + 4360. The model predicts the moose population to be 7710 in 2003.
Explanation:To find a formula for the moose population, we need to determine the rate of change in the population. We can do this by finding the slope of the line that represents the change in population from 1991 to 1999. First, we calculate the change in population: 5880 - 4360 = 1520. Then, we calculate the change in time: 1999 - 1991 = 8. Next, we divide the change in population by the change in time to find the rate of change: 1520/8 = 190. So, the formula for the moose population, P, is P = 190t + 4360, where t represents the years after 1991.
To predict the moose population in 2003, we substitute t = 12 (since 2003 is 12 years after 1991) into the formula: P = 190(12) + 4360 = 7710. Therefore, the model predicts the moose population to be 7710 in 2003.
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points)A password must consist of 16 characters. Each character can be a digit (0-9), an uppercase or lowercase letter (A-Z, a-z) or one out of 10 special characters. How many valid passwords are there? Give your answer in unevaluated form. You don't need to explain it. If you have forgotten your password, but can test 1 trillion passwords per second, how much time would you require to nd the password in the worst-case scenario that your forgotten password is the last one tested? Give the answer in years, rounded to the nearest power of 10.
Answer:
72¹⁶ possible passwords
10¹⁰ years
Step-by-step explanation:
For each of the 16 characters, the number of possible outcomes is 10 numbers, 52 letters, or 10 special characters, totaling 72 possible values. The number of total different 16 characters passwords is:
[tex]n = 72^{16}[/tex]
If you can test 1 trillion passwords per second, the number of passwords per year is:
[tex]P = 10^{12} * 3,600*24*365\\P=3.1536*10^{19}[/tex]
The time in years that would take to test all passwords is:
[tex]T=\frac{72^{16}}{3.1536*10^{19}}\\T = 1.65*10^{10}\ years[/tex]
Rounding to the nearest power of 10, it would take 10¹⁰ years
The question concerns combinatorics in Mathematics, calculating the total possible passwords given 72 character options for a 16-character length (72^16). Given a rate of 1 trillion tests per second, the time it would take to test all these combinations depends on this total, which we express in years.
Explanation:The subject of your question is Combinatorics, which falls under Mathematics. It requires finding the total number of valid passwords that can be comprised of certain types of characters, then finding how long it would take to test all those passwords under a certain rate.
If each character in the password can be one of 10 digits, 52 letters (uppercase and lowercase) or 10 special characters, there are overall 72 possible characters. Given the password length is 16 characters, the total number of possibilities would be 72^16. This represents the total number of valid passwords.
With the ability to test 1 trillion (10^12) passwords per second, to find out how long it would take to test all passwords, you divide the total number of passwords by the testing rate. Expressing this in years (seconds in a year being approximately 3.15 x 10^7), you would have 72^16 divided by (10^12 x 3.15 x 10^7) years. Hence, the time required in the worst-case scenario is ultimately dependent on the total number of valid passwords (72^16).
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A 22-pound child was admitted with acute bronchitis. Her medical orders include Garamycin 2.5 mg/kg q 8h. You receive Garamycin from the pharmacy in a vial labeled 10 mg/ml. Determine the number of milliliters required per dose.
Answer:
2.4948ml
Step-by-step explanation:
First, we change the child's weight into kilograms:
[tex]22pounds=9.9790kgs[/tex]
From the info, the dose is recommended as 2.5mg/kg. Let x be the number of mg administered to the child:
[tex]1kg=2.5mg\\9.9790kg=x\\\\x=2.5\times9.9790\\\\x=24.9475mg[/tex]
#The drug contains 10mg/ml . Let y be the dose size administered, equate and solve for y:
[tex]10mg=1ml\\24.9475mg=y\\\\\thereforey= \frac{1ml\times24.9475mg}{10mg}\\\\y=2.4948[/tex]
Hence, the dose required is 2.4948ml
Answer:
2.5
Step-by-step explanation:
Standard deviation of a normal data distribution is a _______. Group of answer choices
measure of data dispersion
measure of data centrality
measure of data quality
measure of data shape
Answer:
Standard deviation of a normal data distribution is a measure of data dispersion.
Step-by-step explanation:
Standard deviation is used to measure dispersion which is present around the mean data.
The value of standard deviation will never be negative.
The greater the spread, the greater the standard deviation.
Steps-
1. At first, the mean value should be discovered.
2.Then find out the square of it's distance to mean value.
3.Then total the values
4.Then divide the number of data point.
5.the square root have to be taken.
Formula-
SD=[tex]\sqrt{\frac{(\sum{x-x)^2} }{n-1}[/tex]
Advantage-
It is used to measure dispersion when mean is used as measure of central tendency.
Standard deviation of a normal data distribution is a measure of data dispersion.
What is a normal distribution?
A normal distribution is a probability distribution that is symmetric around the mean of the distribution. This means that the there are more data around the mean than data far from the mean. When shown on a graph, a normal distribution is bell-shaped.
What is standard deviation?
Standard deviation is a measure of variation. It measures the dispersion of data from its mean. It can be calculated by determining the value of the square root of variance.
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if one of two supplementary angles has a measure of 121 degrees what is the measure of the other angle?
Answer:
The measure of the other angle is 59°
Step-by-step explanation:
Supplementary Angles
Two angles [tex]\alpha[/tex] and [tex]\beta[/tex] are supplementary when they add up to 180 degrees, i.e.
[tex]\alpha+\beta=180^o[/tex]
One notable property is that together they make a straight angle although they don't have to be together to be supplementary.
We are given one of two supplementary angles with a value of 121 degrees, we can compute the measure of the other angle, say [tex]\alpha[/tex] as
[tex]\alpha=180^o-\beta=180^o-121^o=59^o[/tex]
The measure of the other angle is 59°
A sprint duathlon consists of a 5 km run, a 20 km bike ride, followed by another 5 km run. The mean finish time of all participants in a recent large duathlon was 1.67 hours with a standard deviation of 0.25 hours. Suppose a random sample of 30 participants was taken and the mean finishing time was found to be 1.59 hours with a standard deviation of 0.30 hours. What is the standard error for the mean finish time of 30 randomly selected participants
Answer:
The standard error is 0.0456 for the mean finish time of 30 randomly selected participants.
Step-by-step explanation:
We are given the following in the question:
Population mean, [tex]\mu[/tex] = 1.67 hours
Population standard deviation, [tex]\sigma[/tex] = 0.25 hours
Sample mean, [tex]\bar{x}[/tex] = 1.59 hours
Sample standard deviation, s = 0.30 hours
Sample size, n = 30
We have to find the standard error for the mean finish time of 30 randomly selected participants.
Formula:
[tex]\text{Standard error} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{0.25}{\swqrt{30}} = 0.0456[/tex]
Thus, the standard error is 0.0456 for the mean finish time of 30 randomly selected participants.
The standard error for the mean finish time of 30 randomly selected participants is 0.0549 hours.
Explanation:The standard error for the mean finish time of 30 randomly selected participants can be calculated using the formula:
Standard Error = Standard Deviation / √(Sample Size)
Plugging in the given values, the standard error would be:
Standard Error = 0.30 / √(30) = 0.0549 hours
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You wish to estimate the average weight of a mouse. You obtain 10 mice, sampled uniformly at random and with replacement from the mouse population. Their weights are 21; 23; 27; 19; 17; 18; 20; 15; 17; 22 grams respectively. (a) What is the best estimate for the average weight of a mouse, from this data
Answer:
The best estimate for the average weight of a mouse, from this data is 19.9 grams.
Step-by-step explanation:
The best estime for the weight of a mouse from this data is the sum of all these weights divided by the number of mices.
10 mices
Their weights are 21; 23; 27; 19; 17; 18; 20; 15; 17; 22 grams
So
[tex]M = \frac{21+23+27+19+17+18+20+15+17+22}{10} = 19.9[/tex]
The best estimate for the average weight of a mouse, from this data is 19.9 grams.
In order to estimate the height of all students at your university, let's assume you have measured the height of all psychology majors at the university. The resulting raw scores are called _________. constants data coefficients statistics
Answer:
Data
Step-by-step explanation:
We are given the following in the question:
We want to measure height of all psychology majors at the university.
Thus, the resulting raw scores of each individual are called the data.
Data point:
Height of each psychology majors at the university
Data:
Collection of all heights of all psychology majors at the university
These value are constants but comprises a data.
They are neither coefficients nor statistic because they do not describe a sample.
Thus, the correct answer is
Data
Use the information given to find the appropriate minimum sample size. (Round your answer up to the nearest whole number.)Estimating μ correct to within 3 with probability 0.99. Prior experience suggests that the measurements will range from 8 to 40.
The minimum sample size required to estimate μ within 3 with a confidence level of 0.99, given a standard deviation of 8, is approximately 48. This was determined by plugging the values into the sample size formula and rounding up to the nearest whole number.
Explanation:To find the minimum sample size, we need to use the formula for sample size n, = (Z_α/2 * σ / E)^2. In this problem, you want to estimate μ correct to within 3 with a probability of 0.99. In other words, you want the error E to be 3 and the confidence level to be 0.99.
The Z value corresponding to a confidence level of 0.99 is approximately 2.576 (you can find this value from a standard Z-table). The measurements range from 8 to 40, so we can estimate the standard deviation σ as (40 - 8) / 4 = 8.
Plugging these values into the formula, we get n = (2.576 * 8 / 3)^2 = 47.36. This number must be rounded up to the nearest whole number because the sample size cannot be a fraction. So, the minimum sample size required is 48.
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Consider two x distributions corresponding to the same x distribution. The first x distribution is based on samples of size n = 100 and the second is based on samples of size n = 225. Which x distribution has the smaller standard error? The distribution with n = 100 will have a smaller standard error. The distribution with n = 225 will have a smaller standard error. Explain your answer. Since σx = σ2/√n, dividing by the square root of 100 will result in a small standard error regardless of the value of σ2. Since σx = σ/n, dividing by 100 will result in a small standard error regardless of the value of σ. Since σx = σ/n, dividing by 225 will result in a small standard error regardless of the value of σ. Since σx = σ/√n, dividing by the square root of 100 will result in a small standard error regardless of the value of σ. Since σx = σ/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of σ. Since σx = σ2/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of σ2.
Answer:
The distribution with n = 225 will give a smaller standard error.
Since sigma x = sigma/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of sigma.
Step-by-step explanation:
Standard error is given by standard deviation (sigma) divided by square root of sample size (√n).
The distribution with n = 225 would give a smaller standard error because the square root of 225 is 15. The inverse of 15 multiplied by sigma is approximately 0.07sigma which is smaller compared to the distribution n = 100. Square of 100 is 10, inverse of 10 multiplied by sigma is 0.1sigma.
0.07sigma is smaller than 0.1sigma
x^2-16/(x+4)(x-5) x=-4 x=1 continuous at x=-4?
Answer:
Yes, its continuous
Step-by-step explanation:
We use the formula:
x^2-y^2=(x-y)(x+y),
And we know that 16=4^2, so we have:
[tex]\frac{x^2-16}{(x+4)(x-5)}=\frac{(x-4)(x+4)}{(x+4)(x-5)}=\frac{x-4}{x-5}[/tex]
So for x=-4 we have -8/-9,i.e, it is 8/9, so it is continuous.
I dont know what is x=1, because for x=1 the function has value 3/4.
But function is not continuous in x=5 becaus for that x we will get 1/0, and that is not definite.
:)
Each year, taxpayers are able to contribute money to various charities via their IRS tax forms. The following list contains the amounts of money (in dollars) donated via IRS tax forms by Each year, taxpayers are able to contribute money taxpayers:
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197
(a) For these data, which measures of central tendency take more than one value? Choose all that apply.
Mean
Median
Mode
None of these measures
(b) Suppose that the measurement 197 (the largest measurement in the data set) were replaced by 246. Which measures of central tendency would be affected by the change? Choose all that apply.
Mean
Median
Mode
None of these measures
(c) Suppose that, starting with the original data set, the largest measurement were removed. Which measures of central tendency would be changed from those of the original data set? Choose all that apply.
Mean
Median
Mode
None of these measures
(d) Which of the following best describes the distribution of the original data? Choose only one.
Negatively skewed
Positively skewed
Roughly symmetrical
Answer:
(a) None of these measures
(b) Mean
(c) Mean and Median
(d) Roughly Symmetrical
Step-by-step explanation:
(a)
Mean
Total number in the set = 23
Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+197 = 1804
Mean = Sum of set / total no of set
1804/23 = 78.435
Median is the middle number in the set after it had been arranged from lowest to highest
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197
The Median is 62
Mode the value that appear most
Mode is 62
None of them takes more than one value
(b) If 197 is replaced by 246, the set becomes
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 246
The mean becomes
Total number in the set = 23
Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+246= 1853
Mean = Sum of set / total no of set
1853/23 = 80.565
The Median and Mode remains the same.
(c) When the largest measurements are removed, the number of values in the set reduces and this affects the Mean and the Median. The mode will still remain unchanges since it is a small number and appears the most.
A publisher knows that from all the writers the company published, 20% wrote romantic novels and 40% wrote sci-fi books. If we look at the last 5 years, 40% published 2 books, and 30% published only 1 book. From the writers that did not published in the last 5 years, 20% wrote romantic novels and 40% sci-fi books. From the writers that wrote other types of books, 50% published 2 books. Finally, the number of sci-fi writers that published 1 and the number that published 2 books was the same. 1. What is the average number of books published in the last five years? 2. What proportion of writers are sci-fi writers and published 2 books during the last five years? 3. What is the probability that if we choose a romantic novels writer, he had not published in the last five years? 4. What is the probability that a writer did not publish romantic or sci-fi and did not publish exactly 1 book in the past five years? 5. If we select a writer that had published in the last five years, what is the probability that he writes romantic novels? 6. If we select 5 writers, what is the probability that they didn't publish any books in the past 5 years? 7. What is the probability that a writer is not a romantic novel writer and published more than 1 book in the past five years? 8. Are types of books and the number of books published in the past five years statistically independent? 9. Is writing sci-fi books and not publishing in the last years statistically independent? 10. What is the probability that if we choose 4 writers, 2 of them have published 1 romantic novel in the last 5 years? 11. What proportion of writers did not write sci-fi and published one or two books?
The answers to mentioned questions are conditional probabilities and proportions, requiring more detailed numerical data to provide exact values. But they can be calculated using simple formulas of probability and proportion.
Explanation:This is a statistics problem and we would need more details to fully answer these questions. But here are some general insights:
The average number of books published would require total number of books/total number of writers in the last 5 years.The proportion of sci-fi writers that published 2 books would be the total number of sci-fi writers with 2 books/total number of sci-fi writers.If we choose a romantic novels writer, the probability of them not having published in the last 5 years would be the number of romantic writers who didn't publish in last 5 years/total number of romantic writers.A writer not published romantic or sci-fi and did not publish exactly 1 book in the past five years would be calculated by first calculating the total of such authors and then dividing by total authors.If we know the writer published in the last five years, the probability they write romantic novels would be number of romantic writers that published in last 5 years / total number of writers who published in the last 5 years.Without explicit numbers provided for each category of writer, it's impossible to give numerical solutions to these problems. Instead, we can only provide the formulas to solve them. The same reasoning applies to all subsequent questions.
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Average number of books published in the last five years = 1.1 books
Proportion of sci-fi writers who published 2 books = 0.2
Probability that a romantic novel writer had not published in the last five years = 0.04
Probability a writer who did not publish romantic or sci-fi did not publish exactly 1 book = 0.3
Probability a writer who published in the last years writes romantic novels = 0.2
Probability that 5 writers didn't publish any books = .00243.
Probability that a non-romantic writer published more than 1 book= 0.32
Independence of book types and number published= independent events
Independence of writing sci-fi and not publishing = independent events.
Probability 2 of 4 have published romantic novels= 0.1536
Proportion of writers did not write sci-fi and published 1 or 2 books=0.42
Let's breakdown this complex problem step-by-step.
1. Average number of books published in the last five years
Given that 40% published 2 books and 30% published 1 book, we can find the average as follows:
Average = (0.4 * 2) + (0.3 * 1) + (0.3 * 0) = 0.8 + 0.3 + 0 = 1.1 books
2. Proportion of sci-fi writers who published 2 books
Since the number of sci-fi writers publishing 1 and 2 books is the same, and sci-fi comprises 40% of all writers:
Proportion = 0.4 * 0.5 = 0.2 or 20%
3. Probability that a romantic novel writer had not published in the last five years
20% of total writers wrote romantic novels, and 20% of those did not publish in the last five years:
Probability = 0.2 * 0.2 = 0.04 or 4%
4. Probability a writer who did not publish romantic or sci-fi did not publish exactly 1 book
If 60% did not publish romantic or sci-fi, half of these published 2 books:
Probability = 0.6 * 0.5 = 0.3 or 30%
5. Probability a writer who published in the last years writes romantic novels
40% published 2 books and 30% published 1 book, so 70% published. Romantic novelists comprise 20% of all writers:
Probability = (0.2 * 0.7) / 0.7 = 0.2 or 20%
6. Probability that 5 writers didn't publish any books
The probability that one writer didn't publish is 0.3:
Probability = 0.3^5 = 0.00243 or 0.243%
7. Probability that a non-romantic writer published more than 1 book
80% are non-romantic, and 40% of total published 2 books:
Probability = (0.4 * 0.8) = 0.32 or 32%
8. Independence of book types and number published
We need to see if P(A ∩ B) = P(A)*P(B). Since specific data does not align well, these events are not independent.
9. Independence of writing sci-fi and not publishing
40% wrote sci-fi and 30% did not publish:
P(A ∩ B) = 0.4 * 0.3 = 0.12 or 12%
These are independent events.
10. Probability 2 of 4 have published romantic novels
Using binomial distribution:
Probability P(X = 2) = 4C2 * (0.2)^2 * (0.8)^2 = 0.1536 or 15.36%
11. Proportion of writers did not write sci-fi and published 1 or 2 books
60% did not write sci-fi, and of those published, 70% published 1 or 2 books:
Proportion = 0.6 * 0.7 = 0.42 or 42%
The annual consumption of beef per person was about 64.8 lb in 2000 and about 60.1 lb in 2006. Assume B(t), the annual beef consumption t years after 2000, is decreasing according to the exponential decay model. a) Find the value of k, and write the equation b) Estimate the consumption of beef in 2011 c) In what year (theoretically) will the consumption of beef be 10 lb?
Answer:
a) B(t)= 64.8e^(-0.01255t), k=-0.0125 b) 56.5 c)2150
Step-by-step explanation:
a) B(t)= Ae^(-kt)
at t=0, B(t)=64.8
A=64.8
at t=6, B(t)=60.1
60.1=64.8e^(-6k)
k=0.0125
b) B(t)=64.8e^(-0.0125×11)
B(t)= 56.5
c) 10=64.8e^(-0.0125t)
0.15432=e^(-0.0125t)
-0.0125t=ln(0.15432)
-0.0125t=-1.869
t=149.5 or 150
Year= 2150
a)k ≈ 0.01267 and the equation is [tex]B(t) = 64.8e^{-0.01267t[/tex]
b)B(11) ≈ 56.38 lb
c) t ≈ 143.86
A beef consumption model follows exponential decay where B(t), the annual beef consumption t years after 2000,
can be modeled by the equation: [tex]B(t) = B_0e^{-kt}.[/tex]
Given that [tex]B_0[/tex] = 64.8 lb in 2000 and B(6) = 60.1 lb in 2006,
we first need to find the decay constant k.
a) Find the value of k, and write the equation:
We use the given data points to solve for k.
[tex]B(6) = 64.8e^{-6k} = 60.1[/tex]
[tex]e^{-6k} = 60.1 / 64.8[/tex]
[tex]e^{-6k}[/tex] ≈ 0.9272
Taking the natural log of both sides:
-6k = ln(0.9272)
k ≈ -ln(0.9272) / 6
k ≈ 0.01267
Therefore, the equation for B(t) is:
[tex]B(t) = 64.8e^{-0.01267t[/tex]
b) Estimate the consumption of beef in 2011:
For t = 11 (since 2011 is 11 years after 2000):
[tex]B(11) = 64.8e^{-0.01267 * 11[/tex]
B(11) ≈ 64.8[tex]e^{-0.13937[/tex]
B(11) ≈ [tex]64.8 * 0.8699[/tex]
B(11) ≈ 56.38 lb
c) In what year (theoretically) will the consumption of beef be 10 lb?
We need to solve for t when B(t) = 10 lb:
[tex]10 = 64.8e^{-0.01267t[/tex]
[tex]e^{-0.01267t} = 10 / 64.8[/tex]
[tex]e^{-0.01267t[/tex] ≈ 0.1543
Taking the natural log of both sides:
-0.01267t = ln(0.1543)
t ≈ -ln(0.1543) / 0.01267
t ≈ 143.86
Thus, the theoretical year is approximately 2000 + 144 = 2144.
Find the zeroes and give the multiplicity.
f(x) = 4x4 + 8x3 + 4x2
48 is what 4x4 + 8x3 + 4x2 equals So which means my assumption would definitely be that 'F' is 24 So it would be like- 24x, Like 24 x 2..? I'm so so sorry if i'm wrong but i'm 95.0% sure i'm right! OwO
Answer:
f(x) = 48
Multiplicity: 24 * 2
Step-by-step explanation:
Evaluate the function
Rather Simple
And i believe multiplicity you mean as in the equation you would use to get 48 right?
so that would be 24 * 2
Hope this helps~
54 and 67 use numbers and operations to write each phrase as an expression
The phrases you would like to be written as expressions are not listed. I would nevertheless, explain how to write phrases as expressions so that the same approach could be applied to you own question.
Phrases are dynamic, depending on the problem. They do not necessarily take a particular form.
The constant thing about phrases is the operators connecting the words in the phrases. Theses operators are:
Addition (+), Subtraction (-), Division (÷), and Multiplication (×).
In word problems, it is a matter of interpretation, these operators can be written in many ways.
ADDITION
plus
the sum of
increase
grow
add
profit
And so on.
SUBTRACTION
minus
loss
decrease
reduce
subtract
And so on
MULTIPLICATION
times
multiply
triple
And so on
DIVISION
split
share
divide
distribute
And so on.
Examples
(1) 56 is added to a number to give 100
Interpretation: x + 56 = 100
(2)The difference between Mr. A and Mr. B is 5
Interpretation: A - B = 5
(3) This load (L1) is three times heavier than that one (L2)
Interpretation: L1 = 3L2
(4) Share this orange (P) equally between the three children
Interpretation: P/3
To write expressions using the numbers 54 and 67 with operations, you can add (54 + 67), subtract (54 - 67), multiply (54 x 67), or divide (54 ÷ 67). The order of operations is important and parentheses may be used to dictate the sequence of calculations. Practice with expressions can be enhanced by working through problems symbolically and then substituting numbers.
Explanation:The question asks to write expressions using the numbers 54 and 67 with operations. When creating expressions, operations such as addition (+), subtraction (-), multiplication (x or ×), and division (÷) are typically used. For example, if we want to add 54 and 67, the expression would be 54 + 67.
If we wanted to multiply them, the expression would be 54 x 67 (or 54 × 67). Remember that in writing expressions, the order of operations is important, so if we want to perform different operations, we may need to use parentheses to ensure the correct sequence of calculations.
For more practice with expressions and operations, you can approach problems by covering up one number and solving for it using the remaining information (as mentioned in the given examples). This technique helps you solve various types of problems using the same principle. Moreover, when working with algebraic expressions, it's beneficial to solve the equation symbolically first and then substitute the numbers.
Substituting numbers and parentheses placement can significantly alter the result of the expression, as shown in the examples given in the question. The expressions created through experimentation, such as (74) to the third power, can be solved symbolically or using a calculator to understand how different operations and powers affect the result.
How many zeros are at the end of457 · 885?Explain how you can answer this question without actually computing the number. (Hint:10 = 2 · 5.)When this number is written in ordinary decimal form, each 0 at its end comes from a factor of , or one factor of 2 and one factor of .Since there are factors of 2 and factors of 5, there are exactly factors of 10 in the number. This implies that the number ends with zeroes.
The right format of the number is (45^8)(88^5).
Answer:
There are 8 zeros
Step-by-step explanation:
Using the unique factorization of integers theorem, we can break any integer down into the product of prime integers.
So breaking it down we have;
(45^8) = (3 x 3 x 5)^(8)
(88^5) = (2 x 2 x 2 x 11)^(5)
Now, if we put it back together as separate factors, we'll get;
(3^(16)) x (5^(8) ) x (2^(15)) x (11^(5))
Now let's find the number of zeroes by figuring out how many factors of 10 (which equals 2 x 5) we can make. Thus, we can make 8 factors of 10 so it looks like;
(3^(16)) x (2^(7)) x (11^(5)) x (10^(8))
Thus, we can see that there will be 8 zeros as the end is (10^(8))
The number of trailing zeros in the product of 457 and 885 is determined by the factors of 10 (2 and 5) in these numbers. In this specific case, there are no trailing zeros. This also applies in scientific notation - the number of significant figures after the decimal in scientific notation indicates the quantity of zeros at the end of the number.
Explanation:To determine the number of zeroes at the end of the number 457 · 885, consider the factors in the product. Each zero at the end of a number results from a factor of 10, which contains a factor of 2 and a factor of 5. Looking at the numbers 457 and 885, we notice that neither has a factor of 5, therefore there are no trailing zeroes in the product of 457 and 885.
This approach also applies to the scientific notation. The number of significant figures after the decimal in the scientific notation of a number corresponds to the quantity of zeros at the end of it. In such cases, leading zeros are not significant and only serve as placeholders to locate the decimal point. For instance, in the case of the number 1.300 × 10³, the scientific notation shows that there are three significant figures after the decimal and therefore, three zeros at the end of the number.
Overall, understanding how to find the number of trailing zeroes in a product, such as 457 · 885, without actual computation involves a knowledge of the factors of the numbers being multiplied and the principles of significant figures in scientific notation.
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The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population P0 has doubled in 5 years. Suppose it is known that the population is 8,000 after 3 years. What was the initial population P0? (Round your answer to one decimal place.) P0 = What will be the population in 10 years? (Round your answer to the nearest person.) persons How fast is the population growing at t = 10? (Round your answer to the nearest person.) persons/year
Answer:
5278.0
21112
2927
Step-by-step explanation:
P = Po[2^(t/5)]
8000 = Po(2^⅗)
Po = 5278.0
P = 5278(2^(10/5))
P = 21112
P = Po[2^(t/5)]
ln(P/Po) = (t/5)ln2
ln(P) - ln(Po) = (t/5)ln2
1/P . dP/dt = ln2/5
dP/dt = P(ln2)/5
At t = 10, P = 21112
dP/dt = 2927
Consider the experiment of rolling a pair of dice. Suppose that we are interested in the sum of the face values showing on the dice. (a) How many sample points are possible? (Hint: use the counting rule for multiple-step experiments.) (b) List the sample points. There to sum the face values of a pair of dice to 2. There to sum the face values of a pair of dice to 3. There to sum the face values of a pair of dice to 4. There to sum the face values of a pair of dice to 5. There to sum the face values of a pair of dice to 6. There to sum the face values of a pair of dice to 7. There to sum the face values of a pair of dice to 8. There to sum the face values of a pair of dice to 9. There to sum the face values of a pair of dice to 10. There to sum the face values of a pair of dice to 11. There to sum the face values of a pair of dice to 12. (c) What is the probability of obtaining a value of 5? (d) What is the probability of obtaining a value of 8 or greater? (e) Because each roll has six possible even values (2, 4, 6, 8, 10, and 12) and only five possible odd values (3, 5, 7, 9, and 11), the dice should show even values more often than odd values. Do you agree with this statement? Explain. This statement correct because P(odd) = and P(even) = . (f) What method did you use to assign the probabilities requested? classical method empirical method subjective method relative frequency method
a) 21 sample points
b) Sum 2: (1, 1)
- Sum 3: (1, 2)
- Sum 4: (1, 3), (2, 2)
- Sum 5: (1, 4), (2, 3)
- Sum 6: (1, 5), (2, 4), (3, 3)
- Sum 7: (1, 6), (2, 5), (3, 4)
- Sum 8: (2, 6), (3, 5), (4, 4)
- Sum 9: (3, 6), (4, 5)
- Sum 10: (4, 6), (5, 5)
- Sum 11: (5, 6)
- Sum 12: (6, 6)
c) Probability of obtaining sum of 5 is 2/21
d) Probability of obtaining 8 or greater is 3/7
e) Probability of even is higher that the probability of odd, so even sum are expect to have more appear.
f) classic method
What is probability?
(a) There are 21 possible sample points when rolling a pair of dice.
(b) Here are the sample points for each sum:
- Sum 2: (1, 1)
- Sum 3: (1, 2)
- Sum 4: (1, 3), (2, 2)
- Sum 5: (1, 4), (2, 3)
- Sum 6: (1, 5), (2, 4), (3, 3)
- Sum 7: (1, 6), (2, 5), (3, 4)
- Sum 8: (2, 6), (3, 5), (4, 4)
- Sum 9: (3, 6), (4, 5)
- Sum 10: (4, 6), (5, 5)
- Sum 11: (5, 6)
- Sum 12: (6, 6)
(c) The probability of obtaining a sum of 5 is
: (1, 4), (2, 3)
P(sum of 5) = no of sum of 5 /total number of our
= 2/21
(d) The probability of obtaining a sum of 8 or greater is (2, 6), (3, 5), (4, 4) (3, 6), (4, 5) (4, 6), (5, 5),(5, 6) (6, 6)
P(sum of 8 or greater) = no of sum of 8 and above
P(sum => 8) = 9/21 = 3/7
(e) Yes!
Out of 21 sample points 12 are even while 9 are odd.
P(even) = 12/21 = 4/7
P(odd) = 1 - 4/7 = 3/7
Probability of even is higher that the probability of odd, so even sum are expect to have more appear.
(f) We used the classical method, which involves counting the number of favorable outcomes and dividing by the total number of possible outcomes.
Suppose that a softball team is composed of 15 employees of a furniture store of whom 2 work part-time. What proportion of the team work part-time?
Answer:
Therefore, we conclude that 2/15 of the team work part-time.
Step-by-step explanation:
We know that a softball team is composed of 15 employees of a furniture store of whom 2 work part-time. We calculate what proportion of the team work part-time.
So, we will divide the number of those people who work part-time, by the number of people employed in a softball team. We get:
x=2/15
Therefore, we conclude that 2/15 of the team work part-time.