Answer:
Part 1
(a) 0.28434
(b) 0.43441
(c) 29.9 mm
Part 2
(a) 0.97722
Step-by-step explanation:
There are two questions here. We'll break them into two.
Part 1.
This is a normal distribution problem healthy children having the size of their left atrial diameters normally distributed with
Mean = μ = 26.4 mm
Standard deviation = σ = 4.2 mm
a) proportion of healthy children have left atrial diameters less than 24 mm
P(x < 24)
We first normalize/standardize 24 mm
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (24 - 26.4)/4.2 = -0.57
The required probability
P(x < 24) = P(z < -0.57)
We'll use data from the normal probability table for these probabilities
P(x < 24) = P(z < -0.57) = 0.28434
b) proportion of healthy children have left atrial diameters between 25 and 30 mm
P(25 < x < 30)
We first normalize/standardize 25 mm and 30 mm
For 25 mm
z = (x - μ)/σ = (25 - 26.4)/4.2 = -0.33
For 30 mm
z = (x - μ)/σ = (30 - 26.4)/4.2 = 0.86
The required probability
P(25 < x < 30) = P(-0.33 < z < 0.86)
We'll use data from the normal probability table for these probabilities
P(25 < x < 30) = P(-0.33 < z < 0.86)
= P(z < 0.86) - P(z < -0.33)
= 0.80511 - 0.37070 = 0.43441
c) For healthy children, what is the value for which only about 20% have a larger left atrial diameter.
Let the value be x' and its z-score be z'
P(x > x') = P(z > z') = 20% = 0.20
P(z > z') = 1 - P(z ≤ z') = 0.20
P(z ≤ z') = 0.80
Using normal distribution tables
z' = 0.842
z' = (x' - μ)/σ
0.842 = (x' - 26.4)/4.2
x' = 29.9364 = 29.9 mm
Part 2
Population mean = μ = 65 mm
Population Standard deviation = σ = 5 mm
The central limit theory explains that the sampling distribution extracted from this distribution will approximate a normal distribution with
Sample mean = Population mean
¯x = μₓ = μ = 65 mm
Standard deviation of the distribution of sample means = σₓ = (σ/√n)
where n = Sample size = 100
σₓ = (5/√100) = 0.5 mm
So, probability that the sample mean distance ¯x for these 100 will be between 64 and 67 mm = P(64 < x < 67)
We first normalize/standardize 64 mm and 67 mm
For 64 mm
z = (x - μ)/σ = (64 - 65)/0.5 = -2.00
For 67 mm
z = (x - μ)/σ = (67 - 65)/0.5 = 4.00
The required probability
P(64 < x < 67) = P(-2.00 < z < 4.00)
We'll use data from the normal probability table for these probabilities
P(64 < x < 67) = P(-2.00 < z < 4.00)
= P(z < 4.00) - P(z < -2.00)
= 0.99997 - 0.02275 = 0.97722
Hope this Helps!!!
A small post office has two open windows. Customers arrive according to a Poisson distribution at the rate of 1 every 3 minutes. However, only 80% of them seek service at the windows. The service time per customer is exponential, with a mean of 5 minutes. All arriving customers form one line and access available windows on a FIFO basis.
(a) What is the probability that an arriving customer will wait in line?
(b) What is the probability that both windows are idle?
(c) What is the average length of the waiting line?
(d) Would it be possible to offer reasonable service with only one window? Explain.
Answer:
A) probability that an arriving customer will wait in line is 67%
B)the probability that both windows are idle is 0.33
C) the average length of the waiting line is 1.33
D)it would not be possible to offer a reasonable service with only one window
Step-by-step explanation:
arrival rate: δ = 20 x 0.80 = 16 customers per hour
service rate: μ = 2 × (60/5) = 24 customers/hour
Utilization factor is given as;
Φ = δ/μ
So, Φ = 16/24 ≈ 0.67
A) the probability that an arriving customer will wait in line is;
16/24 x 100% ≈ 67%
B) probability that both windows are idle is;
P(x=0) = 1 - 0.67 = 0.33
C) The average number of customers in the post office will be;
L_s = Φ/(1 - Φ)
L_s = 0.67/(1 - 0.67)
L_s = 0.67/0.33
L_s ≈ 2 customers
Thus, the average length of the waiting line is;
L_w = L_s - Φ
L_w = 2 - 0.67
L_w = 1.33
D) this part demands that we find the utilization factor with only one window.
Thus;
arrival rate: δ = 20 x 0.80 = 16 customers per hour
And
service rate: μ = 1 × (60/5) = 12 customers/hour
Thus, Utilization factor = 16/12 = 1.33
Thus, it would not be possible to offer a reasonable service with only one window
Dr. Potter provides vaccinations against polio and measles. Each polio vaccination consists of 4 doses and each measles vaccination consists of 2 doses. Last year Dr. potter gave a total of 60 vaccinations that consisted of a total of 184 doses. How many polio vaccinations did dr. potter give last year
Answer:
Dr. Potter gave 32 polio vaccinations and 28 measles vaccinations
Step-by-step explanation:
Total of 184 doses
Polio vaccination= 4 doses
Measles vaccination=2 doses
184=4p+2m
92=2p+m
Lets plug in 32+28
92=(2*32)+28
92=64+28
p=32, m=28
At a middle school, 30% of students buy lunch in the cafeteria and the remaining students bring lunch from home. A spinner with 10 equal-sized sections numbered 0-9 will be used to simulate the lunch trend. How can you design a simulation to guess whether the next 20 students buy lunch or bring lunch from home?
Allocating 3 portions to x and 7 portions to y.
Step-by-step explanation:
Given that,
The percentage of students buy lunch at cafeteria: 30% = x = 0.3
Hence, the students bringing the lunch from home would be = y = 1 - 0.3 = 0.7
Now, the spinner has that equal-sized sections. Also, the probability of x and y are 0.3 and 0.7.
After multiplying both the probability by 10, we get
x = 3
y = 7
It shows that for every three students who buy lunch from cafeteria, seven students bring food from home. Hence, we can allocate the side of spinner for simulation in such as way:
Section 0 = y
Section 1 = y
Section 2 = x
Section 3 = y
Section 4 = y
Section 5 = x
Section 6 = y
Section 7 = y
Section 8 = x
Section 9 = y
A guitar had been marked down by 34% and sold for $825.
What is the original price of the guitar?
The original price of the guitar was $1250, calculated by taking the sale price of $825 and dividing it by 0.66, which represents the remaining percentage of the price after a 34% discount.
Explanation:To find the original price of the guitar that has been marked down by 34% and sold for $825, we first need to consider that after the markdown, the guitar's price is equivalent to 100% - markdown percentage of the original price. In this case, it's 100% - 34% = 66% of the original price.
Let's denote the original price by P. Since 66% of P equates to $825, we can set up the following equation:
0.66 imes P = $825
Solving for P:
P = $825 / 0.66
P = $1250
Therefore, the original price of the guitar was $1250.
The Harrisons drove 304.2 miles in 6.25 hours. What was their average speed, to the nearest tenth?
Answer:
The average speed was 48.7 miles per hour.
Step-by-step explanation:
The average speed v is given by the following formula:
[tex]v = \frac{d}{t}[/tex]
In which d is the distance, and t is the time.
The Harrisons drove 304.2 miles in 6.25 hours
This means that [tex]d = 304.2, t = 6.25[/tex]
We have the distance is miles and the time in hours, so the distance is in miles per hour.
So
[tex]v = \frac{304.2}{6.25} = 48.7[/tex]
The average speed was 48.7 miles per hour.
The average rainfall in Phoenix is 8.29 inches per year. The table shows recent data on the difference in annual rainfall from the average. Phoenix Annual Total Rainfall Year Rainfall compared to average yearly rainfall 2008 +6.57 inches 2009 –2.68 inches 2010 +12.26 inches 2011 –4.38 inches 2012 –4.46 inches Which list represents the years from driest to wettest?
Answer:
c
Step-by-step explanation:
Gary has 63 counters.he pura them in an array with 9 columbs how many rows are there
National data indicates that 35% of households own a desktop computer. In a random sample of 570 households, 40% owned a desktop computer. Does this provide enough evidence to show a difference in the proportion of households that own a desktop? Identify the appropriate null and alternative hypotheses.
Answer:
Yes, this provide enough evidence to show a difference in the proportion of households that own a desktop.
Step-by-step explanation:
We are given that National data indicates that 35% of households own a desktop computer.
In a random sample of 570 households, 40% owned a desktop computer.
Let p = population proportion of households who own a desktop computer
SO, Null Hypothesis, [tex]H_0[/tex] : p = 25% {means that 35% of households own a desktop computer}
Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 25% {means that % of households who own a desktop computer is different from 35%}
The test statistics that will be used here is One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of 570 households who owned a desktop computer = 40%
n = sample of households = 570
So, test statistics = [tex]\frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }[/tex]
= 2.437
Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that % of households who own a desktop computer is different from 35%.
Suppose it is known that 10% of all people in Texas have a specific blood type. Suppose we take a random sample of 500 Texas residents. We want to find chance that fewer than 40 Texas residents in this sample have that blood type. In the next 4 questions, find the box model, the average and standard deviation of the box and use these values to find the expected value and standard error. Then calculate the associated chance of having fewer than 40 Texas residents in the sample with that specific blood type. Suppose you calculated EV and SE correctly in the previous two problems. The chance that fewer than 40 Texas residents in this sample have that blood type is the area under the normal curve to the:
Answer:
a) 50
b) 6.71
c) 0.0681
Step-by-step explanation:
check the attached file below
Write two different word problems that can be represented by the following equation 5.25x+7.50=75.75
Answer:
12.75
Step-by-step explanation:
A survey of UF students asked for their employment status and their year in school. The results appear below.
yr in school job no job
Freshman 16 22
Sophomore 24 15
Junior 17 20
Senior 25 19
Super Senior 8 5
What is the distribution of the test statistic under the null hypothesis
Answer:
There is no relationship between your year in school and having a job.
Step-by-step explanation:
In this instance, the chi sq test need to be performed.
Chi sq is used to determine if there is a significant relationship between two categorical variables.
The two variables here are year in school and employment status.
The two variables are independent(no relationship exists)
This implies that there is null hypothesis
Therefore, the Null Hypothesis is
There is no relationship between your year in school and having a job.
Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. For example, enter 1+4+9 instead of 12+22+32 or 14, or enter 1/2+1/2 instead of 0.5+0.5 or 1. The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.
The correct question is:
Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. For example, enter 1 + 4 + 9 instead of 1² + 2² + 3² or 14, or enter 1/2 + 1/2 instead of 0.5 + 0.5 or 1.
The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.
[tex](a) \sum_{k=0}^5 2^k \\ \\(b) \sum_{k=2}^7 \frac{1}{k} \\ \\(c) \sum_{k=1}^5 k^2 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k[/tex]
Answer:
[tex](a) \sum_{k=0}^5 2^k = $1 + 2 + 4 + 8 + 16 + 32$ \\ \\(b) \sum_{k=2}^7 \frac{1}{k} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7} \\ \\(c) \sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]
Step-by-step explanation:
[tex](a) \sum_{k=0}^5 2^k\\For k = 0: 2^k = 2^0 = 1\\For k = 1: 2^1 = 2\\For k = 2: 2^2 = 4\\For k = 3: 2^3 = 8\\For k = 4: 2^4 = 16\\For k = 5: 2^5 = 32\\\sum_{k=0}^5 2^k = 1 + 2 + 4 + 8 + 16 + 32[/tex]
[tex](b) \sum_{k=2}^7 \frac{1}{k}\\For k = 2: 1/2\\For k = 3: 1/3\\For k = 4: 1/4\\For k = 5: 1/5\\For k = 6: 1/6\\For k = 7: 1/7\\ \sum_{k=2}^7 \frac{1}{k} = 1/2 + 1/3 + 1/4 + 1/5 + 1/6+ 1/7[/tex]
[tex](c) \sum_{k=1}^5 k^2\\For k = 1: 1^2 = 1\\For k = 2: 2^2 = 4\\For k = 3: 3^2 = 9\\For k = 4: 4^2= 16\\For k = 5: 5^2 = 25\\\sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25[/tex]
[tex](d) \sum_{k=1}^6 \frac{1}{6}\\For k = 1: 1/6\\For k = 2: 1/6\\For k = 3: 1/6\\For k = 4: 1/6\\For k = 5: 1/6\\For k = 6: 1/6\\ \sum_{k=1}^6 \frac{1}{6} = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6[/tex]
[tex](e) \sum_{k=1}^6 2k\\For k = 1: 2\times1 = 2\\For k = 2: 2\times2 = 4\\For k = 3: 2\times3 = 6\\For k = 4: 2\times4 = 8\\For k = 5: 2\times5 = 10\\For k = 6: 2\times6 = 12\\\sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]
A comparative study of organic and conventionally grown produce was checked for the presence of E. coli. Results are summarized below. The Prevalence of E. Coli in Organic and Conventional Produce Sample Size E. Coli Prevalence Organic 200 5 Conventional 500 25 Is there a significant difference in the proportion of E. Coli in organic vs. conventionally grown produce? Test at α = 0.10. Be sure to report your hypotheses, show all work, and explain the meaning of your answer.
Answer:
The calculated z- value = 1.479 < 1.645 at 0.10 or 90% level of significance.
The null hypothesis is accepted at 90% level of significance.
There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce.
Step-by-step explanation:
Step:-(i)
Given first sample size n₁ = 200
The first sample proportion [tex]p_{1} = \frac{5}{200} = 0.025[/tex]
Given first sample size n₂= 500
The second sample proportion [tex]p_{2} = \frac{25}{500} = 0.05[/tex]
Step:-(ii)
Null hypothesis :H₀:There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
Alternative hypothesis:-H₁
There is significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
level of significance ∝=0.10
Step:-(iii)
The test statistic
[tex]Z =\frac{p_{1} - p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]
where p = [tex]\frac{n_{1} p_{1} + n_{2}p_{2} }{n_{1}+n_{2} }= \frac{200X0.025+500X0.05 }{500+200}[/tex]
p = 0.0428
q = 1-p =1-0.0428 = 0.9572
[tex]Z =\frac{0.025- 0.05}{\sqrt{0.0428X0.9571(\frac{1}{200 }+\frac{1}{500 } } }[/tex]
Z = -1.479
|z| = |-1.479|
z = 1.479
The tabulated value z= 1.645 at 0.10 or 90% level of significance.
The calculated z- value = 1.479 < 1.645 at 0.10 or 90% level of significance.
The null hypothesis is accepted at 90% level of significance.
Conclusion:-
There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
A box is 4 inches wide, 5 inches long and 3 inches tall. What’s the equation that would be used to find the surface area of the box?
Equation used to determine the surface area of the box : 2( lb + bh + lh)
Surface area of the box : 94in²
Given, A box is 4 inches wide, 5 inches long and 3 inches tall.
Formula of surface area of cuboid : 2( lb + bh + lh)
Here,
l = 5in
b = 4in
h = 3in
Substitute the values,
Surface area = 2(5×4 + 4×3 + 5×3)
Surface area = 94in²
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cuanto es dos cuartos mas un octavo
Answer:
0.625
Step-by-step explanation:
What is the solution of a system defined by 5x + 2y = 30 and -5x + 4y = 0?
Answer:
x=4 y=5
Step-by-step explanation:
5x + 2y = 30 and -5x + 4y = 0
Solve by adding the two equations together to eliminate x
5x + 2y = 30
-5x + 4y = 0
----------------------
6y = 30
Divide each side by 6
6y/6 = 30/6
y =5
Now we solve for x
5x + 2y = 30
5x +2(5) = 30
5x+10 = 30
Subtract 10 from each side
5x+10-10 = 30-10
5x = 20
Divide each side by 5
5x/5 = 20/5
x = 4
Find the area of the figure.
5.5 cm
20 cm
The area of the figure is
Answer:
25.5 cm²
Step-by-step explanation:
5.5 cm × 20 cm = 25.5 cm²
Answer:i think the answer is 110 but it would be more helpful if i knew what kind of shape it is
Step-by-step explanation:
to find the area of a square or rectangle(assuming this is a square or rectangle) you multiply the base by the height
Solve the right triangle. Round to two decimal places
A=20 , b=6.00
Answer:14
Step-by-step explanation:
What is the volume of this rectangular prism? 10/3 cm 4/5 cm 1/5 cm
The volume of the rectangular prism with dimensions 10/3 cm, 4/5 cm, and 1/5 cm is 8/15 cm³.
Explanation:The volume of a rectangular prism can be found using the formula: Volume = length x width x height. In this case, the length, width, and height of the prism are given as 10/3 cm, 4/5 cm, and 1/5 cm respectively. Replace these dimensions in the formula:
Volume = (10/3 cm) x (4/5 cm) x (1/5 cm) = 8/15 cm³.
Therefore, the volume of the rectangular prism is 8/15 cm³.
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A total of 58 trophies will be given out at the swim meet each box holds 6 trophies how many boxes are needed to carry the trophies
Answer:
You will need 10 boxes
Step-by-step explanation:
Nine boxes will only hold 54 trophies so you need one more than nine.
Answer:
You will need 348 boxes for the trophies
Step-by-step explanation:
just multiply 58 and 6
Suppose that the quality control manager for a cereal manufacturer wants to ensure that bags of cereal are being filled correctly. The equipment is calibrated to fill bags with a mean of 17 oz of cereal with a standard deviation of 0.2 oz. The quality control inspector selects a random sample of 52 boxes and finds that the mean amount of cereal for these boxes is 17.04 oz. He uses this data to conduct a one-sample z ‑test with a null hypothesis of H 0 : μ = 17 against the alternative hypothesis H 1 : μ ≠ 17 , where μ is the mean amount of cereal in each box. He calculates a z ‑score of 1.44 and a P -value of 0.1499 .
Are these results statistically significant at a significance level of 0.05?
No, these results are not statistically significant because p>0.05,
No, these results are not statistically significant because p < 0.05.
Yes, these results are statistically significant because p < 0.05.
Yes, these results are statistically significant because p > 0.05.
Answer: No, these results are not statistically significant because
p > 0.05
Step-by-step explanation:
The null hypothesis is
H0 : μ = 17
The alternative hypothesis is
H 1 : μ ≠ 17
where μ is the mean amount of cereal in each box.
The p value that he got is 0.1499. This is greater than alpha = 0.05 which is the given level of significance.
If the level of significance is lesser than the p value, we would accept the null hypothesis.
Therefore, the correct option is
No, these results are not statistically significant because p>0.05
Use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form. [one half (cosine StartFraction pi Over 16 EndFraction plus i sine StartFraction pi Over 16 EndFraction )]Superscript 8
Answer:
[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]
Step-by-step explanation:
The complex number given is
[tex]z = (\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2}[/tex]
Now, remember that the DeMoivre's theorem states that
[tex]( \cos(x) + i\sin(x) )^n = \cos(nx) + i\sin(nx)[/tex]
Then for this case we have that
[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]
Suppose you are the CEO of a company that produces sheets of metal that are 1 centimeter thick. This metal is evaluated on the basis of its hardness which is determined by measuring the depth of penetration of a hardened point. Suppose that this depth of penetration is normally distributed with a mean of 1 millimeter and a standard deviation of .02 millimeters.
You are on trial for distributing faulty metal. If the metal is deemed faulty when the depth of penetration is more than 1.3 millimeters, what is the probability you are guilty?
Answer:
0% probability you are guilty
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 = 1, \sigma = 0.02[/tex]
If the metal is deemed faulty when the depth of penetration is more than 1.3 millimeters, what is the probability you are guilty?
This is 1 subtracted by the pvalue of Z when X = 1.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.3 - 1}{0.02}[/tex]
[tex]Z = 15[/tex]
[tex]Z = 15[/tex] has a pvalue of 1
1 - 1 = 0
0% probability you are guilty
We can use the z-score formula to calculate the probability of being guilty of distributing faulty metal based on the depth of penetration. The probability is practically zero.
Explanation:To find the probability that you are guilty of distributing faulty metal, we need to calculate the probability that the depth of penetration is more than 1.3 millimeters. Since the depth of penetration is normally distributed with a mean of 1 millimeter and a standard deviation of 0.02 millimeters, we can use the z-score formula to standardize the value. The z-score is calculated as (x - μ) / σ, where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.
Substituting the values into the formula, we have z = (1.3 - 1) / 0.02 = 15. Therefore, we need to find the probability that the z-score is greater than 15. Using a standard normal distribution table or calculator, we find that this probability is practically zero. Hence, the probability that you are guilty is practically zero.
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A certain circle can be represented by the following equation. x2 + y2 + 10x + 12y + 25 = 0. What is the center of this circle? What is the radius of this circle? Please help!
Let's try to complete the squares.
The x-part starts with [tex]x^2+10x[/tex], which is the beginning of [tex]x^2+10x+25=(x+5)^2[/tex]. So, we'll think of [tex]x^2+10x[/tex] as [tex](x+5)^2-25[/tex]
Similarly, we have that
[tex]y^2+12y = (y+6)^2-36[/tex]
So, the equation becomes
[tex]x^2 + y^2 + 10x + 12y + 25 = 0 \iff (x+5)^2-25 + (y+6)^2-36+25=0 \iff (x+5)^2+ (y+6)^2-36=0 \iff (x+5)^2+ (y+6)^2=36[/tex]
Now we have writte the equation of the circle in the form
[tex](x-k)^2+(y-h)^2=r^2[/tex]
When the equation is in this form, everything is more simple: the center is [tex](k,h)[/tex] and the radius is [tex]r[/tex].
Answer:
Center// (-5,-6)
Radius// 6
The Colorado Mining and Mineral Company has 1000 employees engaged in its mining operations. It has been estimated that the probability of a worker meeting with an accident during a 1-yr period is 0.08. What is the probability that more than 70 workers will meet with an accident during the 1-yr period
Answer:
86.65% probability that more than 70 workers will meet with an accident during the 1-yr period
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value 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]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.08, n = 1000[/tex]
So
[tex]\mu = E(X) = np = 1000*0.08 = 80[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.08*0.92} = 8.58[/tex]
What is the probability that more than 70 workers will meet with an accident during the 1-yr period
Using continuity correction, this is [tex]P(X \geq 70 + 0.5) = P(X \geq 70.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 70.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70.5 - 80}{8.58}[/tex]
[tex]Z = -1.11[/tex]
[tex]Z = -1.11[/tex] has a pvalue of 0.1335
1 - 0.1335 = 0.8665
86.65% probability that more than 70 workers will meet with an accident during the 1-yr period
The probability that more than 70 workers will be involved in an accident is 0.8665
The given parameters are:
[tex]\mathbf{n = 1000}[/tex] --- population
[tex]\mathbf{p = 0.08}[/tex] --- the probability that a worker meets an accident
[tex]\mathbf{x = 70}[/tex] -- the number of workers
Start by calculating the mean and the standard deviation
[tex]\mathbf{\mu = np}[/tex] --- mean
So, we have:
[tex]\mathbf{\mu = 1000 \times 0.08}[/tex]
[tex]\mathbf{\mu = 80}[/tex]
[tex]\mathbf{\sigma = \sqrt{\mu(1 - p)}}[/tex]
So, we have:
[tex]\mathbf{\sigma = \sqrt{80 \times (1 - 0.08)}}[/tex]
[tex]\mathbf{\sigma = \sqrt{73.6}}[/tex]
[tex]\mathbf{\sigma = 8.58}[/tex]
The probability is then represented as
[tex]\mathbf{P(x > 70) = P(x > 70.5)}[/tex] ---- By continuity correction
Calculate the z-score for x = 70.5
[tex]\mathbf{z = \frac{x - \mu}{\sigma}}[/tex]
So, we have:
[tex]\mathbf{z = \frac{70.5 - 80}{8.58}}[/tex]
[tex]\mathbf{z = -1.11}[/tex]
So, we have:
[tex]\mathbf{P(x > 70) = P(z > -1.11)}[/tex]
Using z-scores of probabilities, we have:
[tex]\mathbf{P(x > 70) = 0.8665}[/tex]
Hence, the probability that more than 70 workers will be involved in an accident is 0.8665
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In 2014, the Community College Survey of Student Engagement reported that 32% of the students surveyed rarely or never use academic advising services. Suppose that in reality, 42% of community college students rarely or never use academic advising services at their college. In a simulation we select random samples from this population. For each sample we calculate the proportion who rarely or never use academic advising services. If we randomly sample 200 students from this population repeatedly, the standard error is approximately 3.5%. Is it unusual to see 32% who rarely or never use academic advising services in one of these samples
Answer:
[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples
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.
Z scores below -2 are considered unusually low.
Z scores above 2 are considered unusually high.
For a sample proportion p in a sample of size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem, we have that:
[tex]\mu = 0.42, \sigma = \sqrt{\frac{0.42*0.58}{200}} = 0.0349[/tex]
Is it unusual to see 32% who rarely or never use academic advising services in one of these samples
What is the z-score for X = 0.32?
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.32 - 0.42}{0.0349}[/tex]
[tex]Z = -2.865[/tex]
[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples
No, it is not unusual to see 32% of students rarely or never using academic advising services in one of these samples.
Explanation:To determine if it is unusual to see 32% of students rarely or never using academic advising services in one of these samples, we can compare it to the range of values that would be considered usual. In this case, we can use the 95% confidence interval provided, which states that the true proportion of community college students who rarely or never use academic advising services is between 0.113 and 0.439. If the observed proportion falls within this interval, it would be considered usual; otherwise, it would be considered unusual.
Since 32% falls within the range of 0.113 and 0.439, it is considered a usual value. Therefore, it is not unusual to see 32% of students rarely or never use academic advising services in one of these samples.
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survey on televisions requiring repairs within four years was conducted. Nineteen out of 200 televisions from company A and 25 out of 200 televisions from company B needed repairs. Do these data show that televisions from company A are more reliable than televisions from company B?
Answer:
[tex]z=\frac{0.095-0.125}{\sqrt{0.11(1-0.11)(\frac{1}{200}+\frac{1}{200})}}=-0.959[/tex]
[tex]p_v =P(Z<-0.959)=0.169[/tex]
Comparing the p value with the significance level assumed[tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to to FAIL to reject the null hypothesis, and we can't conclude that company A are more reliable than televisions from company B at 5% of significance.
Step-by-step explanation:
Data given and notation
[tex]X_{1}=19[/tex] represent the number of tvs who need a repair for A
[tex]X_{2}=25[/tex] represent the number of tvs who need a repair for B
[tex]n_{1}=200[/tex] sample 1 selected
[tex]n_{2}=200[/tex] sample 2 selected
[tex]p_{1}=\frac{19}{200}=0.095[/tex] represent the proportion estimated for the sample A
[tex]p_{2}=\frac{25}{200}=0.125[/tex] represent the proportion estimated for the sample B
[tex]\hat p[/tex] represent the pooled estimate of p
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
Concepts and formulas to use
We need to conduct a hypothesis in order to check if company A are more reliable than televisions from company B (that means p1<p2) , the system of hypothesis would be:
Null hypothesis:[tex]p_{1} \geq p_{2}[/tex]
Alternative hypothesis:[tex]p_{1} < p_{2}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{19+25}{200+200}=0.11[/tex]
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.095-0.125}{\sqrt{0.11(1-0.11)(\frac{1}{200}+\frac{1}{200})}}=-0.959[/tex]
Statistical decision
Since is a left sided test the p value would be:
[tex]p_v =P(Z<-0.959)=0.169[/tex]
Comparing the p value with the significance level assumed[tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to to FAIL to reject the null hypothesis, and we can't conclude that company A are more reliable than televisions from company B at 5% of significance.
Final answer:
To determine if televisions from company A are more reliable than televisions from company B, we can perform a hypothesis test.
Explanation:
To determine if televisions from company A are more reliable than televisions from company B, we can perform a hypothesis test. We will compare the proportions of televisions requiring repairs in the two companies.
Step 1: State the hypotheses:
H0: The proportion of televisions requiring repairs in company A is the same as in company B.
HA: The proportion of televisions requiring repairs in company A is less than in company B.
Step 2: Set the significance level, let's say α = 0.01.
Step 3: Calculate the test statistic. We will use the Z-test for comparing proportions.
Step 4: Calculate the p-value.
Step 5: Compare the p-value to the significance level. If the p-value is less than α, we reject the null hypothesis and conclude that televisions from company A are more reliable than televisions from company B.
By performing the above steps, we can determine if the data shows that televisions from company A are more reliable than televisions from company B.
A biologist is trying to determine the average age of a local forest. She cuts down 18 randomly selected trees and counts the number of tree rings, which can be used to estimate the age of the tree. What critical value should she use to construct a 99% confidence interval
Answer:
The critical value of t for 99% confidence interval is 2.898.
Step-by-step explanation:
The complete question is:
A biologist is trying to determine the average age of a local forest. She cuts down 18 randomly selected trees and counts the tree rings. They find the average number of tree rings to be 83 with a variance of 320. What is the critical value for the 99% confidence interval?
The population variance is not known and the sample size is too small. So a t-confidence interval will be used to estimate the population mean age of a local forest.
The (1 - α)% confidence interval for population mean is:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]
The information provided is:
n = 18
(1 - α)% = 99%
The degrees of freedom of the critical value of t is:
n - 1 = 18 - 1 = 17
Compute the critical value of t as follows:
[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (18-1)}=t_{0.005, 17}=2.898[/tex]
*Use a t-table.
Thus, the critical value of t for 99% confidence interval is 2.898.
A florist charges $12.00 for delivery plus an additional $1.50 per mile from the flower shop. The florist pays the delivery driver $0.75 per mile and $4.50 for gas per delivery. If x is the number of miles a delivery location is from the flower shop, what expression models the amount of money the florist earns for each delivery?
The amount the florist earns for each delivery, with 'x' being the miles away from the flower shop, can be modeled with the equation: Earnings = (7.5 + 0.75x). This represents the fixed net income of $7.5 and $0.75 per mile after paying the driver.
Explanation:The florist charges $12.00 for delivery and an additional $1.50 per mile from the flower shop. However, the florist also has costs to cover, namely $0.75 per mile to pay the driver, and $4.50 for gas per delivery. The net earning per delivery, with 'x' representing the number of miles a delivery location is from the flower shop, can be modeled by the following algebraic expression: Earnings = (12 + 1.5x) - (0.75x + 4.5).
This actually simplifies to: Earnings = (7.5 + 0.75x). The 7.5 is the fixed net income for each delivery (gross earnings minus the gasoline cost) and 0.75x is the per-mile net income after the driver is paid.
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At a large university, the mean amount spent by students for cell phone service is $58.90 per month with a standard deviation of $3.64 per month. Consider a group of 44 randomly chosen university students. What is the probability that the mean amount of their monthly cell phone bills is more than $60?
Answer:
2.28% probability that the mean amount of their monthly cell phone bills is more than $60
Step-by-step explanation:
To solve this question, we need 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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 58.90, \sigma = 3.64, n = 44, s = \frac{3.64}{\sqrt{44}} = 0.54875[/tex]
What is the probability that the mean amount of their monthly cell phone bills is more than $60?
This is 1 subtracted by the pvalue of Z when X = 60. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{60 - 58.90}{0.54875}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that the mean amount of their monthly cell phone bills is more than $60