Answer: 0.0026
Step-by-step explanation:
Let p denotes the proportion of students plan to go into general practice.
As per given , we have
Alternative hypothesis : [tex] H_a: p>0.28[/tex]
Since the alternative hypothesis [tex](H_a)[/tex] is right-tailed so the test is a right-tailed test.
Also , it is given that ,
i.e. sample size : = 130
x= 490
[tex]\hat{p}=0.39[/tex]
Test statistic(z) for population proportion :
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
, where p=population proportion.
[tex]\hat{p}[/tex]= sample proportion
n= sample size.
[tex]z=\dfrac{0.39-0.28}{\sqrt{\dfrac{0.28(1-0.28)}{130}}}\\\\=\dfrac{0.11}{0.0393798073988}=2.79330975101\approx2.79[/tex]
P-value for right-tailed test = P(z>2.79)=1-P(z≤ 2.79) [∵P(Z>z)=1-P(Z≤z)]
=1- 0.9974=0.0026 [using z-value table]
Hence, the P-Value for a test of the school's claim = 0.0026
To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%. Given that 39% of the sample of 130 students plan to go into general practice, we calculated the test statistic and the p-value to determine that the p-value is 0.3078.
Explanation:To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%.
Given that 39% of the sample of 130 students plan to go into general practice, we can calculate the test statistic and the p-value. Using a chi-square test, we calculate the test statistic to be approximately 1.307. The degrees of freedom for this test is 1.
Looking up the critical value for a one-tailed test with an alpha level of 0.05 and 1 degree of freedom, the critical value is approximately 3.841. Since the test statistic is less than the critical value, we fail to reject the null hypothesis. Therefore, the p-value is greater than 0.05.
Therefore, the correct answer for the P-value is 0.3078.
Learn more about Hypothesis testing here:https://brainly.com/question/34171008
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Peter and Amanda collect seashells on a beach. They both had 0 seashells when they started. Peter collects 5 seashells every minute, and Amanda collects 8 seashells every minute. Which graph and explanation correctly represent the number of shells Peter and Amanda have collected?
Just wondering I know the equation but I don't get the graphs??? which graph would it be
Answer:
;-; I belive it's the 1st or 2nd one
Step-by-step explanation:
hope this helps you.
ask me for explanation if you want and I'll try to explane it the best I Can.
(P.S;please mark me as brainlyest,ty in advaince)
Given a line passing through points (1, 0) and (4,9),
what is the slope of the line?
Answer:
3
Step-by-step explanation:
The slope (m) is the ratio of the difference in y-values to the corresponding difference in x-values:
m = (y2 -y1)/(x2 -x1)
m = (9 -0)/(4 -1) = 9/3
m = 3
The slope of the line is 3.
What is unit price ? Use in your own words .
Unit price means the cost per unit.
I'll provide an example.
If 4 bars of soap cost $2.35, find the unit price. Round to the nearest cent if necessary.
Remember that the unit price is the cost per unit which in this case is the cost per bar of soap. Since $2.35 is the cost for 4 bars of soap, to find the cost per bar of soap, we divide 4 into $2.35.
When we do this, we get an answer of 0.587 which rounds up to 59 cents.
Therefore, the unit price for soap is 59 cents.
Point C is 13 of the way from point A to point B. Which statement is NOT true?
A. Point C divides segment AB so that AC:CB is 1:3.
B. If AC = 5, then CB = 10.
C. Point C is 2/3 of the way from point B to point A.
D. Point C divides segment AB in a ratio of 1:2.
Answer:
A. Point C divides segment AB so that AC:CB is 1:3
Step-by-step explanation:
Like many multiple-choice questions, you don't actually need to know how to work the problem. You just need to know what the question means.
Comparing answers, you see that choices A and D cannot both be right. You also note that choice D says the same thing as choice B.
Since there is only one False answer, it must be choice A.
__
When the division is into segments that are 1/3 the length and 2/3 the length, the segments have the ratio ...
(1/3) : (2/3) = 1 : 2
The incorrect statement is B. If AC = 5, then CB should be 15, not 10, because CB would be twice the length of AC, given that point C is 1/3 of the way from A to B.
Explanation:The question asks us to identify which statement about point C being 1/3 of the way from point A to point B on a line segment is NOT true. Let's analyze each statement.
A. Point C divides segment AB so that AC:CB is 1:3. Since point C is 1/3 of the way from A to B, this statement is true because for every 1 part AC, there are 3 parts CB, making the ratio of AC to CB 1:3.B. If AC = 5, then CB = 10. This statement is false because if AC represents 1/3 of the total distance, then CB should be twice as long as AC (since it would represent 2/3), making CB = 15 if AC = 5.C. Point C is 2/3 of the way from point B to point A. This is true if we consider the direction from B to A, as C would be 1/3 of the way from A to B, or 2/3 from B to A.D. Point C divides segment AB in a ratio of 1:2. This statement is also false because based on the given condition, it should divide AB in a ratio of 1:3, not 1:2.The answer is statement B. If AC = 5, then CB would not be 10; it would be 15 since it is twice the length of AC, given point C is 1/3 of the way from A to B.
Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test the hypothesis that the proportion of one-person households exceeds 0.27. A random sample of 125 households found that 43 consisted of one person. To conduct the hypothesis test, what distribution would you use to calculate the critical value and the p-value?
Answer:
To conduct the hypothesis test I would use the standard normal distribution to calculate the critical value and the p-value.
Step-by-step explanation:
To conduct the hypothesis test I would use the standard normal distribution because there is a large sample size of n = 125 households. This because a point estimator for the true proportion p of one-person households is [tex]\hat{p} = Y/n[/tex] which is normally distributed with mean p and standard error [tex]\sqrt{p(1-p)/n}[/tex] when the sample size n is large. Here Y is the random variable that represents the number of one-person households observed. Then the test statistic is [tex]Z = \frac{\hat{p}-0.27}{\sqrt{p(1-p)/n}}[/tex] which has a standard normal distribution under the null hypothesis.
A data set includes data from student evaluations of courses.
The summary statistics are n=80, x overbar =4.39, s =2.29.
Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.50.
Assume that a simple random sample has been selected.
Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Answer:
We fail to reject the null hypothesis at the significance level of 0.05.
Step-by-step explanation:
We have a larga sample size of n = 80, [tex]\bar{x} = 4.39[/tex] and s = 2.29. We want to test
[tex]H_{0}: \mu = 4.50[/tex] vs [tex]H_{1}: \mu \neq 4.50[/tex] (two-tailed alternative)
Because we have a large sample, our test statistic is
[tex]Z = \frac{\bar{X}-4.50}{s/\sqrt{n}}[/tex] which is normal standard approximately. We have the observed value
[tex]z_{0} = \frac{4.39-4.50}{2.29/\sqrt{80}} = -0.4296[/tex].
The p-value is given by 2P(Z < -0.4296) = (2)(0.3337) = 0.6674 (because of the simmetry of the normal density)
With the significance level [tex]\alpha = 0.05[/tex], we fail to reject the null hypothesis because the p-value is greater than 0.05.
We are going to test if the mean is equals to 4.50, thus, the null hypothesis is:
[tex]H_0: \mu = 4.5[/tex]
At the alternative hypothesis, we test if the mean is different to 4.50, that is:
[tex]H_1: \mu \neq 4.50[/tex]
Since we have the standard deviation for the sample, the t-distribution is used. The value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
For this problem:
[tex]t = \frac{4.39 - 4.5}{\frac{2.29}{\sqrt{80}}}[/tex]
[tex]t = -0.43[/tex]
We are testing if the mean is different from a value, thus, the p-value of test is found using a two-tailed test, with [tex]t = -0.43[/tex] and 80 - 1 = 79 df.
Using a t-distribution calculator, the p-value is of 0.6684.
The p-value is of 0.6684 > 0.05, which means that we can conclude that the population of student course evaluations has a mean equal to 4.50.
A similar problem is given at https://brainly.com/question/24989605
Quadrilateral EFGH is on a coordinate plane. Which statement is true?
Answer:
see below
Step-by-step explanation:
Opposite sides are parallel in a parallelogram, so if they have different slope, the figure will not be a parallelogram.
_____
Comments on other answer choices
If the slope of diagonal EG is perpendicular to that of diagonal FH, it only proves the figure is some sort of kite. The figure may or may not be a parallelogram.
Adjacent sides being different lengths does not prove anything (except that the figure is not a rhombus).
Proving angle F is a right angle does not prove anything else about the shape of the figure. The figure may or may not be a parallelogram. (If it is a parallelogram, it is also a rectangle.)
Answer:
Have no fear, the answer is top left GIVE BRANLIEST PLZ
Step-by-step explanation:
Circle the best answer. The choice between a z-test and a t-test for a population mean depends primarily on: a. the sample size. b. the level of significance. c. whether a one- or two-tailed test is indicated. d. whether the given standard deviation is from the population or the sample. e. a z-test should never be used. 3
Answer:
a)The sample size
Step-by-step explanation:
t - student distribution should be used whe we are facing a Normal Distribution population andwhen a sample size n is below 30.
t- student distribution is also a associated to a bell shape but is more flat and the values are more spread, tails are much more wide.
t- student use the concept of degree of fredom ( each degree of fredom correspont to an specific curve). As degree of fredom increase the curves become more close to a bell shape. For n = 30 t-student curve is a bell shape one
7) Lindsay has some yellow and red flowers in a vase. The ratio of yellow to red flowers is 4:7. The number of yellow flowers in Lindsay’s vase is shown. How many red flowers are in the vase?
Answer:
[tex]\frac{7}{4}[/tex]×x
Step-by-step explanation:
given ratio of yellow to red flowers is 4:7.
let number of yellow flowers in Lindsay’s vase is x
let number of red flowers in Lindsay’s vase is y
[tex]\frac{x}{y}[/tex]=[tex]\frac{4}{7}[/tex]
number of red flowers such that x is known is y= [tex]\frac{7}{4}[/tex]×x
A lakefront resort is planning for its summer busy season. It wishes to estimate with 95% confidence the average number of nights each guest will stay for a consecutive visit. Using a sample of guests who stayed last year, the average number of nights per guest is calculated at 5 nights. The standard deviation of the sample is 1.5 nights. The size of the sample used is 120 guests and the resort desires a precision of plus or minus .5 nights. What is the standard error of the mean in the lakefront resort example? Within what range below can the resort expect with 95% confidence for the true population means to fall? Show the calculation; otherwise, the answer will not be accepted.
Answer:
[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]
The 95% confidence interval would be given by (4.729;5.271)
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the "range of values below and above the sample statistic in a confidence interval".
The standard error of a statistic is "the standard deviation of its sampling distribution or an estimate of that standard deviation"
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
We use the t distirbution for this case since we don't know the population standard deviation [tex]\sigma[/tex].
Where the standard error is given by: [tex]SE=\frac{s}{\sqrt{n}}[/tex]
And the margin of error would be given by: [tex]ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=120-1=119[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. And we see that [tex]t_{\alpha/2}=1.98[/tex]
The standard error would be given by:
[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]
Now we have everything in order to replace into formula (1) and calculate the interval:
[tex]5-1.98\frac{1.5}{\sqrt{120}}=4.729[/tex]
[tex]5+1.98\frac{1.5}{\sqrt{120}}=5.271[/tex]
So on this case the 95% confidence interval would be given by (4.729;5.271)
Final answer:
The standard error of the mean (SE) is calculated to be 0.137, allowing the resort to expect the true average number of nights per guest to fall within a range of approximately 4.73 to 5.27 nights with 95% confidence.
Explanation:
The standard error of the mean (SE) is calculated by dividing the sample standard deviation by the square root of the sample size. In this case, the standard error is 1.5 / sqrt(120) = 0.137. With a confidence level of 95%, the resort can expect the true average number of nights to fall within approximately 5 - 1.96 * 0.137 nights to 5 + 1.96 * 0.137 nights, which is roughly between 4.73 and 5.27 nights.
Of the total population of american households, including older americans and perhaps some not so old, 17.3 % recieve retiremet income. in a random sampple of 120 hoseholds, what is the probability that more than 20 household but fewer than 35 household recieve a retirement income?
Answer:20 hoseholds, what is the probability that more than
Step-by-step explanation:
Determine whether the function is a linear transformation. T: P2 → P2, T(a0 + a1x + a2x2) = (a0 + a1 + a2) + (a1 + a2)x + a2x2.
Answer with Step-by-step explanation:
We are given that a function
[tex]T:P_2\rightarrow P_2[/tex]
[tex]T(a_0+a_1x+a_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2[/tex]
We have to determine the given function is a linear transformation.
If a function is linear transformation then it satisfied following properties
[tex]1.T(x+y)=T(x)+T(y)[/tex]
2.[tex]T(ax)=aT(x)[/tex]
[tex]T(a_0+a_1x+a_2x^2+b_0+b_1x+b_2x^2)=T((a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2)=(a_0+b_0+a_1+b_1+a_2+b_2)+(a_1+b_1+a_2+b_2)x+(a_2+b_2)x^2[/tex]
[tex]T(a_0+a_1x+a_2x^2+b_0+b-1x+b_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2+(b_0+b_1+b_2)+(b_1+b_2)x+b_2x^2[/tex]
[tex]T(a_0+a_1x+a_2x^2+b_0+b-1x+b_2x^2)=T(a_0+a_1x+a_2x^2)+T(b_0+b_1x+b_2x^2)[/tex]
[tex]T(a(a_0+a_1x+a_2x^2))=T(aa_0+aa_1x+aa_2x^2)[/tex]
[tex]T(a(a_0+a_1x+a_2x^2))=(aa_0+aa_1+aa_2)+(aa_1+aa_2)x+(aa_2)x^2[/tex]
[tex]T(a(a_0+a_1x+a_2x^2))=a(a_0+a_1+a_2)+a(a_1+a_2)x+aa_2x^2=a((a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2)=aT(a_0+a_1x+a_2x^2)[/tex]
Hence, the function is a linear transformation because it satisfied both properties of linear transformation.
[tex]\( T \)[/tex] satisfies both additivity and scalar multiplication, [tex]\( T \)[/tex] is indeed a linear transformation.
To determine if the function [tex]\( T: P2 \to P2 \)[/tex], defined by [tex]\( T(a_0 + a_1 x + a_2 x^2) = (a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2 \)[/tex], is a linear transformation, we need to check two properties:
1. Additivity: [tex]\( T(u + v) = T(u) + T(v) \) for all \( u, v \in P2 \)[/tex].
2. Scalar Multiplication: [tex]\( T(cu) = cT(u) \) for all \( u \in P2 \) and \( c \in \mathbb{R} \)[/tex].
Let's verify these properties:
Additivity Check
Let [tex]\( u = a_0 + a_1 x + a_2 x^2 \) and \( v = b_0 + b_1 x + b_2 x^2 \)[/tex].
Compute [tex]\( T(u + v) \)[/tex]:
[tex]\[u + v = (a_0 + b_0) + (a_1 + b_1)x + (a_2 + b_2)x^2\][/tex]
[tex]\[T(u + v) = [(a_0 + b_0) + (a_1 + b_1) + (a_2 + b_2)] + [(a_1 + b_1) + (a_2 + b_2)]x + (a_2 + b_2)x^2\][/tex]
[tex]\[T(u + v) = [(a_0 + a_1 + a_2) + (b_0 + b_1 + b_2)] + [(a_1 + a_2) + (b_1 + b_2)]x + (a_2 + b_2)x^2\][/tex]
Compute [tex]\( T(u) + T(v) \)[/tex]:
[tex]\[T(u) = (a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2\][/tex]
[tex]\[T(v) = (b_0 + b_1 + b_2) + (b_1 + b_2)x + b_2 x^2\][/tex]
[tex]\[T(u) + T(v) = [(a_0 + a_1 + a_2) + (b_0 + b_1 + b_2)] + [(a_1 + a_2) + (b_1 + b_2)]x + (a_2 + b_2)x^2\][/tex]
Since [tex]\( T(u + v) = T(u) + T(v) \)[/tex], the function [tex]\( T \)[/tex] satisfies additivity.
Scalar Multiplication Check
Let [tex]\( u = a_0 + a_1 x + a_2 x^2 \)[/tex] and [tex]\( c \in \mathbb{R} \)[/tex].
Compute [tex]\( T(cu) \)[/tex]:
[tex]\[cu = c(a_0 + a_1 x + a_2 x^2) = (ca_0) + (ca_1)x + (ca_2)x^2\][/tex]
[tex]\[T(cu) = [(ca_0) + (ca_1) + (ca_2)] + [(ca_1) + (ca_2)]x + (ca_2)x^2\][/tex]
[tex]\[T(cu) = c[(a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2]\][/tex]
Compute [tex]\( cT(u) \)[/tex]:
[tex]\[T(u) = (a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2\][/tex]
[tex]\[cT(u) = c[(a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2]\][/tex]
Since [tex]\( T(cu) = cT(u) \)[/tex], the function [tex]\( T \)[/tex] satisfies scalar multiplication.
In a survey of 40 Clemson students, it was found that the mean age (in years) when they would like to get married is 27.4 with a standard deviation of 6. How many Clemson students would need to be surveyed to estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence?
Answer:
Step-by-step explanation:
To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to survey at least 92 Clemson students.
Explanation:To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to calculate the sample size needed for this level of precision. The formula to determine the sample size is:
n = (z * σ / E)^2
where n is the required sample size, z is the z-score corresponding to the desired confidence level (in this case 90% confidence level which corresponds to a z-score of 1.645), σ is the standard deviation, and E is the desired margin of error (1.5 years).
Plugging in the values, we get:
n = (1.645 * 6 / 1.5)^2
n = 91.154
Rounding up to the nearest whole number, we need to survey at least 92 Clemson students in order to estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence.
The package of a particular brand of rubber band says that the bands can hold a weight of 7 lbs. Suppose that we suspect this might be an overstatement of the breaking weight So we decide to take a random sample of 36 of these rubber bands and record the weight required to break each of them. The mean breaking weight of our sample of 36 rubber bands is 6.6 lbs. Assume that the standard deviation of the breaking weight for the entire population of these rubber bands is 2 lbs.Finding a random sample with a mean this low in a population with mean 7 and standard deviation 2 is very unlikely. a. True b. False
Answer:
False this value is very likely to find in this distribution
Step-by-step explanation:
With mean 7 ( μ ) and standad deviation (σ ) 2 we can observe, value 6.6 is close to the lower limit of the interval
μ ± 0,5 σ 7 ± 1 in which we should find 68,3 % of all values
(just 6 tenth to the left)
And of course 6.6 is inside the interval
μ ± 1 σ where we find 95.7 % of th value
We conclude this value is not unlikely at all
Final answer:
To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we can use hypothesis testing. The calculated z-score is less than the critical z-value, indicating that it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in the population. Therefore, the statement is true.
Explanation:
To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we need to use hypothesis testing. We can set up the null and alternative hypotheses as follows:
Null hypothesis (H0): The population mean breaking weight is equal to 7 lbs.
Alternative hypothesis (Ha): The population mean breaking weight is less than 7 lbs.
Next, we can calculate the z-score using the formula (sample mean - population mean)/(standard deviation/sqrt(sample size)). Substituting the values, we get (6.6 - 7)/(2/sqrt(36)) = -0.6/ (2/6) = -0.6/ (1/3) = -1.8.
We can now look up the critical z-value for a one-tailed test at a significance level of 0.05. The critical z-value is -1.645. Since the calculated z-score (-1.8) is less than the critical z-value (-1.645), we can reject the null hypothesis. Therefore, it is true that finding a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs is very unlikely.
Membership in an elite organization requires a test score in the upper 30% range. If the mean is equal to 115 and the standard deviation is equal to 12, find the lowest acceptable score that would enable a candidate to apply for membership. Assume the variable is normally distributed. (Show Work)
Answer:
Lowest acceptable score = 121.3
Step-by-step explanation:
Mean test score (μ) = 115
Standard deviation (σ) = 12
The z-score for any given test score 'X' is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
In this situation, the organization is looking for people who scored in the upper 30% range, that is, people at or above the 70-th percentile of the normally distributed scores. At the 70-th percentile, the corresponding z-score is 0.525 (obtained from a z-score table). The minimum score, X, that would enable a candidate to apply for membership is:
[tex]0.525=\frac{X-115}{12}\\X=121.3[/tex]
A right triangle whose hypotenuse is StartRoot 7 EndRoot7 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.
Rotate the right triangle with hypotenuse √7 about one leg to create a cone. Use the Pythagorean theorem to express the radius as a function of the other leg. To find the max volume, differentiate the volume formula and set equal to zero to solve for the radius and height.
Explanation:The problem presents a right triangle with hypotenuse of √7 meters. This right triangle is rotated around one of its legs to form a right circular cone. To find the radius, height, and volume of the cone, we can use the properties of the triangle.
Imagine that the triangle is rotated about the shorter leg. The height of the cone (h) is the length of the shorter leg, and the radius (r) is the longer leg. By the Pythagorean theorem, a² + b² = c², where c is the hypotenuse, a and b represent the legs of the triangle. Assume that a is the longer leg and b is the shorter leg.
From the Pythagorean theorem we get a² = (√7)² - b² = 7 - b² and so a = √(7 - b²).
The volume of the cone V = (1/3)πr²h = (1/3)π(√(7 - b²))²b = (1/3)π(7 - b²)b. Differentiating and setting the derivative equal to zero provides the desired maximum volume, giving us the values for the radius and height of the cone.
Learn more about Right Circular Cone here:
https://brainly.com/question/29767724
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In professional basketball games during 2009-2010, when Kobe Bryant of the Los Angeles Lakers shot a pair of free throws, 8 times he missed both, 152 times he made both, 33 times he only made the first, and 37 times he made the second. Is it plausible that the successive free throws are independent?
Answer:
Is plausible that the successive throws are independent
Step-by-step explanation:
1) Table with info given
The observed values are given by the following table
__________________________________________________
First shot Made Second shot missed Total
__________________________________________________
Made 152 33 185
Missed 37 8 45
__________________________________________________
Total 189 41 230
2) Calculations and test
We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:
Null hypothesis: Independence between two successive free throws
Alternative hypothesis: No Independence between two successive free throws
_____________________________________________________
First shot Made Second shot missed
_____________________________________________________
Made 189(185)/230=152.0217 41(185)/230=32.9783
Missed 189(45)/230=36.9783 41(45)/230=8.0217
_____________________________________________________
On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.
3) Calculate the chi square statistic
The statistic for this case is given by:
[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]
Where O represent the observed values and E the expected values. Replacing the values that we got we have this
[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]
Now with the calculated value we can find the degrees of freedom
[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.
Now we can calculate the p value
[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]
On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.
Can someone please help me with that one problem!
Answer:
AAS method can be used to prove that the two triangles are congruent.
Step-by-step explanation:
According to the question for the two triangles one pair of opposite angles are equal. One another pair of angles are equal for the two and one pair of sides are also equal of the two.
Hence, the two given triangles are congruent by AAS rule.
Hence, AAS method can be used to prove that the two triangles are congruent.
A dormitory has 40 students---12 sophomores, 8 juniors, and 20 seniors. Which of the following is equal to the number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left, all 8 juniors in the middle, and all 20 seniors on the right?
Answer:
The number of ways is equal to [tex]12!8!20![/tex]
Step-by-step explanation:
The multiplication principle states that If a first experiment can happen in n1 ways, then a second experiment can happen in n2 ways ... and finally a i-experiment can happen in ni ways therefore the total ways in which the whole experiment can occur are
n1 x n2 x ... x ni
Also, given n-elements in which we want to put them in a row, the total ways to do this are n! that is n-factorial.
For example : We want to put 4 different objects in a row.
The total ways to do this are [tex]4!=4.3.2.1=24[/tex] ways.
Using the multiplication principle and the n-factorial number :
The number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left,all 8 juniors in the middle, and all 20 seniors on the right are : The total ways to put all 12 sophomores in a row multiply by the ways to put the 8 juniors in a row and finally multiply by the total ways to put all 20 senior in a row ⇒ [tex]12!8!20![/tex]
There are 1.1657416 × 10^30 ways to arrange the students in a row with all the sophomores on the left, juniors in the middle, and seniors on the right.
Explanation:To find the number of ways to arrange the students in a row with all the sophomores on the left, juniors in the middle, and seniors on the right, we need to calculate the permutations of each group and then multiply them together.
The number of ways to arrange the 12 sophomores is 12!, which is 479,001,600.
The number of ways to arrange the 8 juniors is 8!, which is 40,320.
The number of ways to arrange the 20 seniors is 20!, which is 2,432,902,008,176,640,000.
Multiplying these three numbers together, we get a total of 1.1657416 × 10^30 ways to arrange all the students in a row for a picture.
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Express the sum of the polynomial 5x^2+6x−17 and the square of the binomial (x+6) as a polynomial in standard form.
Answer:
6x^2 + 18x + 19.
Step-by-step explanation:
5x^2 + 6x - 17 + (x + 6)^2
= 5x^2 + 6x - 17 + x^2 + 12x + 36
= 6x^2 + 18x + 19.
Answer:
6x^2 + 18x + 19
Step-by-step explanation:
The standard form of a polynomial depends on the degree of the polynomial. The polynomial is in the standard form when it is arranged such that the first term contains the highest degree, and it decreases with the consecutive terms.
The square of the binomial is (x+6)^2
(x+6)^2 = (x+6)(x+6) = x^2 + 6x + 6x + 36 = x^2 + 12x + 36
The sum of the two polynomials will be 5x^2+6x−17 + x^2 + 12x + 36
Collecting like terms,
5x^2 + x^2 + 6x + 12x + 36 -17
= 6x^2 + 18x + 19
which graph represents this function?
f(x) = 1/2 x – 5
O
Answer: Slope 1/2
y-int -5
graph (0,-5),(1,-9/2)
Step-by-step explanation:
A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with sequals17.6 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.
Answer: 95% confidence interval would be (175.04,194.96).
Step-by-step explanation:
Since we have given that
Mean = 185 mg
Standard deviation = 17.6 mg
At 95% confidence level, z = 1.96
So, Interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=185\pm 1.96\times \dfrac{17.6}{\sqrt{12}}\\\\=185\pm 9.958\\\\=(185-9.958,185+9.958)\\\\=(175.042,194.958)\\\\=(175.04,194.96)[/tex]
Hence, 95% confidence interval would be (175.04,194.96).
Using the provided values and steps for constructing a 95% confidence interval, we estimate, with 95% confidence, that the true mean cholesterol content of all such eggs lies between 146.5 and 223.5 milligrams.
Explanation:To construct a 95% confidence interval for the mean cholesterol content of all such eggs, we will use the provided sample mean (185 milligrams), standard error (17.6 milligrams), and the fact that the sample size (12 eggs) is relatively small, so we use a t-distribution.
Firstly, we need to find the t-score for a 95% confidence level. Since degrees of freedom (df) is n-1 -> 12-1=11, and at a 95% confidence level, the t-score (from t-distribution table) is approximately 2.201 for a two-tailed test.
Next, we use the formula for confidence interval:
Lower bound = Sample mean - (t-score * standard error) Upper bound = Sample mean + (t-score * standard error)
Calculating these gives:
Lower bound = 185 - (2.201 * 17.6) ≈ 146.5 milligrams Upper bound = 185 + (2.201 * 17.6) ≈ 223.5 milligrams
So, we estimate with 95 percent confidence that the true mean cholesterol content of all such eggs is between 146.5 and 223.5 milligrams.
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Black perch are ocean fish found on reefs near shore. A graduate student at CSULB is interested in testing whether black perch from two different populations behave differently, on average. In her experiment, she measured routine swimming velocity (to the nearest 0.001 cm/s) for 18 fish from one population, and 21 fish from another. a) Which statistical test would you use to analyze these data?
One sample t-test
Step-by-step explanation:
In this statistical test, you will be able to test if a sample mean, significantly differs from a hypothesized value.Here you can test if the average swimming velocity differs significantly from an identified value in the hypothesis.Then you can conclude whether the group of 18 fish or that of 21 fish has a significantly higher or lower mean velocity than the one in the hypothesis.
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Keywords: population, velocity, statistical test,average
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Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting out x apartments is given by the following function. P(x) = -11 x^2 + 1804 x - 43,000 To maximize the monthly rental profit, how many units should be rented out?
units What is the maximum monthly profit realizable? $
Answer:
82 apartments should be rented.
Maximum profit realized will be $30964.
Step-by-step explanation:
Monthly profit realized from renting out x apartments is modeled by
P(x) = -11x² + 1804x - 43000
To maximize the profit we will take the derivative of the function P(x) with respect to x and equate it to zero.
P'(x) = [tex]\frac{d}{dx}(-11x^{2}+1804x-43000)[/tex]
= -22x + 1804
For P'(x) = 0,
-22x + 1804 = 0
22x = 1804
x = 82
Now we will take second derivative,
P"(x) = -22
(-) negative value of second derivative confirms that profit will be maximum if 82 apartments are rented.
For maximum profit,
P(82) = -11(82)² + 1804(82) - 43000
= -73964 + 147928 - 43000
= $30964
Therefore, maximum monthly profit will be $30964.
The number of units that should be rented out to maximize the monthly rental profit is 82 units.
The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
To find the number of units that should be rented out to maximize the monthly rental profit and the maximum monthly profit, we need to analyze the given profit function:
[tex]\[ P(x) = -11x^2 + 1804x - 43,000 \][/tex]
This is a quadratic function of the form [tex]\( P(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -11 \), \( b = 1804 \)[/tex], and c = -43,000 .
For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] with a < 0 (which opens downwards), the maximum value occurs at the vertex. The x-coordinate of the vertex for the function P(x) is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting a = -11 and b = 1804 :
[tex]\[ x = -\frac{1804}{2(-11)} = \frac{1804}{22} = 82 \][/tex]
Therefore, the number of units that should be rented out to maximize the profit is x = 82 .
Next, we calculate the maximum monthly profit by substituting x = 82 back into the profit function:
[tex]\[ P(82) = -11(82)^2 + 1804(82) - 43,000 \][/tex]
Calculating step by step:
[tex]\[ 82^2 = 6724 \]\[ -11(6724) = -73,964 \]\[ 1804(82) = 147,928 \]\[ P(82) = -73,964 + 147,928 - 43,000 \]\[ P(82) = 30,964 \][/tex]
So, the maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
In summary:
- The number of units that should be rented out to maximize the monthly rental profit is 82 units.
- The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
According to a survey, 50% of Americans were in 2005 satisfied with their job.Assume that the result is true for the current proportion of Americans. A. Find the mean and standard deviation of the proportion for a sample of1000.
Answer:
[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]
Step-by-step explanation:
The probability distribution of sampling distribution [tex]\hat{p}[/tex] is known as it sampling distribution.
The mean and standard deviation of the proportion is given by :-
[tex]\mu_{\hat{p}}=p\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]
, where p =population proportion and n= sample size.
Given : According to a survey, 50% of Americans were in 2005 satisfied with their job.
i.e. p = 50%=0.50
Now, for sample size n= 1000 , the mean and standard deviation of the proportion will be :-
[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{0.50(1-0.50)}{1000}}=\sqrt{0.00025}\\\\=0.0158113883008\approx0.0158[/tex]
Hence, the mean and standard deviation of the proportion for a sample of 1000:
[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]
A tire company measures the tread on newly-produced tires and finds that they are normally distributed with a mean depth of 0.98mm and a standard deviation of 0.35mm. Find the probability that a randomly selected tire will have a depth less than 0.70mm. Would this outcome warrant a refund (meaning that it would be unusual)?
Answer:
0.212 is the probability that a randomly selected tire will have a depth less than 0.70 mm.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.98 mm
Standard Deviation, σ = 0.35 mm
We are given that the distribution of tire tread is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(depth less than 0.70 mm)
P(x < 0.70)
[tex]P( x < 0.70) = P( z < \displaystyle\frac{0.70 - 0.98}{0.35}) = P(z < -0.8)[/tex]
Calculating from normal z table, we have:
[tex]P(z<-0.8) = 0.212[/tex]
[tex]P(x < 0.70) = 0.212 = 21.2\%[/tex]
Thus, this event is not unusual and will not warrant a refund.
Verify The Sum & Difference Identity:
cos(x + y) / sin(x - y) = 1 - cotxcoty / cotx - coty
I’ve struggled with this problem over the last couple of days. Any help is appreciated!
The sum and difference identity cos(x + y) / sin(x - y) = 1 - cotxcoty / cotx - coty is verified
Solution:
Given expression is:
[tex]\frac{\cos (x+y)}{\sin (x-y)}=\frac{1-\cot x \cot y}{\cot x-\cot y}[/tex]
Let us first solve L.H.S
[tex]\frac{\cos (x+y)}{\sin (x-y)}[/tex] ------ EQN 1
We have to use the sum and difference formulas
cos(A + B) = cosAcosB – sinAsinB
sin(A - B) = sinAcosB – cosAsinB
Applying this in eqn 1 we get,
[tex]=\frac{\cos x \cos y-\sin x \sin y}{\sin x \cos y-\sin y \cos x}[/tex]
[tex]\text { Taking sinx } \times \text { siny as common }[/tex]
[tex]=\frac{\sin x \sin y\left(\frac{\cos x \cos y}{\sin x \sin y}-1\right)}{\sin x \sin y\left(\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}\right)}[/tex]
[tex]\begin{array}{l}{=\frac{\frac{\cos x}{\sin x} \times \frac{\cos y}{\sin y}-1}{\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}}} \\\\ {=\frac{\cot x \times \cot y-1}{\cot y-\cot x}} \\\\ {=\frac{\cot x \cot y-1}{\cot y-\cot x}}\end{array}[/tex]
Taking -1 as common from numerator and denominator we get,
[tex]\begin{array}{l}{=\frac{-(1-\cot x \cot y)}{-(\cot x-\cot y)}} \\\\ {=\frac{(1-\cot x \cot y)}{(\cot x-\cot y)}}\end{array}[/tex]
= R.H.S
Thus L.H.S = R.H.S
Thus the given expression has been verified using sum and difference identity
At 6:00 PM, a flagpole that is 35 feet tall casts a shadow that is 50 feet long. At the same time, how long will a person's shadow be if they are 4 feet tall?
Please help ASAP!!! :(
Answer:
The length of the person’s shadow is 5.7ft
Explanation:
Length of the flagpole =a= 35ft
Length of the shadow of the flagpole= b=50ft
Length of the person=c= 4ft
Suppose the length of the person’s shadow is=d
According to the rules of trigonometry
[tex]\frac{\text { Length of the flagpole }}{\text { Length of the shadow of the flagpole }}=\frac{\text { Length of the person }}{\text { Length of the person's shadow }}[/tex]
[tex]\frac{a}{b}=\frac{c}{d}[/tex]
[tex]\frac{35}{50}=\frac{4}{d}[/tex]
35d=200
d=[tex]\frac{200}{35}[/tex]
d=5.7ft
Hence, The length of the person’s shadow is 5.7ft.
A survey of all medium and large-sized corporations showed that 64% of them offer retirement plans to their employees. Let P be the proportion in a random sample of 50 such corporations that offer retirement plans to their employees. Find the probabilty that the value of P will be: a. less than 0.57 b. greater than 0.71
Answer:
a) 0.9292
b)0.1515
Step-by-step explanation:
Explained in attachment.
Answer:
mwah, thank you so much for posting this question! you just saved my life!! :,)
Step-by-step explanation:
thank you thank you.
Mwah!!!!
Two particles move in the xy-plane. At time t, the position of particle A is given by x(t)=5t−5 and y(t)=2t−k, and the position of particle B is given by x(t)=4t and y(t)=t2−2t−1.
(a) If k=−6, do the particles ever collide?
(b) Find k so that the two particles are certain to collide.
k=
(c) At the time the particle collide in (b), which is moving faster?
A. particle A
B. particle B
C. neither particle (they are moving at the same speed)
Answer:
Part A) Not collide
Part B) k = 4
Part C) Particle B is moving fast.
Step-by-step explanation:
Two particles move in the xy-plane. At time, t
Position of particle A:-
[tex]x(t)=5t-5[/tex]
[tex]y(t)=2t-k[/tex]
Position of particles B:-
[tex]x(t)=4t[/tex]
[tex]y(t)=t^2-2t+1[/tex]
Part A) For k = -6
Position particle A, (5t-5,2t+6)
Position of particle B, [tex](4t,t^2-2t-1)[/tex]
If both collides then x and y coordinate must be same
Therefore,
For x-coordinate:5t - 5 = 4t
t = 5
For y-coordinate:[tex]2t+6=t^2-2t-1[/tex]
[tex]t^2-4t-7=0[/tex]
[tex]t=-1.3,5.3[/tex]
The value of t is not same. So, k = -6 A and B will not collide.
Part B) If both collides then x and y coordinate must be same
For x-coordinate:5t - 5 = 4t
t = 5
For y-coordinate:[tex]2t-k=t^2-2t-1[/tex]
Put t = 5
[tex]10-k=25-10-1[/tex]
[tex]k=4[/tex]
Hence, if k = 4 then A and B collide.
Part C)
Speed of particle A, [tex]\dfrac{dA}{dt}[/tex]
[tex]\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]
[tex]\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4[/tex]
Speed of particle B, [tex]\dfrac{dB}{dt}[/tex]
[tex]\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]
[tex]\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}[/tex]
At t = 5
[tex]\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2[/tex]
Hence, Particle B moves faster than particle A
To determine if the particles collide, we set up equations with their x-coordinates and y-coordinates. Part (a) asks if they collide when k = -6. Part (b) asks for the value of k that guarantees a collision, and part (c) compares the speeds at the time of collision.
Explanation:To determine if the particles collide, we need to find out if their x-coordinates and y-coordinates are equal at any given time. Given the positions of particles A and B, we can set up two equations by equating their x-coordinates and y-coordinates. For part (a), when k = -6 we can solve the equations to find if they intersect. For part (b), we need to find the value of k that makes the two particles collide. Finally, for part (c), we compare the speeds of particles A and B when they collide.
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