Final answer:
By setting up a system of equations based on the given information, we find that 200 Basic plans and 80 Standard plans were purchased by the new subscribers.
Explanation:
A cable network offers two types of plans: a Basic plan for $7.31 per month and a Standard plan which costs $3.00 more per month, totaling $10.31. To find out how many Basic and Standard plans were purchased by 280 new subscribers, who altogether paid $2286.80, we can set up a system of equations. Let x represent the number of Basic plans and y represent the number of Standard plans.
The total amount of plans purchased: x + y = 280
The total amount paid: 7.31x + 10.31y = 2286.80
Solving this system of equations:
Step 1: Multiply the first equation by 7.31 to make the x coefficients equal: 7.31x + 7.31y = 2046.80
Step 2: Subtract this equation from the second equation to eliminate x and solve for y: 3y = 240, thus y = 80. Therefore, 80 Standard plans were purchased.
Step 3: Substitute y back into one of the original equations to solve for x: x + 80 = 280, thus x = 200. Therefore, 200 Basic plans were purchased.
So, there were 200 Basic plans and 80 Standard plans purchased by the new subscribers.
An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273–279) "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique" showed that only 25 out of 37 tears (67.6%) located between 3 and 6 mm from the meniscus rim were healed.
(a) Calculate a 99% two-sided confidence interval on the proportion of such tears that will heal. Round the answers to 3 decimal places.
(b) Calculate a 99% lower confidence bound on the proportion of such tears that will heal. Round the answer to 3 decimal places.
Answer:
63
Step-by-step explanation:
The lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)
How to find the confidence interval for population proportion from large sample?Suppose we're given that:
Favorable Cases X (in count, in sample)Sample Size N Level of significance = [tex]\alpha[/tex]Then, the sample proportion of favorable cases is:
[tex]\hat{p} = \dfrac{X}{N}[/tex]
The critical value at the level of significance [tex]\alpha[/tex] is [tex]Z_{1- \alpha/2}[/tex]
The corresponding confidence interval is:
[tex]CI = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right)[/tex]
We need to construct the 99% confidence interval for the population proportion.
We have been provided with the following information about the number of favorable cases(the number of tears located between 3 and 6 mm from the meniscus rim were healed):
Favorable Cases X =25Sample Size N =37The sample proportion is computed as follows, based on the sample size N = 37 and the number of favorable cases X = 25
[tex]\hat p = \displaystyle \frac{X}{N} = \displaystyle \frac{25}{37} = 0.676[/tex]
The critical value for [tex]\alpha = 0.01[/tex] is [tex]z_c = z_{1-\alpha/2} = 2.576[/tex]
The corresponding confidence interval is computed as shown below:
[tex]\begin{array}{ccl} CI & = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right) \\\\ \\\\ & = & \displaystyle \left( 0.676 - 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}}, 0.676 + 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}} \right) \\\\ \\\\ & = & (0.477, 0.874) \end{array}[/tex]
Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.477<p<0.874, which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.477, 0.874).
Thus, the lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)
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A group of climbers begin climbing at an elevation of 5000 feet and ascend at a steady rate of 1500 vertical feet per hour. This situation can be modeled by a linear function. Identify the independent and dependent variables.
Final answer:
The independent variable is time and the dependent variable is elevation in the context of a linear function modeling climbers ascending a mountain at a rate of 1500 feet per hour.
Explanation:
The situation described by the student can be modeled by a linear function in which the independent variable is time and the dependent variable is elevation.
As the group of climbers begins at an elevation of 5000 feet and climbs at a steady rate of 1500 feet per hour, the relationship between the time spent climbing and the elevation gained is direct.
The function that models this situation is f(t) = 1500t + 5000, where 't' represents the time in hours, and 'f(t)' represents the elevation in feet.
The top three countries in oil consumption in a certain year are as follows: the United States, China, and Japan. In millions of barrels per day, the three top countries consume 32.2 million barrels of the world’s oil. The United States consumes 12 million more barrels a day than China. China consumes 3.2 million barrels a day than Japan. How many barrels of the world oil consumption did the United States, Japan, and China consume
Answer: Consumption of,
Japan = 4.6 million barrel,
China = 7.8 million barrel,
The US = 19.8 million barrel.
Step-by-step explanation:
Let x be the quantity of oil per day consumed by Japan,
∵ China consumes 3.2 million barrels a day than Japan.
So, the consumption of China = ( x + 3.2) million Barrel,
Also, The United States consumes 12 million more barrels a day than China.
So, the consumption of the US = (x + 3.2 + 12) = (x + 15.2) million barrel,
Thus, the total consumption of these three countries
= x + x + 3.2 + x + 15.2
= (3x + 18.4) million barrel,
According to the question,
3x + 18.4 = 32.2
3x = 32.2 - 18.4
3x = 13.8
⇒ x = 4.6
Hence, Japan, China and the US consume 4.6 million barrel, 7.8 million barrel and 19.8 million barrel respectively.
Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91.The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The day" subscript refers to the statistics day students. The "night subscript refers to the statistics night students. Assume that the standard deviations are equal. A concluding statement is:
a. There is sufficient evidence to conclude that statistics night students' mean on Exam 2 is better than the statistics day students' mean on Exam 2.
b. There is insufficient evidence to conclude that the statistics day students' mean on Exam 2 is better than the statistics night students' mean on Exam 2.
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
d. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
Answer:
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
Step-by-step explanation:
Given that a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2.
Group Group One Group Two
Mean 75.8600 75.4100
SD 16.9100 19.7300
SEM 2.8583 3.2436
N 35 37
*SEM is std error/sqrt n
Mean difference = 0.4500
[tex]H_0: \bar x = \bar y\\H_a: \bar x \neq \bar y[/tex]
(two tailed test)
Std error for difference = 4.342
Test statistic t = [tex]\frac{0.45}{4.342} \\=0.1036[/tex]
df =70
p value = 0.9178
Since p >0.05 we accept H0
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
Final answer:
The concluding statement is that there is insufficient evidence to conclude a significant difference in the means of statistics day students and statistics night students on Exam 2.
Explanation:
The hypothesis test being conducted in this scenario is a two-sample t-test. The null hypothesis, denoted as H0, states that there is no significant difference between the mean scores of statistics day students and statistics night students on Exam 2. The alternative hypothesis, denoted as Ha, states that there is a significant difference between the means of the two groups. To determine whether there is sufficient evidence to support the alternative hypothesis, we can compare the t-statistic with the critical t-value from a t-distribution table.
In this case, since the standard deviations are assumed to be equal, we can calculate the pooled standard deviation and use it to calculate the t-statistic. With the given sample means, standard deviations, and sample sizes, the calculated t-statistic is -0.246. The critical t-value for a two-tailed test with a significance level of 0.05 and 70 degrees of freedom is approximately 1.994. Since the calculated t-statistic (-0.246) falls within the range between -1.994 and 1.994, we fail to reject the null hypothesis.
Therefore, the concluding statement is:
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2.
The diameter of bushings turned out by a manufacturing process is a normally distributed random variable with a mean of 4.035 mm and a standard deviation of 0.005 mm. A sample of 25 bushings is taken once an hour. (a) Within what interval should 95 percent of the bushing diameters fall? (Round your answers to 4 decimal places.) The 95% confidence interval is from to (b) Within what interval should 95 percent of the sample means fall? (Round your answers to 5 decimal places.) The 95% confidence interval is from to (c-1) What conclusion would you reach if you saw a sample mean of 4.020? The sample came from a population that a population mean equal to 4.035. (c-2) What conclusion would you reach if you saw a sample mean of 4.055? The sample came from a population that a population mean equal to 4.035.
A) the z score for 95% is 1.96
Multiply by the deviation:
1.96 x 0.005 = 0.0098
Now add and then subtract that from the mean:
4.035 - 0.0098 = 4.0252
4.035 + 0.0098 = 4.0448
The interval is (4.0252, 4.0448)
B) 4.035 +/- 1.96 x sqrt( 0.005/sqrt(25))
= 4.035 +/-0.00196
Answer: ( 4.033, 4.037)
C) the conclusion is that both 4.020 and 4.055 are out of the range.
4.020 is below the lowest range and 5.055 is higher than highest range.
The 95% confidence interval for the sample means is; CI = (4.033, 4.037)
What is the Confidence Interval?A) We are given;
Mean; x' = 4.035 mm
standard deviation; σ = 0.005 mm
sample size; n = 25
Formula for confidence interval is;
CI = x' ± z(σ/√n)
The z score for 95% confidence level is 1.96. Thus;
CI = 4.035 ± 1.96(0.005)
CI = 4.035 ± 0.0098
CI = (4.025, 4.045)
B) From the formula for confidence interval earlier stated, we have;
CI = 4.035 ± 1.96(0.005/√(25))
CI = 4.035 ± 0.00196
CI = (4.033, 4.037)
C)i) The conclusion is that both 4.020 is out of the range of the confidence interval.
ii) The conclusion is that 4.055 is higher than the confidence interval
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A large manufacturing firm tests job applicants who recently graduated from college. The test scores are normally distributed with a mean of 500 and a standard deviation of 50. Management is considering placing a new hire in an upper level management position if the person scores in the upper 6 percent of the distribution. What is the lowest score a college graduate must earn to qualify for a responsible position?
Answer:
The lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 50
We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.06.
P(X > x) = 6% = 0.06
[tex]P( X > x) = P( z > \displaystyle\frac{x - 500}{50})=0.06[/tex]
[tex]= 1 - P( z \leq \displaystyle\frac{x - 500}{50})=0.06 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 500}{50})=1-0.06=0.94 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z < 1.555) = 0.94[/tex]
[tex]\displaystyle\frac{x - 500}{50} = 1.555\\x = 577.75[/tex]
Hence, the lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.
graph f(x), -f(x) and y= 4x ^ 2 cos(x)
Answer:
See attached image for the graph of the function
Step-by-step explanation:
Notice that this is the product of a power function ([tex]4x^2[/tex]) times the trigonometric and periodic function cos(x). So the zeros (crossings of the x axis will be driven by the values at which they independently give zero. That is the roots of the power function (only x=0) and the many roots of the cos function: [tex]x= \frac{\pi}{2} , \frac{3\pi}{2} ,...[/tex], and their nagetiva values.
Notice that the blue curve in the graph represents the original function f(x), with its appropriate zeros (crossings of the x-axis), while the orange trace is that of "-f(x)". Of course for both the zeroes will be the same, while the rest of the curves will be the reflection over the x-axis since one is the negative of the other.
The value given below is discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. Probability of fewer than 6 passengers who do not show up for a flight Choose the correct answer below. A. The area between 5.5 and 6.5 B. The area to the left of 6.5 C. The area to the right of 6.5 D. The area to the right of 5.5 E. The area to the left of 5.5
Answer:
E. The area to the left of 5.5 and D. The area to the right of 5.5
Step-by-step explanation:
Please see attachment .
Find a recurrence relation for the number of sequences of 1s, 3s, and 5s whose terms sum to n. (b) Repeat part (a) with the added condition that no 5 can be followed by a 1. (c) Repeat part (a) with the condition of no subsequence of 135
Answer:
The solution has been given in the following attachment .
Step-by-step explanation:
Lee is running around the perimeter of a circular track at a rate of 10 ft/sec. The track has a radius of 100 yards. After 10 seconds, Lee turns and runs along a radial line to the center of the circle. Once he reaches the center, he turns and runs along a radial line to his starting point on the perimeter. Assume Lee does not slow down when he makes these two turns.
(a) Sketch a picture of the situation.
(b) How far has Lee traveled once he returns to his starting position? circuit? enclosed by Lee's path.
(c) How much time will elapse during Lee's
(d) Find the area of the pie shaped sector.
The picture is given below. The perimeter, total time, and area of the circuit will be 700 feet, 70 seconds, and 100 square meters, respectively.
What is the arc length of the sector?Let r is the radius of the sector and θ be the angle subtended by the sector at the center.
Then the arc length of the sector of the circle will be
Arc = (θ/2π) 2πr
Then the area of the sector of the circle will be
Arc = (θ/2π) 2πr
We know that 100 yards = 300 feet.
The perimeter covered in 10 seconds will be given as,
10 x 10 = (θ/2π) 2π(300)
θ = 19.1°
The perimeter of the circuit will be given as,
P = 100 + 300 + 300
P = 700 feet
The total time is given as,
Time = 700 / 10
Time = 70 seconds
The area of the shape will be given as,
A = (19.1/360) 2π(300)
A = 100 square feet
The picture is given below. The perimeter, total time, and area of the circuit will be 700 feet, 70 seconds, and 100 square meters, respectively.
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The lengths of nails produced in a factory are normally distributed with a mean of 6.13 centimeters and a standard deviation of 0.06 centimeters.Find the two lengths that separate the top 7% and the bottom 7%.These lengths could serve as limits used to identify which nails should be rejected.Round your answer to the nearest hundredth, if necessary.
Answer:
Bottom 7%: L= 6.04 cm
Top 7%: L= 6.22 cm
Step-by-step explanation:
Mean length of the population (μ) = 6.13 cm
Standard deviation (σ) = 0.06 cm
The z-score for any given length 'X' is:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
What we want to know is the length at the 7-th percentile and at the 93-rd percentile.
According to a z-score table, the 7-th percentile has a correspondent z-score of -1.476 and the 93-rd percentile has a z-score of 1.476. Therefore, the bottom 7% and top 7% are separated by the following lengths:
[tex]z(X_B)=\frac{X_B-\mu}{\sigma}\\-1.476=\frac{X_B-6.13}{0.06}\\X_B = 6.04\\z(X_T)=\frac{X_T-\mu}{\sigma}\\1.476=\frac{X_T-6.13}{0.06}\\X_T = 6.22[/tex]
Bottom 7%: L= 6.04 cm.
Top 7%: L= 6.22 cm.
To find the lengths that separate the top 7% and the bottom 7% of nails produced in the factory, we can use the z-score formula. The z-scores can be calculated using the inverse normal distribution function, and then the lengths can be found using the formula x = μ + zσ. The lengths that separate the top 7% and the bottom 7% are approximately 6.22 cm and 6.04 cm, respectively.
Explanation:To find the lengths that separate the top 7% and the bottom 7% of nails produced in the factory, we can use the z-score formula. The z-score represents how many standard deviations a value is from the mean. For the top 7%, we can find the z-score by using the formula: z = invNorm(1 - 0.07), where invNorm is the inverse normal distribution function. Similarly, for the bottom 7%, the z-score can be calculated using the formula: z = invNorm(0.07). Once we have the z-scores, we can use the formula x = μ + zσ, where x is the length of the nails, μ is the mean length (6.13 cm), z is the z-score, and σ is the standard deviation (0.06 cm).
Using a calculator or software, we can find the z-scores:
For the top 7%: z = invNorm(1 - 0.07) = invNorm(0.93) ≈ 1.48For the bottom 7%: z = invNorm(0.07) ≈ -1.48Substituting the values into the x = μ + zσ formula:
For the top 7%: x = 6.13 + 1.48 * 0.06 ≈ 6.22 cmFor the bottom 7%: x = 6.13 - 1.48 * 0.06 ≈ 6.04 cmTherefore, the two lengths that separate the top 7% and the bottom 7% are approximately 6.22 cm and 6.04 cm, respectively.
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An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was = 119.6 ounces. Suppose the standard deviation is known to be σ = 6.5 ounces. Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean μ. Based on the 25 recent birth records, the sampling distribution of the sample mean can be represented by:
A. N(μ, 6.5).
B. N(μ, 1.30).
C. N(119.6, 1.30).
D. N(119.6, 6.5).
Answer:
B. N(μ, 1.30).
Step-by-step explanation:
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]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was = 119.6 ounces. Suppose the standard deviation is known to be σ = 6.5 ounces.
Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean μ.
This means that for the sampling distribution, the mean is the mean of the weight of all babies born, so [tex]\mu[/tex] and [tex]s = \frac{6.5}{\sqrt{25}} = 1.30[/tex].
So the correct answer is
B. N(μ, 1.30).
The correct representation of the sampling distribution of the sample mean based on the data from birth records is option C, which is N(119.6, 1.30). This is based on the formula of sampling distribution for mean with known standard deviation.
Explanation:The question is asking about the sampling distribution of the sample mean which is a statistical concept used in inferential statistics. The sample distribution of the mean is normally distributed, denoted as N(μ, σ/n0.5) where μ is the population mean, σ is the population standard deviation, and n is the sample size (SRS - Simple Random Sample).
Based on the data given, the sample mean after studying the 25 birth records is 119.6 ounces and the standard deviation is known to be 6.5 ounces. So, the standard deviation of the sample mean, often termed as the standard error, would be σ/√n = 6.5/√25 = 1.3 ounces. Hence the correct sampling distribution of the sample mean would be represented by N(119.6, 1.30), which is option C in your question.
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Shelby bought a 2-ounce tube of blue paint. She used 2/3 ounce to paint the water, 3/5 ounce to paint the sky, and some to paint a flag. After that, she has 2/15 ounce left. How much paint did Shelby use to paint her flag?
Answer:
The amount of ounce used for paint is [tex]\frac{9}{15} ounce[/tex]
Step-by-step explanation:
used amount for water is [tex]\frac{2}{3}[/tex] ounce.
used amount for sky is [tex]\frac{3}{5}[/tex] ounce.
thus, the total amount used is [tex]\frac{2}{3}[/tex] + [tex]\frac{3}{5}[/tex]
= [tex]\frac{19}{15}[/tex]
Thus the total unused amount is 2 - [tex]\frac{19}{15}[/tex] = [tex]\frac{11}{15}[/tex] .
But the remaining amount is [tex]\frac{2}{15}[/tex].
Thus used amount for flag is [tex]\frac{11}{15}[/tex] - [tex]\frac{2}{15}[/tex]
= [tex]\frac{9}{15}[/tex]
Suppose that we wanted to estimate the true average number of eggs a queen bee lays with 95 percent confidence. The margin of error we are willing to accept is 0.5. Suppose we also know that s is about 10. At minimum, what sample size should we use ?
Answer:
At least 1537 samples needed to estimate the true average number of eggs a queen bee lays with 95 percent confidence
Step-by-step explanation:
Minimum sample size required can be found using the formula
N≥[tex](\frac{z*s}{ME} )^2[/tex] where
N is the sample sizez is the corresponding z-score of 95% confidence level (1.96)s is the standard deviation (10)ME is the margin of error (0.5)then N≥[tex](\frac{1.96*10}{0.5} )^2[/tex] =1536.64
Then, at least 1537 samples needed to estimate the true average number of eggs a queen bee lays with 95 percent confidence
The minimum sample size needed is 1537 for a margin of error of 0.5.
What is margin of error?Margin of error is used to determine by what value there is deviation from the real value. Margin of error (E) is given by:
[tex]E=Z_\frac{\alpha}{2} *\frac{standard\ deviation}{\sqrt{sample\ size} }[/tex]
The 95% confidence level have a z score of 1.96. Hence for E = 0.5, standard deviation = 10. hence:
[tex]0.5=1.96*\frac{10}{\sqrt{sample\ size}}\\ \\sample\ size=1537[/tex]
The minimum sample size needed is 1537 for a margin of error of 0.5.
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In a little over a month, from June 5, 1879, to July 2, 1879, Albert Michelson measured the velocity of light in air 100 times (Stigler, Annals of Statistics, 1977). Today we know that the true value is 299,734.5 km/sec. Michelson’s data have a mean of 299,852.4 km/sec with a standard deviation of 79.01. (a) Find a two-sided 95% confidence interval for the true mean (the true value of the speed of light). (b) What does the confidence interval say about the accuracy of Michelson’s measurements?
Answer:
(a) According to the Albert Michelson's measurements, two-sided 95% confidence interval for velocity of light in the air would be 299,852.4±15.5
(b) Since true velocity of light in the air is not included in the 95% confidence interval, we can say that Michelson’s measurements were not accurate.
Step-by-step explanation:
Confidence Interval can be calculated using M±ME where
M is the sample mean velocity of light in air (299,852.4)ME is the margin of error from the meanAnd margin of error (ME) from the mean can be calculated using the formula
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the corresponding statistic in 95% confidence level (1.96)s is the sample standard deviation (79.01)N is the sample size (100)Then ME=[tex]\frac{1.96*79.01}{\sqrt{100} }[/tex] ≈ 15.5
According to the Albert Michelson's measurements, 95% confidence interval for velocity of light in the air would be 299,852.4±15.5
High Tech, Inc randomly tests its employees about company policies. Last year in the 400 random tests conducted, 14 employees failed the test.A. Develop a 99% confidence interval for the proportion of applicants that fail the test. (round answers to 3 decimal places) Confidence interval for the proportion mean is between _____ and ____B. Would it be reasonable to conclude 5% of the employees cannot pass the employee test? Yes or No
Answer:
Confidence interval for the proportion mean is between 0.0113 and 0.0587. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test.
Step-by-step explanation:
We have a large sample size of n = 400 random tests conducted. Let p be the true proportion of employees who failed the test. A point estimate of p is [tex]\hat{p} = 14/400 = 0.035[/tex], we can estimate the standard deviation of [tex]\hat{p}[/tex] as [tex]\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.035(1-0.035)/400}=0.0092[/tex]. A [tex]100(1-\alpha)%[/tex] confidence interval is given by [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n[/tex], then, a 99% confidence interval is [tex]0.035\pm z_{0.005}0.0092[/tex], i.e., [tex]0.035\pm (2.5758)(0.0092)[/tex], i.e., (0.0113, 0.0587). [tex]z_{0.005} = 2.5758[/tex] is the value that satisfies that there is an area of 0.005 above this and under the standard normal curve. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test, because this inverval contain 0.05.
The 99% confidence interval for the proportion of employees that fail the test is between 0.016 and 0.054. Since 5% is within this range, it is reasonable to conclude that 5% of the employees cannot pass the test.
Explanation:To compute a 99% confidence interval for the proportion of employees that fail the test, we first need to calculate the sample proportion (p). Here, 14 employees failed the tests out of 400, so p = 14/400 = 0.035. The 99% confidence interval requires Z-score of 2.576 (as 99% of the data lies within 2.576 standard deviations of the mean in a normal distribution).
The estimation error (E) can be calculated using the formula E = Z * √( (p*(1-p)) / n), where n is the total number of tests. Substituting the values, E = 2.576 * √(0.035 * (1 - 0.035) / 400) = 0.019
So, the 99% confidence interval is (p - E, p + E) = (0.035 - 0.019, 0.035 + 0.019) = (0.016, 0.054). Thus, we are 99% confident that the true proportion of employees that fail the test is between 0.016 and 0.054.
Given that 5% (or 0.05) is within the 99% confidence interval we calculated, it would be reasonable to conclude that 5% of the employees cannot pass the employee test.
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An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1.3 years of the population mean. Assume the population of ages is normally distributed.(a) Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 1.5 years.
Answer: 4
Step-by-step explanation:
For given Population standard deviation[tex](\sigma)[/tex] , the formula to sample size is given by :-
[tex]n=(\dfrac{\sigma\cdot z*}{E})^2[/tex]
, where z* = Two-tailed critical value.
E = Margin of error.
Given : [tex]\sigma=1.5[/tex] years
We know that , Critical value for 90% confidence interval : z* = 1.645
Margin of error : E=1.3 years
Then, the minimum sample size required to construct a 90% confidence interval for the population mean will be :-
[tex]n=(\dfrac{1.5\cdot 1.645}{1.3})^2\\\\=(1.89807692308)^2\\\\=3.60269600593\approx4[/tex] [Rounded to the nearest whole number.]
Hence, the minimum sample size required = 4
The minimum sample size required to construct a 90% confidence interval for the population mean age of students at a college, given a population standard deviation of 1.5 years and desired margin of error of 1.3 years, is 3 students.
Explanation:The admissions director is trying to estimate the mean age of all students enrolled at the college within 1.3 years of the actual population mean with a 90% confidence level. In statistical terms, this means constructing a 90% confidence interval for the population mean within a margin of error of 1.3 years.
The formula to compute sample size for a confidence interval when the population standard deviation is known is:
n = (Zσ/E)^2
where:
n is the sample sizeZ is the z-score for the desired confidence level (for 90% confidence level, Z = 1.645)σ is the population standard deviation (here, 1.5 years)E is the desired margin of error (here, 1.3 years)
Substituting the given values into the formula, we have:
n = (1.645*1.5/1.3)^2
Solving for n, we get approximately 2.41. Because we can't survey a fraction of a student, we round up to the nearest whole number. Thus, the minimum sample size required is 3 students.
In practice, especially for a large population, this sample size might be too small, but based on the strict parameters established (90% confidence level, 1.3 years margin of error, and 1.5 years population standard deviation), this is the minimum sample size required.
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You have a standard deck of 52 cards (i.e., 4 aces, 4 twos, 4 threes, …, 4 tens, 4 jacks, 4 queens, and 4 kings) that contains 4 suits (hearts, clubs, spades, and diamonds). We draw one card from the deck. What is the probability that the card is NEITHER a face card (jack, king, or queen) NOR a heart? 27/52
Answer:
30/52 or 0.5769 or 57.69%
Step-by-step explanation:
In a standard deck of 52 cards, the number of face cards (F) and the number of hearts (H) is given by:
[tex]F=4+4+4 =12\\H=\frac{52}{4}=13[/tex]
Out of all hearts, three of them are face cards (jack, king, and queen). Therefore, the probability of a card being EITHER a face card or a heart is:
[tex]P(F \cup H) = P(F) +P(H) - P(F \cap H) \\P(F \cup H)=\frac{12+13-3}{52} =\frac{22}{52}[/tex]
Therefore, the probability of card being NEITHER a face card NOR a heart is:
[tex]P=1-P(F \cup H) \\P=1-\frac{22}{52}=\frac{30}{52}\\\\P=0.5769\ or\ 57.69\%[/tex]
Translating mathematical statements in English into logical expressions. info About Consider the following statements in English. Write a logical expression with the same meaning. The domain is the set of all real numbers. (a) There is a number whose cube is equal to 2. (b) The square of every number is at least 0. (c) There is a number that is equal to its square. (d) Every number is less than or equal to its square.
Answer:
See below
Step-by-step explanation:
(a) There is a number whose cube is equal to 2.
[tex]\large \exists x \in \mathbb{R}\;|\;x^3=2[/tex]
(b) The square of every number is at least 0.
[tex]\large \forall x\in \mathbb{R},\;x^2\geq 0[/tex]
(c) There is a number that is equal to its square.
[tex]\large \exists x \in \mathbb{R}\;|\;x^2=x[/tex]
(d) Every number is less than or equal to its square.
[tex]\large \forall x\in \mathbb{R},\;x\leq x^2[/tex]
(By the way, this last statement is not true when 0 < x < 1)
In humans wavy hair(W) is dominant over straight(w). Dark hair(D) dominates red(d).A wavy dark haired male(heterozygous dominant for both traits) married a wavy red haired woman(homozygous recessive for both traits).What is the probability of having a wavy red haired child ?
Answer:
1/4
Step-by-step explanation:
Parents WwDd wwdd
Gametes WD Wd wD wd wd wd wd wd
Offspring possibilities: WwDd WwDd WwDd WwDd
Wwdd Wwdd Wwdd Wwdd
wwDd wwDd wwDd wwDd
wwdd wwdd wwdd wwdd
P (wavy red hair) =4/16 = 1/4
Answer:
The probability is one half (1/2)
But it seems to be a mistake in the statement .."married a wavy red haired woman(homozygous recessive for both traits)". If the woman has wavy hair, the trait can't be homozygous recessive, although she certainly is homozygous recessive for the second trait because is red hair.
Step-by-step explanation:
As wavy hair is dominant over straight hair, it means only one allele is necessary to be visible: WW (homozygous), and Ww (heterozygous) produce the wavy hair trait. An homozygous recessive (ww) doesn't show the wavy trait character; he or she would have straight hair
The woman, if she's homozygous and wavy hair, would be (WW). As she is homozygous for red hair, she has red hair.
If the statement "A wavy dark haired male(heterozygous dominant for both traits) married a wavy red haired woman(homozygous recessive for both traits)" is correct, meaning that she is not wavy hair but "STRAIGHT", the probability of having a wavy red hair changes, being only 1/4
Calculation can be performed with Punnett squares
For example for the wavy dark hair male (heterozygous for both traits), the genotype is WwDd and the possible allele combinations would be: WD, Wd, wD, and wd
For the woman, if she's wavy red hair (homozygous), the genotype would be WWdd and the possible allele combinations would be only Wd.
Then you need cross the male allele combinations against Wd
If the woman is straight red hair (homozygous for both traits), the genotype would be wwdd and the possible allele combinations would be only wd.
Then you need cross the male allele combinations against wd, and obtain the proportions produced from the cross
An example of Punnett square below
In the US court system, a defendant is assumed innocent until proven guilty. Suppose that you regard a court system as a hypothesis test with these null and alternative hypotheses: H0: Defendant is innocent Ha: Defendant is guilty There are 2 possible decisions regarding H0 and 2 possible truths as to the innocence or guilt of the defendant, making 4 possible combinations. What are those 4 combinations? Identify the two correct decisions.
Answer:
Step-by-step explanation:
A test hypothesis by definition is a test based on probabilities, therefore it always be possible to make errors when decision is made. According to that we always have 4 possibilities, two right and to wrong (4 possiblities)
For instance in our particular case
H₀ = Defendant is innocent Hₐ = defendant is guilty
If we arrive to conclusion that H₀ is right, and he is really innocent we took a correct decision, And the result will be correct; if we take the decision of reject H₀ when the defendant is guilty again we took the right decision. These are the two correct decision in that case.
On the other hand what happens if we take the decision of rejecting H₀ (accepting Hₐ ) and the defendant is innocent, we are sending the defendant to jail and he is innocente (we are making I type error) and the defendant will pay for it. Finally if we accept H₀ and this decision is not right we will make the defendant be free and he is really guilty
What is the probability that a randomly selected tire will fail before the 35,000 mile warranty mileage stated?
Probabilities are used to determine how likely, or often an event is, to happen. The probability that the selected tire fails before 35000-mile warranty is 0.11702
From the complete question, we have:
[tex]n = 41[/tex] --- number of tires
[tex]Mileage: 33095\ 34589\ 39411\ 42386\ 37886\ 33096\ 44185\ 38273\ 42387\ 36117[/tex]
[tex]44373\ 39896\ 42758\ 34028\ 39768\ 44392\ 35826\ 44945\ 41756\ 41087[/tex]
[tex]43716\ 33478\ 41430\ 39397\ 39517\ 38068\ 42216\ 43447\ 33372\ 42631[/tex]
[tex]42215\ 44367\ 33186\ 41567\ 38534\ 33873\ 43484\ 39761\ 35531\ 40926\ 38348[/tex]
First, we calculate the mean
[tex]\mu = \frac{\sum x}{n}[/tex]
This gives:
[tex]\mu = \frac{33095+ 34589 +.............+40926 +38348}{41}[/tex]
[tex]\mu = \frac{1619318}{41}[/tex]
[tex]\mu = 39496[/tex]
Next, calculate the standard deviation
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]
This gives:
[tex]\sigma = \sqrt{\frac{(33095 - 39496)^2 + (34589 - 39496)^2 +.......+ (40926 - 39496)^2 + (38348- 39496)^2}{41-1}}[/tex]
[tex]\sigma = \sqrt{\frac{572531448}{40}}[/tex]
[tex]\sigma = \sqrt{14313286.2}[/tex]
[tex]\sigma = 3783[/tex]
The probability a tire will fail before 35000 is represented as:
[tex]P(x < 35000)[/tex]
Calculate the z score
[tex]z = \frac{x - \mu}{\sigma}[/tex]
This gives
[tex]z = \frac{35000 - 39496}{3783}[/tex]
[tex]z = \frac{-4496}{3783}[/tex]
[tex]z = -1.19[/tex]
So, we have:
[tex]P(x < 35000) = P(z < -1.19)[/tex]
From z table of values:
[tex]P(z < -1.19) = 0.11702[/tex]
Hence:
[tex]P(x < 35000) = 0.11702[/tex]
So, the probability that the selected tire fails before 35000-mile warranty is 0.11702
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The probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412.
Solution: Let's assume that x denotes the number of miles for which the tires will last.
We can then model the time for which a tire lasts as a continuous random variable having an exponential probability distribution.
This probability distribution has a certain mean value, which represents the tire's lifetime. On average, the lifetime of a tire is 44,000 miles.
This implies that the tire's decay parameter, lambda (λ), is given by: λ=1/44000Therefore, the probability that the tire will last less than 35,000 miles can be calculated by integrating the probability density function from 0 to 35,000,
which is given by:[tex]f(x)=λe^(-λx)[/tex]Here's the calculation:[tex]:$$\begin{aligned} P(X \le 35,000) &= \int_0^{35,000} f(x) dx \\ &= \int_0^{35,000} \lambda e^{-\lambda x} dx \\ &= \left[-e^{-\lambda x}\right]_0^{35,000} \\ &= -e^{-\lambda 35,000} + e^{-\lambda 0} \\ &= -e^{(-1/44,000)\cdot 35,000} + e^{(-1/44,000)\cdot 0} \\ &= -e^{-0.795} + e^{0} \\ &= 0.3184 + 1 \\ &= 1.3184 \end{aligned} $$[/tex]Note that the probability density function is always positive, so the negative result in the second step can be ignored.
As a result, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is: P(X ≤ 35,000) = 0.3184 - 1 = -0.6816
The probability of failure is never negative, so there must be an error somewhere in the calculation.
Therefore, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412, which is the complement of P(X > 35,000):P(X > 35,000) = 1 - P(X ≤ 35,000) = 1 - 0.3184 = 0.6816P(X ≤ 35,000) = 0.3184P(X < 35,000) = 1 - P(X > 35,000) = 1 - 0.6816 = 0.3184P(X < 35,000) = 0.3184
Therefore, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412.
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The probable question is :-
Question 5 10 pts What is the probability that a randomly selected tire will fail before the 35,000 mile warranty mileage stated?
O 0.0500
O 0.0412
O 0,09218
O 0.0885
Why i got low score?
Answer:
There is exactly one real root.
Step-by-step explanation:
There are two parts to arrive at the solution.
(i) The polynomial has at least one real root.
(ii) The polynomial has exactly one real root.
We prove (i) using Intermediary Value Theorem.
f(x) = x³ + x - 1 = 0 is a polynomial. So, it is continuous.
At x = 1, f(1) = 1
At x = o, f(0) = -1
Since, there is a change of sign it should have crossed through zero.
Now, to prove there is exactly one real root we use Rolle's theorem.
Let us assume there are two real roots to the polynomial, say 'a' and 'b'.
Then f(a) = 0 and f(b) = 0.
⇒ f(a) = f(b)
To use Rolle's theorem we need the function to be continuous, differentiable and for any two points a,b f(a) = f(b) there should exist a 'c' such that f'(c) = 0.
Now, f'(x) = 3x² + 1
Note that f'(x) is always greater than equal to 1.
It can never be zero for any c. This contradicts Rolle's Theorem. o, our assumption that two real roots exist must be wrong.
Hence, we conclude that there is exactly one real root to the polynomial.
A pizza parlor offers a choice of 12 different toppings. How many 4-topping pizzas are possible? (no double orders of toppings are allowed)
In this situation, does the order of toppings matter? Yes/No?
There are ____ possible 4-topping pizzas.
9514 1404 393
Answer:
No; 495
Step-by-step explanation:
For many folks, the order of application of toppings will not matter. For those folks, the number of possible 4-topping pizzas is
C(12,4) = 495 = 12!/(8!×4!)
There are 495 possible 4-topping pizzas.
Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 7, and the y-axis, oriented counterclockwise starting from the origin. Label the edges of the boundary as C_1, C_2, C_3 starting from the bottom edge going counterclockwise. Give each edge a constant speed parametrization with domain 0 lessthanorequalto t lessthanorequalto 1: edge C_1 x_1(t) = y_1(t) = edge C_2 x_2(t) = y_2(t) = edge C_3 x_3(t) = y_3 = integral_C y^2xdx + x^2ydy = integral_C1 y^2xdx + x^2ydy + integral_C2 y^2xdx + x^2ydy + integral_C3 y^2xdx + x^2ydy 10pt = Applying Green's theorem, integral_C y^2 xdx + x^2 ydy = dxdy The vector field F = y^2x i + x^2y j is:
The question asks for the parametrization of boundary edges of a region in the 1st quadrant and the computation of a line integral over it. The region is bounded by the x-axis, quarter-circle of radius 7, and the y-axis. This is done by parameterizing each edge and applying Green's theorem to calculate the related integral.
Explanation:This exercise is essentially setting up and evaluating a line integral over closed path, which tells you about the interaction between a vector field and a curve in its domain. The region in question is in the first quadrant, and is bounded by the x-axis (C_1), a quarter circle (arc) of radius 7 (C_2), and the y-axis (C_3). The oriented counterclockwise is a convention means going from the origin along the x-axis, then following the quarter-circle (arc) around, and then starting back along the y-axis towards the origin.
The parametrization of the constant-speed edge C_1 is x_1(t)=7t, y_1(t)=0, for 0 <= t <= 1. For edge C_2 is x_2(t)=7cos(π/2t), y_2(t)=7sin(π/2t), for 0 <= t <= 1 and for C_3 is x_3(t)=0, y_3(t)=7(1-t), for 0 <= t <= 1. As the integral of a scalar function over C (the quarter-circle location) is equal to the sum of integrals over the three parts (C1, C2, C3), we can break it into three simpler parts and integrate separately.
The Vector Field F related to the integral can be represented by F = y^2xi + x^2yj based on the given equation. To find the resulting integral via Green's theorem, you would just take the divergence of the field F, yielding a double integral over the region.
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The graph of f(x) = x2 is translated to form
g(x) = (x – 2)2 – 3.
On a coordinate plane, a parabola, labeled f of x, opens up. It goes through (negative 2, 4), has a vertex at (0, 0), and goes through (2, 4).
Which graph represents g(x)?
On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).
On a coordinate plane, a parabola opens up. It goes through (negative 3, 4), has a vertex at (negative 2, 3), and goes through (negative 1, 4).
On a coordinate plane, a parabola opens up. It goes through (1, 4), has a vertex at (2, 3), and goes through (3, 4).
On a coordinate plane, a parabola opens up. It goes through (negative 4, 1), has a vertex at (negative 2, negative 3), and goes through (0, 1).
The correct answer is: A,
On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).
Step-by-step explanation:
Figure represents graph of [tex]f(x) = x^{2}[/tex] and [tex]g(x) = (x-2)^{2} -3[/tex]
Here, [tex]f(x) = x^{2}[/tex] is " Translated " or " Transformation " to [tex]g(x) = (x-2)^{2} -3[/tex]
In process of transformation,
You should remember that shape of curve remain same and only changes we get in vertex shift
Now, Vertex can be shift in two direction, we are going to discuss both the cases
(A). Shifting of Vertex in X-Axis:
A new function g(x) = f(x - c) represents to X-axis shift and In graph of f(x), Curve is shifted c units along right side of the X-axis
(B). Shifting of Vertex in Y-axis:
A new function g(x) = f(x) + b represents to Y-axis shift and In graph of f(x), Curve is shifted b units along the upward direction of Y-axis
Looking at the figure, You can see that vertex of f(x) is shifted 2 Units in X-axis and Negative 3 units in Y-axis and result into g(x)
Now, [tex]g(x) = (x-2)^{2} -3[/tex] = f(x-2) + (-3)
Therefore, new vertex we get is (2,-3)
Also, [tex]g(x) = (x-2)^{2} -3[/tex]
[tex]g(0) = (0-2)^{2} -3[/tex]
[tex]g(0) = (4} -3[/tex]
[tex]g(0) = 1 [/tex]
So. g(x) passes through (0,1)
The correct answer is: A On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).
Answer:
Graph A is the correct answer
Step-by-step explanation:
I juust took the test and got it right :D hope this helps
Find the Laplace transform of the given function; a and b are real constants. f(t) = eat sinh bt Your answer should be an expression in terms of a, b and s. L{f(t)}(s) = F(s) =
The Laplace transform of the function f(t) = eat sinh bt is F(s) = (e^a)/((s - a)^2 + b^2) after manipulating and simplifying using the standard Laplace transform rules.
Explanation:The Laplace transform of the function f(t) = eat sinh bt is given by the formula:
F(s) = L{f(t)}(s) = Int0->infinity[ e^(at - s)*sinh(bt) dt]
By splitting the hyperbolic sine function, sinh(bt), into exponentials, we get:
F(s) = 0.5*Int0->infinity [ (e^(-s + a + b)*t - e^(-s + a - b)*t) dt]
As a result, after applying Laplace transform rules and simplifying, we get:
F(s) = (e^a)/((s - a)^2 + b^2).
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Use the model for projectile motion, assuming no air resistance.
A baseball, hit above the ground, leaves the bat at an angle of 45 degrees and is caught by an outfielder 3 feet above the ground and 300 feet from home plate.
What is the initial speed of the ball, and how high does it rise?
Answer:
s=29.93m/s
h=22.88m
Step-by-step explanation:
we must find the initial speed, we will determine its position (x-y).
x component [tex]s=0+v_{0}cos\alpha.t+0=v_{0}cos\alpha.t[/tex]
y component [tex]h=0+v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}\\[/tex] since the ball is caught at the same height then h=0
[tex]h=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0\\v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0;v_{0}sin\alpha.t=\frac{1}{2}gt^{2}\\t=\frac{2v_{0}sin\alpha}{g}\\[/tex]
where t= flight time;[tex]s=v_{0}cos\alpha.t[/tex], replacing t:
[tex]v_{0}=\sqrt{\frac{sg}{sin2\alpha}}[/tex]
[tex]s=v_{0}cos\alpha(\frac{2sin\alpha }{g})=\frac{v_{0} ^{2}2sin\alpha.cos\alpha}{g}=\frac{v_{0} ^{2}sin(2\alpha)}{g}]
: the values must be taken to the same units
[tex]300ft*0.3048m/ft=91.44m[/tex]
[tex]v_{0}=\sqrt{\frac{91.44m*9.8\frac{m}{s^{2}}}{sin2(45)}}=\sqrt{895.112(\frac{m}{s} )^{2} }=29.93\frac{m}{s}[/tex]
To calculate the height you should know that this is achieved when its component at y = 0
[tex]v_{y}=v_{0}sin\alpha-gt=0;gt=v_{0}sin\alpha\\\\ t=\frac{v_{0}sin\alpha }{g}\\h=v_{0}sin\alpha .t-\frac{1}{2}gt^{2}[/tex]
replacing t;[tex]h=v_{0}sin\alpha(\frac{v_{0}sin\alpha}{g})-\frac{1}{2}g(\frac{v_{0sin\alpha}}{g}) ^{2}\\[/tex]
finally
[tex]h=\frac{(v_{0}sin\alpha)^{2}}{2g}=\frac{(29.95*sin45)^{2}}{2*9.8}=22.88m[/tex]
A study was interested in determining if eating milk chocolate lowered someone's cholesterol levels. Ten people's cholesterol was measured. Then, each of these individuals were told to eat 100g of milk chocolate every day and to eat as they normally did. After two weeks, their cholesterol levels were measured again. Is there evidence to support that their cholesterol levels went down? How should we write the alternative hypothesis? (mud = the population mean difference= before - after)
A. Ha: mud = 0B. Ha: mud > 0C. Ha: mud < 0D. Ha: mud does not equal 0
The alternative hypothesis whereby mud = the population mean difference is expressed as; Ha: mud > 0
What is the Alternative hypothesis?
An alternative hypothesis is defined as one in which the observers or researchers anticipate a difference (or an effect) between two or more variables.
Now, in this case, the null hypothesis is that eating milk chocolate lowered someone's cholesterol levels. This means the alternate hypothesis is that eating milk chocolate increases someone's cholesterol levels.
Thus, alternative hypothesis in this case is expressed as;
Ha: mud > 0
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Please help asap 20 points plus brainlest
5\6 of a pound + 3\4 of a pound = ??
Answer: 19/12 pounds or 1.583 pounds
Step-by-step explanation:
Let a pound be represented by x. So x = 1 pound.
5/6 of a pound is the same as multiplying 5/6 × x. It becomes 5/6 × x = 5x/6
3/4 of a pound is the same as multiplying 3/4 × x. It becomes 3/4 × x = 3x/4
We want to look for 5/6 of a pound + 3/4 of a pound.
It becomes
5x/6 + 3x/4
Taking lowest common multiple(LCM) of 12, it becomes
= (2×5x + 3×3x) / 12
= (10x +9x)/12
= 19x /12
Substituting x = 1 pound,
It becomes 19/12 pounds
Expressing in decimal, 19/12 = 15.83
We can also solve from the beginning in terms of decimals
5/6 pounds = 0.833 pounds
3/4pounds = 0.75 pounds
5/6 pounds + 3/4 pounds = 0.833 + 0.75 = 1.583 pounds