a. Null hypothesis (H0): Mean cost for the new production method is $215 per hour.
Alternative hypothesis (H1): Mean cost for the new production method is less than $215 per hour.
b. Type I error: Rejecting the null hypothesis when it is true; consequence: Implementing the new method incorrectly and incurring unnecessary costs.
c. Type II error: Failing to reject the null hypothesis when it is false; consequence: Missing the opportunity to adopt a cost-effective production method and continuing with higher operating costs.
We have,
a. The appropriate null and alternative hypotheses would be:
Null Hypothesis (H0): The mean cost for the new production method is equal to or greater than $215 per hour.
Alternative Hypothesis (H1): The mean cost for the new production method is less than $215 per hour.
b. The Type I error in this situation would be rejecting the null hypothesis when it is actually true.
In other words, it would mean concluding that the new production method reduces the mean operating cost per hour when it actually doesn't.
The consequence of making this error is that resources, time, and effort might be invested in implementing the new method based on incorrect information, potentially leading to unnecessary costs and inefficiencies.
c. The Type II error in this situation would be failing to reject the null hypothesis when it is actually false.
In other words, it would mean failing to conclude that the new production method reduces the mean operating cost per hour when it actually does.
The consequence of making this error is that the opportunity to adopt a more cost-effective production method would be missed, potentially leading to continued higher operating costs and missed efficiency gains.
Thus,
a. Null hypothesis (H0): Mean cost for the new production method is $215 per hour.
Alternative hypothesis (H1): Mean cost for the new production method is less than $215 per hour.
b. Type I error: Rejecting the null hypothesis when it is true; consequence: Implementing the new method incorrectly and incurring unnecessary costs.
c. Type II error: Failing to reject the null hypothesis when it is false; consequence: Missing the opportunity to adopt a cost-effective production method and continuing with higher operating costs.
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The null and alternative hypotheses are set as μ ≥ 215 and μ < 215 respectively. A Type I error, rejecting the null hypothesis when it's true, could lead to unnecessary changes. A Type II error, failing to reject the null hypothesis when it's false, could result in missed cost savings.
Explanation:a. In this situation, the null hypothesis (H0) would state that the mean cost of the new production method is equal to or greater than 215. We can denote this as: H0: μ ≥ 215. The alternative hypothesis (Ha) posits that the mean cost of the new production method is less than 215, denoted as: Ha: μ < 215.
b. A Type I error would occur if we incorrectly reject the null hypothesis, concluding that the new method reduces the cost when, in fact, it does not. The consequence could be implementing a production method that doesn't actually provide the desired savings, which could lead to unnecessary changes and associated costs.
c. A Type II error would happen if we fail to reject the null hypothesis when it's actually false, meaning the new method does reduce the cost but we fail to recognize it. The consequence of this error could be missing out on an opportunity to reduce operating costs, potentially leading to higher than necessary expenses.
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Researchers conducted a study of obesity in children. They measured body mass index (BMI), which is a measure of weight relative to height. High BMI is an indication of obesity. Data from a study published in the Journal of the American Dietetic Association shows a fairly strong positive linear association between mother’s BMI and daughter’s BMI (r = 0.506). This means that obese mothers tend to have obese daughters.
1. Based on this study, what proportion of the variation in the daughter BMI measurements is explained by the mother BMI measurements?
2. What are some of the other variables that explain the variability in the daughter BMI?
Answer:
Part a
For this case after do the operations we got a value for the correlation coeffcient of:
[tex] r =0.506[/tex]
With this value we can find the determination coefficient:
[tex] r^2 = 0.506^2 = 0.256[/tex]
And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.
Part a
Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)
Step-by-step explanation:
Previous concepts
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
Solution to the problem
Part a
For this case after do the operations we got a value for the correlation coeffcient of:
[tex] r =0.506[/tex]
With this value we can find the determination coefficient:
[tex] r^2 = 0.506^2 = 0.256[/tex]
And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.
Part a
Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)
Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 millimeters and a standard deviation of 2 millimeters. What is the probability that a measurement exceeds 13 milliamperes
Answer: the probability that a measurement exceeds 13 milliamperes is 0.067
Step-by-step explanation:
Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = current measurements in a strip.
µ = mean current
σ = standard deviation
From the information given,
µ = 10
σ = 2
We want to find the probability that a measurement exceeds 13 milliamperes. It is expressed as
P(x > 13) = 1 - P(x ≤ 13)
For x = 13,
z = (13 - 10)/2 = 1.5
Looking at the normal distribution table, the probability corresponding to the z score is 0.933
P(x > 13) = 1 - 0.933 = 0.067
Final answer:
To find the probability that a current measurement exceeds 13 milliamperes in a normally distributed set with mean 10 mA and SD 2 mA, calculate the Z-score and use a normal distribution table or software.
Explanation:
The student's question seems to mistakenly mix units (millimeters instead of milliamperes), but assuming the intent was to refer to electrical current and not physical measurements of wire, we will proceed on the basis that the actual question is about the probability of a current measurement exceeding 13 milliamperes.
To calculate the probability that a measurement exceeds 13 milliamperes when the current measurements in a strip of wire are normally distributed with a mean of 10 milliamperes and standard deviation of 2 milliamperes, we need to use the Z-score formula:
Z = (X - μ) / σ
where X is the value in question (13 milliamperes), μ is the mean (10 milliamperes), and σ is the standard deviation (2 milliamperes).
Plugging in the values:
Z = (13 - 10) / 2 = 1.5
After finding the Z-score, we would look up this value in a standard normal distribution table or use a statistical software to find the probability that Z exceeds 1.5, which gives us the probability that a measurement exceeds 13 milliamperes. For this Z-score, the probability is approximately 6.68%.
The weights of newborn children in the United States vary according to the Normal distribution with mean 7.5 pounds and standard deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.5 pounds. (a) What is the probability that a baby chosen at random weighs less than 5.5 pounds at birth?(b) You choose three babies at random. What is the probability that their average birth weight is less than 5.5 pounds?
Answer:
a) 5.48% probability that a baby chosen at random weighs less than 5.5 pounds at birth
b) 0.28% probability that their average birth weight is less than 5.5 pounds
Step-by-step explanation:
To solve this question, the normal probability distribution and the central limit theorem are used.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means n of at least 30 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 = 7.5, \sigma = 1.25[/tex]
(a) What is the probability that a baby chosen at random weighs less than 5.5 pounds at birth?
This is the pvalue of Z when X = 5.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.5 - 7.5}{1.25}[/tex]
[tex]Z = -1.6[/tex]
[tex]Z = -1.6[/tex] has a pvalue of 0.0548
5.48% probability that a baby chosen at random weighs less than 5.5 pounds at birth
(b) You choose three babies at random. What is the probability that their average birth weight is less than 5.5 pounds?
[tex]n = 3, s = \frac{1.25}{\sqrt{3}} = 0.7217[/tex]
This is the pvalue of Z when X = 5.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
This is the pvalue of Z when X = 5.5. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{5.5 - 7.5}{0.7217}[/tex]
[tex]Z = -2.77[/tex]
[tex]Z = -2.77[/tex] has a pvalue of 0.0028
0.28% probability that their average birth weight is less than 5.5 pounds
A sampling distribution refers to the distribution of:
A. a sample
B. a population
C. a sample statistic
D. a population parameter
E. repeated samples
F. repeated populations
Answer:
The answer is a population parameter.
Step-by-step explanation:
Population can include people, but other examples include objects, event, businesses, and so on. Population is the entire pool from which statistical sample is drawn.
A parameter is a value that describes a characteristics of an entire population, such as population mean, because you can almost never measure an entire population, you usually don't know the real value of a parameter.
Consider all possible sample of size N that can be drawn from a given population (either with or without replacement). For example, we can compute a statistics (such as the mean and the standard deviation ) that will vary from sample to sample. In this manner we obtain a distribution of statistics that is called Sampling distribution.
In statistics, a sampling distribution is the theoretical distribution of a sample statistic that arises from drawing all possible samples of a specific size from a population. It helps to quantify the variability and predictability of sample statistics when used as estimates for population parameters.
Explanation:A sampling distribution refers to the "distribution of a sample statistic". This is option C from your list. This term describes the probability distribution of a statistic based on a random sample. For example, if we study random samples of a certain size from any population, the mean score will form a distribution. This is the sampling distribution of the mean. Similarly, variance, standard deviations and other statistics also have sampling distributions. The purpose of a sampling distribution is to quantify the variation and uncertainty that arises when we use sample statistics (like the mean) to estimate population parameters (like the population mean).
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A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
[tex]F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}[/tex]
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.
[tex]F(X) = P ( 3=<X =< 5) = [1 - \frac{1}{(x+1)*(x+2)}]\limits^5_3\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(5+1)*(5+2)} + \frac{1}{(3+1)*(3+2)}}\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(42)} + \frac{1}{(20)}}]\\\\F(X) = P ( 3=<X =< 5) = 0.02619[/tex]
- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )
[tex]E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}[/tex]
E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
The average heights of a random sample of 400 people from a city is 1.75 m. It is known that the heights of the population are random variables that follow a normal distribution with a variance of 0.16.
Determine the interval of 95% confidence for the average heights of the population.
Answer:
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
The standard deviation is the square root of the variance. So
[tex]\sigma = \sqrt{0.16} = 0.4[/tex]
Then
[tex]M = 1.96*\frac{0.4}{\sqrt{400}} = 0.0392[/tex]
The lower end of the interval is the mean subtracted by M. So it is 1.75 - 0.0392 = 1.7108m
The upper end of the interval is the mean added to M. So it is 1.75 + 0.0392 = 1.7892m
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Answer:
The interval of 95% confidence for the average heights of the population is = [tex](1.7108, 1.7892)[/tex]Step-by-step explanation:
mean x = [tex]1.75[/tex]
Variance [tex]\rho^2 = 0.16[/tex]
standard deviation [tex](\rho) = \sqrt{0.16} = 0.4[/tex]
n = 400
[tex]95\%[/tex] confidence :
[tex]\alpha = 100\% - 95\% = 5\%\\\\\frac{\alpha}{2} = 2.5\% = 0.025[/tex]
From standard normal distribution table,
[tex]Z_\frac{\alpha}{2} = Z_{0.025} = 1.96[/tex]
Margin of error, [tex]E = Z_\frac{\alpha}{2} * \frac{\rho}{\sqrt{n}}[/tex]
[tex]E = 1.96 * \frac{0.4}{\sqrt{400}}\\\\E = 0.0392[/tex]
Lower limit: x - E
[tex]= 1.75 - 0.0392\\\\= 1.7108[/tex]
Upper limit: x + E
[tex]= 1.75 + 0.0392\\\\= 1.7892[/tex]
[tex]Limits : (1.7108, 1.7892)[/tex]
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Suppose that the warranty cost of defective widgets is such that their proportion should not exceed 5% for the production to be profitable. Being very cautious, you set a goal of having 0.05 as the upper limit of a 90% confidence interval, when repeating the previous experiment. What should the maximum number of defective widgets be, out of 1024, for this goal to be reached.
Answer: 63 defective widgets
Step-by-step explanation:
Given that the proportion should not exceed 5%, that is:
p< or = 5%.
So we take p = 5% = 0.05
q = 1 - 0.05 = 0.095
Where q is the proportion of non-defective
We need to calculate the standard error (standard deviation)
S = √pq/n
Where n = 1024
S = √(0.05 × 0.095)/1024
S = 0.00681
Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.
UCL = p + Za/2(n-1) × S
Where a is alpha of confidence interval = 100 -90 = 10%
a/2 = 5% = 0.05
UCL = p + Z (0.05) × 0.00681
Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.
Z a/2 = 1.64
UCL = 0.05 + 1.64 × 0.00681
UCL = 0.0612
Since the UCL in this case is a measure of proportion of defective widgets
Maximum defective widgets = 0.0612 ×1024 = 63
Alternatively
UCL = p + 3√pq/n
= 0.05 + 3(0.00681)
= 0.05 + 0.02043 = 0.07043
UCL =0.07043
Max. Number of widgets = 0.07043 × 1024
= 72
The weight of adobe bricks for construction is normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 25 bricks is chosen.
a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places
b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.
Answer:
a) 0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds
b) 2.88 pounds
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 are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
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 = 3, \sigma = 0.25, n = 25, s = \frac{0.25}{\sqrt{25}} = 0.05[/tex]
a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places
This is the pvalue of Z when X = 3.10. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3.1 - 3}{0.05}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds
b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.
This is the value of X when Z has a pvalue of 1-0.99 = 0.01. So X when Z = -2.33.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-2.33 = \frac{X - 3}{0.05}[/tex]
[tex]X - 3 = -2.33*0.05[/tex]
[tex]X = 2.88[/tex]
The null hypothesis in ANOVA is that all means of all groups are the same. The alternative is that at least one pair of means is different. We compute an F-statistic to explore sources of variability in our data to conduct the omnibus ANOVA. Question: what do you expect to happen when the null hypothesis is true?
A. More between group variability
B. Less between group variability
Answer:
Correct option: B. Less between group variability
Step-by-step explanation:
The Analysis of Variance (ANOVA) test is performed to determine whether there is a significant difference between the different group mean.
The hypothesis is defined as:
H₀: There is no difference between the group means, i.e. μ₁ = μ₂ = ... = μₙ
Hₐ: At least one of the mean is different from the others, i.e. μ[tex]_{i}[/tex] ≠ 0.
The test statistic is defined as:
[tex]F=\frac{SS_{between}}{SS_{within}}[/tex]
If the null hypothesis is true then the test statistic will be small and if it is false then the test statistic will be large.
In this case it is provided that the null hypothesis is true.
This implies that:
[tex]SS_{between}<SS_{within}[/tex]
Implying that the sum of squares for between group variability is less than within group variability.
Thus, if the null hypothesis is true there will be less between group variability.
A certain paper suggested that a normal distribution with mean 3,500 grams and a standard deviation of 560 grams is a reasonable model for birth weights of babies born in Canada.
One common medical definition of a large baby is any baby that weighs more than 4,000 grams at birth.
What is the probability that a randomly selected Canadian baby is a large baby?
The probability that a randomly selected Canadian baby is a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.
Explanation:To find the probability that a randomly selected Canadian baby is a large baby, we need to calculate the area under the normal distribution curve to the right of 4,000 grams. First, we calculate the z-score using the formula: z = (x - mean) / standard deviation. Plugging in the values, we get z = (4000 - 3500) / 560 = 0.8929.
Next, we need to find the area under the curve to the right of this z-score using a standard normal distribution table or a calculator. The cumulative probability from the table or calculator is approximately 0.187. This means that the probability of a randomly selected Canadian baby being a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.
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Use false-position method to determine the drag coefficient needed so that an 95-kg bungee jumper has a velocity of 46 m/s after 9 s of free fall. Note: The acceleration of gravity is 9.81 m/s2. Start with initial guesses of xl = 0.2 and xu = 0.5 and iterate until the approximate relative error falls below 5%.
Answer:
solution attached below
Step-by-step explanation:
To determine the drag coefficient using the false-position method, start with initial guesses and iterate until the approximate relative error falls below 5%.
Explanation:To determine the drag coefficient needed for a bungee jumper to have a velocity of 46 m/s after 9 s of free fall using the false-position method, we can follow these steps:
Start with initial guesses of xl = 0.2 and xu = 0.5.Calculate the velocity at 9 s using the false-position method.If the calculated velocity is greater than 46 m/s, update xu with the calculated drag coefficient. If the calculated velocity is less than 46 m/s, update xl with the calculated drag coefficient.Repeat steps 2 and 3 until the approximate relative error falls below 5%.The final value of the drag coefficient will be the approximate solution.Learn more about Drag coefficient here:
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Let P(n) be the statement that a postage of n cents can be formed using just 4-cent and 7-cent stamps. Use strong induction to prove that P(n) is true for all integers greater than or equal to some threshold x.
Answer:
True for n = 18, 19, 20, 21
Step-by-step explanation:
[tex]P(n) =[/tex] a postage of [tex]n[/tex] cents; where [tex]P(n) = 4x + 7y[/tex]. ( [tex]x[/tex] are the number of 4-cent stamps and [tex]y[/tex] are the number of 7-cent stamps)
For [tex]n=18, P(18)[/tex] is true.
This is a possibility, if [tex]x= 1 \ and \ y=2[/tex]
[tex]P(18) = 4(1) + 7(2) = 4 + 14 = 18[/tex]
Similarly for [tex]P(19)[/tex]:
[tex]P(19) = 4(3) + 7(0) = 19[/tex]
[tex]P(20) = 4(5) + 7(0) = 20\\P(21) = 4(0) + 7(3) = 21[/tex]
Solve the equation M=7r2h/19 for r in terms of M and h. Assume r, M and h are all positive.
Answer:
[tex]r=\frac{19M}{14h}[/tex]
Step-by-step explanation:
The equation is given as:
[tex]M=\frac{7r2h}{19}[/tex]
Assuming all the unknown variables are positive, we can make [tex]r[/tex] the subject of the formula to obtain it in terms of M & h:
[tex]M=\frac{7r2h}{19}\\M\times19=7r2h\\\\\frac{19M}{2h}=7r\\\\r=\frac{19M}{2h\times7}\\\\r=\frac{19M}{14h}[/tex]
or [tex]r=1.3571M/h[/tex]
Hence, r as in terms of M& H is given as
[tex]r=\frac{19M}{14h} \ or \ 1.3571M/h[/tex]
A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. How many gallons of the 20% solution and 50%
solution should be mixed?
Answer: 60 gallons of the 20% solution and 120 gallons of the 50% solution should be mixed.
Step-by-step explanation:
Let x represent the number of gallons of 20% solution that should be mixed.
Let y represent the number of gallons of 50% solution that should be mixed.
A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. This means that
0.2x + 0.5y = 0.4×180
0.2x + 0.5y = 72- - - - - - - - - - - -1
Since the total number of gallons is 180, it means that
x + y = 180
Substituting x = 180 - y into equation 1, it becomes
0.2(180 - y) + 0.5y = 72
36 - 0.2y + 0.5y = 72
- 0.2y + 0.5y = 72 - 36
0.3y = 36
y = 36/0.3
y = 120
x = 180 - y = 180 - 120
x = 60
To get a 40% solution by mixing a 20% and 50% solution, you need 60 gallons of the 20% solution and 120 gallons of the 50% solution.
Explanation:To solve this problem, we can use the concept of mixtures. Let's represent the amount of 20% solution as x gallons and the amount of 50% solution as y gallons. From the given information, we can set up the following equations:
x + y = 180 (equation 1)
0.2x + 0.5y = 0.4 * 180 (equation 2)
Simplifying equation 2 gives us 0.2x + 0.5y = 72. To eliminate decimals, we can multiply both sides by 10 to get 2x + 5y = 720.
Now we have a system of equations. We can solve it by substitution or elimination. Let's use elimination:
Multiplying equation 1 by 2 gives us 2x + 2y = 360. Subtracting this from equation 2 gives us 3y = 360. Dividing both sides by 3 gives us y = 120. Substituting this value into equation 1 gives us x + 120 = 180. Subtracting 120 from both sides gives us x = 60.
Therefore, we need 60 gallons of the 20% solution and 120 gallons of the 50% solution.
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A tortoise and a hare are competing in a 1600-meter race. The arrogant hare decides to let the tortoise have a 510-meter head start. When the start gun is fired the hare begins running at a constant speed of 9 meters per second and the tortoise begins crawling at a constant speed of 5 meters per second.
a. Define a function f to represent the tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race.
b. Solve f(t)=0for t.
c. Define a function g to represent the hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race.
d. Solve g(t)=0 for t
e. Who won the race?
Answer:
(a)f(t)=1090-5t
(b)f(0)=1090metres
(c)g(t)=1600-9t
(d)g(0)=1600metres
(e)The Hare
Step-by-step explanation:
Total Distance =1600 metres
(a)The tortoise has a 510m headstart and a speed of 5m/s
Distance=Speed X Time
Distance at 5m/s = 5t
Total Distance covered by the tortoise at any time t= 510+5t
Therefore, The tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:
f(t)=1600-(510+5t)
f(t)=1090-5t
(b)f(0)=1090-(5X0)
=1090metres
(c)The hare has a speed of 9m/s
Distance=Speed X Time
Distance at 9m/s = 9t
Total Distance covered by the hare at any time t= 9t
Therefore, The hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:
g(t)=1600-9t
(d)
g(0)=1600-(9X0)=1600metres
(e)The race is finished when the distance from the finish line=0
For the Tortoise
f(t)=1090-5t=0
1090=5t
t=218seconds
For the Hare
g(t)=1600-9t=0
9t=1600
t=177.8seconds
The hare takes a shorter time to reach the finish line so he won the race.
Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 79 and a standard deviation of 6. The top 15% of all scores have been designated As. Your score is 89. Did you earn an A
Answer:
You earned an A.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 79, \sigma = 6[/tex]
The top 15% of all scores have been designated As.
This means that if Z for the score has a pvalue of 1-0.15 = 0.85 or higher, the score is designated as A.
Your score is 89. Did you earn an A?
We have to find the pvalue of Z when X = 89. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89 - 79}{6}[/tex]
[tex]Z = 1.67[/tex]
[tex]Z = 1.67[/tex] has a pvalue of 0.9525. So yes, you earned an A.
It is often said that your chances of winning the lottery if you buy a ticket are just slightly higher than if you don't buy one! Suppose a Lotto game consists of picking 6 of 48 numbers.
What is the probability of winning with the very first Lotto ticket you purchase?
Answer:
1/48 % or 6/48 % chance
Step-by-step explanation:
The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.
What is a permutation?A permutation is a process of calculating the number of ways to choose a set from a larger set in a particular order.
If we want to choose a set of r items from a set of n items in a particular order, we find the permutation nPr = n!/(n-r)!.
What is a combination?A combination is a process of calculating the number of ways to choose a set from a larger set in no particular order.
If we want to choose a set of r items from a set of n items in no particular order, we find the combination nCr = n!/{(r!)(n-r)!}.
How do we solve the given question?In the question, we are asked to determine the probability of winning a lottery by picking 6 numbers from 48 numbers with the first ticket we purchase.
First, we need to calculate the number of combinations of choosing 6 numbers from 48 numbers. As we need to consider no particular order, we will use combinations,
48C6 = 48!/{(6!)(48-6)!} = 48!/(6!*42!) = (43*44*45*46*47*48)/(1*2*3*4*5*6*) (As 48! = 42!*43*44*45*46*47*48, and 42! cancels itself from the numerator and the denominator).
or, 48C6 = 12,271,512.
So, we get the number of combinations = 12,271,512.
We know that we will choose only one particular set of 6 numbers.
∴ The probability of winning on the very first ticket = 1/12,271,512 ≈ 0.0000000815
∴ The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.
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The average daily high temperature in June in LA is 77 degree F with a standard deviation of 5 degree F. Suppose that the temperatures in June closely follow a normal distribution. What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June? How cold are the coldest 10% of the days during June in LA?
Answer:
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 77, \sigma = 5[/tex]
What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June?
This probability is 1 subtracted by the pvalue of Z when X = 83. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{83 - 77}{5}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a pvalue of 0.8849.
1 - 0.8849 = 0.1151
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
How cold are the coldest 10% of the days during June in LA?
High temperatures of X or lower, in which X is found when Z has a pvalue of 0.1, so whn Z = -1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 77}{5}[/tex]
[tex]X - 77 = -1.28*5[/tex]
[tex]X = 70.6[/tex]
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
A) The probability of observing a temperature ≥ 83°F in LA during a randomly chosen day in June is;
p(observing a temperature ≥ 83°F) = 11.507%
B) The coldest 10% of the days during June in LA have temperatures;
Less than or equal to 70.592 °F
This question involves z-distribution which is given by the formula;
z = (x' - μ)/σ
We are given;
Average daily temperature; μ = 77 °F
Standard deviation; σ = 5 °F
Since the temperatures follow a normal distribution, then if we want to find the probability of observing a temperature ≥ 83°F, then;
x' = 83 °F
Thus;
z = (83 - 77)/5
z = 6/5
z = 1.2
Thus;
from online z-score calculator, p-value = 0.11507
Thus, p(observing a temperature ≥ 83°F) = 11.507%
B) We want to find out how cold the coldest 10% of the days during June in LA;
Thus, it means that p = 10% = 0.1
z-score at p = 0.1 from z-score tables is;
z = -1.28155
Thus;
-1.28155 = (x' - 77)/5
-1.28155*5 = x' - 77
-6.40775 = x' - 77
x' = 77 - 6.40775
x' ≈ 70.592 °F
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A rocket is launched upward from a launching pad and the function h determines the rocket's height above the launching pad (in miles) given a number of minutes t since the rocket was launched.
What does the equality h(4) = 516 convey about the rocket in this context? Select all that apply.
(A) The rocket travels 516 miles every 4 minutes
(B) 14 minutes after the rocket was launched, the rocket is 516 miles above the ground
(C) The rocket is currently 516 miles above the ground
(D) 516 minutes after the rocket was launched, the rocket is 4 miles above the ground
Answer:
(B) 4 minutes after the rocket was launched, the rocket is 516 miles above the ground
Step-by-step explanation:
The function h(t) represents the height of the rocket above the launchpad after a time t minutes.
h(4) = 516 means that when t = 4 minutes, the height h is 516 miles above the launch pad. Note that the time t is measured from when the rocket is launched.
The first option indicates a rate of change, which is a velocity. This is not indicated in the question because velocity is a time derivative of the the height function.
Option C implies that the rocket is 516 miles currently but we do not know what time currently is from the time of launch.
The fourth option has reversed the roles of the variable by implying the time is 516 minutes while the height is 4 miles which is not what the function means.
The probability that an archer hits her target when it is windy is 0.4; when it is not windy, her probability of hitting the target is 0.7. On any shot, the probability of a gust of wind is 0.3. Find the probability that a. on a given shot there is a gust of wind and she hits her target.
b. she hits the target with her first shot.
c. she hits the target exactly once in two shots.
d. there was no gust of wind on an occasion when she missed.
Answer:
a) the probability is 0.12 (12%)
b) the probability is 0.61 (61%)
c) the probability is 0.476 (47.6%)
a) the probability is 0.538 (53.8%)
Step-by-step explanation:
a) denoting the event H= hits her target and G= a gust of wind appears hen
P(H∩G) = probability that a gust of wind appears * probability of hitting the target given that is windy = 0.3* 0.4 = 0.12 (12%)
b) for any given shot
P(H)= probability that a gust of wind appears*probability of hitting the target given that is windy + probability that a gust of wind does not appear*probability of hitting the target given that is not windy = 0.3*0.4+0.7*0.7 = 0.12+0.49 = 0.61 (61%)
c) denoting P₂ as the probability of hitting once in 2 shots and since the archer can hit in the first shot or the second , then
P₂ = P(H)*(1-P(H))+ (1-P(H))*P(H) = 2*P(H) *(1-P(H)) = 2*0.61*0.39= 0.476 (47.6%)
d) for conditional probability we can use the theorem of Bayes , where
M= the archer misses the shot → P(M) = 1- P(H) = 0.39
S= it is not windy when the archer shots → P(S) = 1- P(G) = 0.7
then
P(S/M) = P(S∩M)/P(M) = 0.7*(1-0.7)/0.39 = 0.538 (53.8%)
where P(S/M) is the probability that there was no wind when the archer missed the shot
Past records indicate that the probability of online retail orders
that turn out to be fraudulent is 0.08. Suppose that, on a given
day, 20 online retail orders are placed. Assume that the number of
online retail orders that turn out to be fraudulent is distributed as a
binomial random variable.
a. What are the mean and standard deviation of the number of online
retail orders that turn out to be fraudulent?
b. What is the probability that zero online retail orders will turn
out to be fraudulent?
c. What is the probability that one online retail order will turn out
to be fraudulent?
d. What is the probability that two or more online retail orders
will turn out to be fraudulent?
Answer:
a) Mean = 1.6, standard deviation = 1.21
b) 18.87% probability that zero online retail orders will turn out to be fraudulent.
c) 32.82% probability that one online retail order will turn out to be fraudulent.
d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The mean of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]p = 0.08, n = 20[/tex]
a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?
Mean
[tex]E(X) = np = 20*0.08 = 1.6[/tex]
Standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21[/tex]
b. What is the probability that zero online retail orders will turn out to be fraudulent?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887[/tex]
18.87% probability that zero online retail orders will turn out to be fraudulent.
c. What is the probability that one online retail order will turn out to be fraudulent?
This is P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282[/tex]
32.82% probability that one online retail order will turn out to be fraudulent.
d. What is the probability that two or more online retail orders will turn out to be fraudulent?
Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 1) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X \leq 1)[/tex]
In which
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
From itens b and c
[tex]P(X \leq 1) = 0.1887 + 0.3282 = 0.5169[/tex]
[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831[/tex]
48.31% probability that two or more online retail orders will turn out to be fraudulent.
The probability is an illustration of a binomial distribution.
The mean and the standard deviationThe given parameters are:
n = 20
p = 0.08
The mean is calculated as:
[tex]\bar x = np[/tex]
So, we have:
[tex]\bar x = 20 * 0.08[/tex]
[tex]\bar x = 1.6[/tex]
The standard deviation is calculated as:
[tex]\sigma = \sqrt{\bar x * (1 - p)[/tex]
This gives
[tex]\sigma = \sqrt{1.6 * (1 - 0.08)[/tex]
[tex]\sigma = 1.21[/tex]
Hence, the mean is 1.6 and the standard deviation is 1.21
The probability that zero online retail orders will turn out to be fraudulentThis is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(0) = ^{20}C_0 * 0.08^0 * (1 - 0.08)^{20 - 0}[/tex]
[tex]P(0) =0.1887[/tex]
The probability that zero online retail orders will turn out to be fraudulent is 0.1887
The probability that one online retail order will turn out to be fraudulentThis is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(1) = ^{20}C_1 * 0.08^1 * (1 - 0.08)^{20 - 1}[/tex]
[tex]P(1) =0.3281[/tex]
The probability that one online retail orders will turn out to be fraudulent is 0.3281
The probability that two or more online retail orders will turn out to be fraudulentThis is calculated as:
[tex]P(x\ge 2) = 1 - P(0) - P(1)[/tex]
So, we have:
[tex]P(x\ge 2) = 1 - 0.1887 - 0.3281[/tex]
[tex]P(x\ge 2) = 0.4832[/tex]
The probability that two or more online retail orders will turn out to be fraudulent is 0.4832
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g Each year the density of 7 species of Odonata (dragonflies and damselflies) is monitored in a wetland preserve. If the density of each species is to be compared with the density of every other species, how many comparisons must be made
Answer:
There are 21 comparisons to be made.
Step-by-step explanation:
The number of species of Odonata monitored every year is, n = 7.
It is provided that the density of each species is compared with each other.
The number of ways to compare the species (N) without repetition is:
[tex]N=\frac{n(n-1)}{2}\\=\frac{7(7-1)}{2}\\=\frac{7\times6}{2}\\=21[/tex]
Thus, there are 21 comparisons.
The 21 comparisons are as follows:
Specie 1 is compared with the remaining 6.
Specie 2 has already with he 1st so it is compared with the remaining 5.
Specie 3 has already with he 1st and 2nd so it is compared with the remaining 4.
Specie 4 has already with he 1st, 2nd and 3rd so it is compared with the remaining 3.
Specie 5 has already with he 1st, 2nd, 3rd and 4th so it is compared with the remaining 2.
Specie 6 has already with he 1st, 2nd, 3rd, 4th and 5th so it is compared with the remaining 1.
And the specie 7 has already been compared with the others.
Total number of comparisons = 6 + 5 + 4 + 3 + 2 + 1 = 21.
A survey of 8 adults employed full-time was taken. Here are their reported numbers of hours worked per week: 50, 53, 46, 46, 49, 43, 41, 41 (a) What is the mean of this data set? If your answer is not an integer, round your answer to one decimal place. (b) What is the median of this data set? If your answer is not an integer, round your answer to one decimal place. (c) How many modes does the data set have, and what are their values? Indicate the number of modes by clicking in the appropriate circle, and then indicate the value(s) of the mode(s), if applicable. zero modes one mode: two modes:
Answer:
a) 46.1
b) 46
c) Two modes: 46, 41
Step-by-step explanation:
We are given the following sample of hours per week:
50, 53, 46, 46, 49, 43, 41, 41
a) mean of this data set
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{369}{8} = 46.1[/tex]
b) Median of data set
[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]
Sorted data:
41, 41, 43, 46, 46, 49, 50, 53
Median =
[tex]=\dfrac{4^{th}+5^{th}}{2} = \dfrac{46+46}{2} = 46[/tex]
The median of data is 46.
c) Mode of the data set
Mode is the most frequent observation in the data.
The mode of the data are 46 and 41 as they appeared two times.
Thus, there are two modes.
The mean of the data set is 46.1, the median is 46, and there are two modes: 41 and 46.
To answer the questions based on the provided data set of hours worked by eight adults, we need to perform some basic statistical calculations.
(a) Mean
The mean is calculated by summing all the data points and then dividing by the number of data points:
Mean = (50 + 53 + 46 + 46 + 49 + 43 + 41 + 41) / 8 = 369 / 8 = 46.1
(b) Median
First, we need to arrange the data in ascending order: 41, 41, 43, 46, 46, 49, 50, 53. As there are 8 data points (even number), the median is the average of the 4th and 5th values:
Median = (46 + 46) / 2 = 46
(c) Mode
The mode is the number that appears most frequently in the data set. Here, 41 and 46 both appear twice:
There are two modes: 41 and 46 and the mean of the data set is 46.1.
Consider the parameterization of the unit circle given by x=cos(3t^2-t), y=sin(3t^2-t) for t in (-infinity, infinity). Describe in words and sketch how the circle is traced out, and use this to answer the following questions.
(a) When is the parameterization tracing the circle out in a clockwise direction? _________?
(Give your answer as a comma-separated list of intervals, for example, (0,1), (3,Inf)). Put the word None if there are no such intervals.
(b) When is the parameterization tracing the circle out in a counter-clockwise direction? ______?
(Give your answer as a comma-separated list of intervals, for example, (0,1), (3,Inf)). Put the word None if there are no such intervals.
(c) Does the entire unit circle get traced by this parameterization?
A. yes
B. no
(d) Give a time t at which the point being traced out on the circle is at (10):
t= ___________?
Answer and Step-by-step explanation:
The answer is attached below
In this exercise we have to use the knowledge of parameterization and calculate the direction and direction of the equation, so we have to:
A) Clockwise: [tex]t \in [ -\infty, 1/6][/tex]
B) Counter-clockwise: [tex]t \in [ 1/6, \infty][/tex]
C) [tex]\theta \in [ 0, 2 \pi][/tex]
D) [tex]t= 0 \ or \ t=1/3[/tex]
For this exercise, the following equations were informed:
[tex]x= cos(3t^2-t)\\y= sin(3t^2-t)\\t \in [ -\infty, \infty][/tex]
taking the parameterization we have that:
[tex]\phi = 3t^2 - t= t(3t-1)[/tex]
As t increases from [tex][ -\infty, \infty][/tex] [tex]\phi[/tex] decreases, after 0 it becomes negative and after 1/3, goes on increasing. Also:
[tex]\frac{d\phi}{dt} = (6t-1)\\t= 1/6[/tex]
a) For clockwise begin [tex]\phi[/tex] must be decreasing, so:
[tex]t \in [ -\infty, 1/6][/tex]
b) For counter-clockwise [tex]\phi[/tex] must be increasing, so:
[tex]t \in [ 1/6, \infty][/tex]
c) Entise circle gets traced out. For we know:
[tex]x= cos\theta\\y= sin\theta[/tex]
Circle gets traced out once for:
[tex]\theta \in [ 0, 2 \pi][/tex]
d) When point (1, 0) so:
[tex]1= cos(3t^2-t)\\0= sin(3t^2-t)\\t= 0 \or \ t=1/3[/tex]
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The rate at which a professional tennis player used carbohydrates during a strenuous workout was found to be 1.7 grams per minute. If a line were graphed showing time (in minutes) on the horizontal axis and carbohydrates used (in grams) on the vertical axis, what would be the slope of the line?
How many carbohydrates (in grams) would the athlete use in 40 minutes?
Answer:
m=1.7
C=68 gr
Step-by-step explanation:
Function Modeling
We are given a relationship between the carbohydrates used by a professional tennis player during a strenuous workout and the time in minutes as 1.7 grams per minute. Being C the carbohydrates in grams and t the time in minutes, the model is
[tex]C=1.7t[/tex]
The slope m of the line is the coefficient of the independent variable, thus m=1.7
The graph of C vs t is shown in the image below.
To find how many carbohydrates the athlete would use in t=40 min, we plug in the value into the equation
[tex]C=1.7\cdot 40=68\ gr[/tex]
The slope of the line representing the rate of carbohydrate usage is 1.7. Multiply this rate (1.7 grams per minute) by the time (40 minutes) to find the total carbohydrates used, which is 68 grams.
Explanation:The rate at which the tennis player uses carbohydrates is 1.7 grams per minute. In the context of a graph, this rate would represent the slope of the line. So, the slope of the line would be 1.7. Slope, in mathematics, is defined as the change in the y-value (vertical axis) divided by the change in the x-value (horizontal axis). Here, the rate of carbohydrate usage (1.7 grams per minute) is the change in the y-value (carbohydrates used) per change in x-value (time).
Now, you also want to know how many carbohydrates the athlete would use in 40 minutes. We know that the rate of carbohydrate usage is 1.7 grams per minute. So, to find the total amount of carbohydrates used in 40 minutes, you'd simply multiply the rate by the time:
1.7 grams/minute * 40 minutes = 68 grams
So, the athlete would use 68 grams of carbohydrates in 40 minutes.
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The Wall Street Journal reported that Walmart Stores Inc. is planning to lay off 2300 employees at its Sam's Club warehouse unit. Approximately half of the layoffs will be hourly employees (The Wall Street Journal, January 25-26, 2014). Suppose the following data represent the percentage of hourly employees laid off for 15 Sam's Club stores. 55 56 44 43 44 56 60 62 57 45 36 38 50 69 65 a. Compute the mean and median percentage of hourly employees being laid off at these stores. b. Compute the first and third quartiles. c. Compute the range and interquartile range. d. Compute the variance and standard deviation. e. Do the data contain any outliers? f. Based on the sample data, does it appear that Walmart is meeting its goal for reducing the number of hourly employees?
Answer:
(a) The mean is 52 and the median is 55.
(b) The first quartile is 44 and the third quartile is 60.
(c) The value of range is 33 and the inter-quartile range is 16.
(d) The variance is 100.143 and the standard deviation is 10.01.
(e) There are no outliers in the data set.
(f) Yes
Step-by-step explanation:
The data provided is:
S = {55, 56, 44, 43, 44, 56, 60, 62, 57, 45, 36, 38, 50, 69, 65}
(a)
Compute the mean of the data as follows:
[tex]\bar x=\frac{1}{n}\sum x\\=\frac{1}{15}[55+ 56+ 44+ 43+ 44+ 56+ 60+ 62+ 57+ 45 +36 +38 +50 +69+ 65]\\=\frac{780}{15}\\=52[/tex]
Thus, the mean is 52.
The median for odd set of values is the computed using the formula:
[tex]Median=(\frac{n+1}{2})^{th}\ obs.[/tex]
Arrange the data set in ascending order as follows:
36, 38, 43, 44, 44, 45, 50, 55, 56, 56, 57, 60, 62, 65, 69
There are 15 values in the set.
Compute the median value as follows:
[tex]Median=(\frac{15+1}{2})^{th}\ obs.=(\frac{16}{2})^{th}\ obs.=8^{th}\ observation[/tex]
The 8th observation is, 55.
Thus, the median is 55.
(b)
The first quartile is the middle value of the upper-half of the data set.
The upper-half of the data set is:
36, 38, 43, 44, 44, 45, 50
The middle value of the data set is 44.
Thus, the first quartile is 44.
The third quartile is the middle value of the lower-half of the data set.
The upper-half of the data set is:
56, 56, 57, 60, 62, 65, 69
The middle value of the data set is 60.
Thus, the third quartile is 60.
(c)
The range of a data set is the difference between the maximum and minimum value.
Maximum = 69
Minimum = 36
Compute the value of Range as follows:
[tex]Range =Maximum-Minimum\\=69-36\\=33[/tex]
Thus, the value of range is 33.
The inter-quartile range is the difference between the first and third quartile value.
Compute the value of IQR as follows:
[tex]IQR=Q_{3}-Q_{1}\\=60-44\\=16[/tex]
Thus, the inter-quartile range is 16.
(d)
Compute the variance of the data set as follows:
[tex]s^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{15-1}[(55-52)^{2}+(56-52)^{2}+...+(65-52)^{2}]\\=100.143[/tex]
Thus, the variance is 100.143.
Compute the value of standard deviation as follows:
[tex]s=\sqrt{s^{2}}=\sqrt{100.143}=10.01[/tex]
Thus, the standard deviation is 10.01.
(e)
An outlier is a data value that is different from the remaining values.
An outlier is a value that lies below 1.5 IQR of the first quartile or above 1.5 IQR of the third quartile.
Compute the value of Q₁ - 1.5 IQR as follows:
[tex]Q_{1}-1.5QR=44-1.5\times 16=20[/tex]
Compute the value of Q₃ + 1.5 IQR as follows:
[tex]Q_{3}+1.5QR=60-1.5\times 16=80[/tex]
The minimum value is 36 and the maximum is 69.
None of the values is less than 20 or more than 80.
Thus, there are no outliers in the data set.
(f)
Yes, the data provided indicates that the Walmart is meeting its goal for reducing the number of hourly employees
Answer:
Step-by-step explanation:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Write function that take a number n and return the sum of all the multiples of 3 or 5 below n.
Answer:
The function is provided below.
Step-by-step explanation:
Running the program in Python, the code for this problem is as follows:
def func(n):
count = 0 **initializing the sum with 0**
for i in range (1, n):
if i%3 == 0 or i%5 == 0:
count = count + 1
return (count)
When the program is executed enter value 1 to 10 one by one and the result will be 23, the sum of all the multiples of 3 and 5.
In previous years, the average number of sheets recycled per bin was 59.3 sheets, but they believe this number may have increase with the greater awareness of recycling around campus. They count through 79 randomly selected bins from the many recycle paper bins that are emptied every month and find that the average number of sheets of paper in the bins is 62.4 sheets. They also find that the standard deviation of their sample is 9.86 sheets.
What is the value of the test-statistic for this scenario? Round your answer to 3 decimal places.
What are the degrees of freedom for this t-test?
Answer:
There is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 59.3
Sample mean, [tex]\bar{x}[/tex] = 62.4
Sample size, n = 79
Alpha, α = 0.05
Sample standard deviation, s = 9.86
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 59.3\text{ sheets}\\H_A: \mu > 59.3\text{ sheets}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{62.4 - 59.3}{\frac{9.86}{\sqrt{79}} } = 2.7945[/tex]
Degree of freedom = n - 1 = 78
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 78 degree of freedom } = 1.6646[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.
Long-run classical model from Chapter 3. You must provide properly labeled graphs to get full credit!!!!!!! 3) A) Suppose there is a permanent increase in the labor force (L). a) What will be the impact on the real wage (W/P) and the real rental price of capital (R/P)
Answer:
Below the classical model, economic growth is necessarily achieved because of stability in the wage level. For instance, one case of unemployment predominates at a real wage (W / P)1.
Currently, the excessive labor supply would lower the actual wage level before labor supply equals its demand. Eventually, real wage rates would decline to (W / P)F, whereby aggregate labor demand is perfectly matches by aggregate labor supply.
Only the supply side of the production market for products defines the quantity of output & jobs in the classical model.
As the classical method is supply-determined, it states that equiproportional increases (or declines) will not alter the supply of labor in both the rate of money wage and the price level.
Consider the following homogeneous differential equation. y dx = 2(x + y) dy Use the substitution x = vy to write the given differential equation in terms of only y and v.
Answer:
[tex]ydv = (v +2)dy\\[/tex]
Step-by-step explanation:
We are given the following differential equation:
[tex]y dx = 2(x + y) dy[/tex]
We have to substitute
[tex]x = vy[/tex]
Differentiating we get,
[tex]\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}[/tex]
Putting value in differential equation, we get,
[tex]y dx = 2(x + y) dy\\\\y\dfrac{dx}{dy}=2(x+y)\\\\y(v+y\dfrac{dv}{dy}) = 2(vy + y)\\\\vy + y^2\dfrac{dv}{dy} = 2vy +2y\\\\y^2\dfrac{dv}{dy}=vy +2y\\\\y^2dv = y(v+2)dy\\ydv = (v +2)dy\\[/tex]
is the differential equation after substitution.
The given homogeneous differential equation y dx = 2(x + y) dy can be rewritten in terms of y and v using the substitution x = vy. The result is the differential equation y dv/dy = v.
Explanation:The given differential equation is y dx = 2(x + y) dy. To write his equation in terms of y and v using the substitution x = vy, we must first differentiate both sides of x = vy with respect to x to get 1 = v + y dv/dx. We rearrange this to get dx/dy = 1 / (v + y dv/dy). The original equation can now be rewritten after substituting these values, you will get y / (v + y dv/dy) = 2(v + y), simplifying, we get v = 2v + 2y, and after rearranging, we get y dv/dy = v. This is the differential equation in terms of v and y.
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