Answer:
a) [tex]0.81-2.62\frac{0.34}{\sqrt{110}}=0.725[/tex]
[tex]0.81+2.62\frac{0.34}{\sqrt{110}}=0.895[/tex]
So on this case the 99% confidence interval would be given by (0.725;0.895)
b) [tex]n=(\frac{2.58(0.34)}{0.1})^2 =76.95 \approx 77[/tex]
So the answer for this case would be n=77 rounded up to the nearest integer
Step-by-step explanation:
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.
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=0.81[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=0.34 represent the sample standard deviation
n=110 represent the sample size
Part a
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)
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=110-1=109[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,119)".And we see that [tex]t_{\alpha/2}=2.62[/tex]
Now we have everything in order to replace into formula (1):
[tex]0.81-2.62\frac{0.34}{\sqrt{110}}=0.725[/tex]
[tex]0.81+2.62\frac{0.34}{\sqrt{110}}=0.895[/tex]
So on this case the 99% confidence interval would be given by (0.725;0.895)
Part b
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigmas}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =0.1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
We can assume the following estimator for the population deviation [tex]\hat \sigma =s =0.34[/tex]
The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.58/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.58(0.34)}{0.1})^2 =76.95 \approx 77[/tex]
So the answer for this case would be n=77 rounded up to the nearest integer
Complete the recursive formula of the geometric sequence 300, 60, 12, 2.4
D(1)=
D(n)= d(n-1)•
Answer:
D(1)=300
D(n)=d(n-1)× 1/5
Step-by-step explanation:
The 1st term is 300 and the common ratio is 1/5. You have to multiply times 1/5 to get to the next number.
Answer:
300
1/5
Step-by-step explanation:
2e=2 solve the question
E=
Answer: E=1
Step-by-step explanation:
because 2 times 1 is 2 or any number multiplied by 1 is itself
Answer:
e=1, e= -1
Step-by-step explanation:
Case 1:
2e=2
e=2/2
e=1
Case 2:
2e= -2
e= -2/2
e= -1
Hope this helped!!!
The titanium content in an aircraft-grade alloy is an important determinant of strength. A sample of 20 test coupons reveals the following titanium content (in percent). 8.32 8.05 8.93 8.65 8.25 8.46 8.52 8.35 8.36 8.41 8.42 8.30 8.71 8.75 8.60 8.83 8.50 8.38 8.29 8.46 Use the normal approximation for the sign test to test the claim that the median titanium content is equal to 8.5%. Use α=0.05. Calculate the observed number of plus differences. r+= Enter your answer; r+ Calculate to 2 decimal places the test statistic. z0= Enter your answer; z0 Calculate to 3 decimal places the P-value. P-value = Enter your answer; P-value H0 : μ˜=8.5 versus H1: μ˜≠8.5. Conclusion: Choose your answer; Conclusion
Answer:
attached
Step-by-step explanation:
attached
To test the claim that the median titanium content is 8.5%, we use the normal approximation for the sign test. We calculate the observed number of plus differences, the test statistic, and the P-value. We fail to reject the null hypothesis because the P-value is larger than the significance level.
Explanation:To test the claim that the median titanium content is equal to 8.5%, we can use the normal approximation for the sign test. The sign test is used when the data is not normally distributed. In this case, we have 20 test coupons with titanium content in percent. To calculate the observed number of plus differences (r+), we count the number of values greater than 8.5% and the number of values less than 8.5% in the sample. In this case, there are 7 values greater than 8.5% and 13 values less than 8.5%. Therefore, r+ = 7.
To calculate the test statistic (z0), we use the formula z0 = (r+ - n/2) / sqrt(n/4), where n is the sample size. In this case, n = 20. Plugging in the values, we get z0 = (7 - 20/2) / sqrt(20/4) = -1.5.
To calculate the P-value, we need to find the probability of observing a test statistic as extreme as -1.5 or more extreme in a standard normal distribution. We can use a statistical table or software to find the P-value. In this case, the P-value is 0.133, calculated using a standard normal distribution table.
Since the P-value (0.133) is larger than the significance level α (0.05), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the median titanium content is different from 8.5%.
4 divided by 1/2 as a fraction
4 divided by 1/2 as a fraction is equal to 8/1 or simply 8.
How to simplify the fractionTo divide a number by a fraction, you can multiply the number by the reciprocal of the fraction.
In this case, we have 4 divided by 1/2. The reciprocal of 1/2 is 2/1 (or simply 2).
So, we can rewrite the expression as:
4 * (2/1)
Multiplying these numbers, we get:
8/1
Therefore, 4 divided by 1/2 as a fraction is equal to 8/1 or simply 8.
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Statistics students at a community college wonder whether the cars belonging to students are, on average, older than the cars belonging to faculty. They select a random sample of 58 cars in the student parking lot and find the average age to be 8.5 years with a standard deviation of 6.2 years. A random sample of 41 cars in the faculty parking lot have an average age of 5.1 years with a standard deviation of 3.5 years. Note: The degrees of freedom for this problem is df = 93.033791. Round all results to 4 decimal places
Answer:
There is evidence to support the claim that the cars belonging to students are, on average, older than the cars belonging to faculty.
Step-by-step explanation:
We have to perform a hypothesis test on the difference between means.
The claim is that the cars belonging to students are, on average, older than the cars belonging to faculty.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]
being μ1: population mean age of students cars, μ2: population mean age of faculty cars.
The significance level is assumed to be 0.05.
The information about the students cars sample is:
Mean M1: 8.5 years.
Standard deviation s1: 6.2 years.
Sample size n1: 58 cars.
The information about the faculty cars sample is:
Mean M2: 5.1 years.
Standard deviation s2: 3.5 years.
Sample size n2: 41 cars.
The difference between means is:
[tex]M_d=M_1-M_2=8.5-5.1=3.4[/tex]
The standard error of the difference between means is:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}} =\sqrt{\dfrac{6.2^2}{58}+\dfrac{3.5^2}{58}}=\sqrt{ 0.6628 +0.2988 }=\sqrt{0.9615}\\\\\\s_{M_d}=0.9806[/tex]
Now we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{3.4-0}{0.9806}=3.4673[/tex]
The degrees of freedom are 93.033791, so the P-value for this right tail test is:
[tex]P-value=P(t>3.4673)=0.0004[/tex]
As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected.
There is evidence to support the claim that the cars belonging to students are, on average, older than the cars belonging to faculty.
Peter is a revenue manager of a 300-room hotel. Over this past weekend (including both Saturday and Friday evenings) he sold 450 of his guestrooms with an average daily rate (ADR) of $300.00. What as his hotel's RevPAR for thepast weekend?
Answer:
$225
Step-by-step explanation:
-RevPar is defined as the room revenue divided by the number of rooms available:
[tex]RevPar=\frac{Room \ Revenue}{Rooms \ Available}[/tex]
-The average number of rooms sold per day in the two days is:
[tex]mean =\frac{450}{2}\\\\=225\ rooms[/tex]
The PevPar can then be calculated as:
[tex]RevPar=\frac{Room \ Revenue}{Rooms \ Available}\\\\=\frac{ADR\times mean occupancy}{Rooms \ Available}\\\\=\frac{300\times 225}{300}\\\\=225[/tex]
Hence, the PevPar is $225
Twenty eight is 40% of
Answer:
11.20 is maybe the answer
Answer:
70
Step-by-step explanation:
Assume the unknown value is 'Y'
28 = 40% x Y
28 = 40/100 x Y
Multiplying both sides by 100 and dividing both sides of the equation by 40 we will arrive at:
Y = 3 x 100/40
Y = 70%
You have prepared 10 types of treats for your 5 cats. You don’t know which treat each of your cats will go for, so you have bought for each type enough treats for all your cats. Assume that each cat is equally likely to choose any type of treats, and let X be the number of pairs of cats that will choose the same type of treats. Compute E(X) and Var(X).
Answer:
E(X) = 1.5
Var(X) = 2.325
Step-by-step explanation:
X - the number of pairs of cats (out of the 5 cats) to choose the same type of treat, can take values of:
X = (0, 1, 2, 3, 4, 5)
Also probability of choosing a treat, since they are all equally likely is: f(x) = 1/10
E(X) - expectation of x, is given by:
E(X) = Summation [X*f(x)]
E(X) = 0x(1/10) + 1x(1/10) + 2x(1/10) + 3x(1/10) + 4x(1/10) + 5x(1/10)
= (1/10) x (1+2+3+4+5)
E(X) = 3/2 = 1.5
Also, variance is:
Var(X) = Summation f(x)*[X - E(X)]^2
= (1/10)x(0-1.5)^2 + (1/10)x(1-1.5)^2 +(1/10)x(2-1.5)^2 +(1/10)x(3-1.5)^2 +(1/10)x(4-1.5)^2 +(1/10)x(5-1.5)^2
= (1/10)x[2.25 + 0.25 + 2.25 + 6.25 + 12.25]
Var(X) = 2.325
Moira said, "If a right triangle has one side with a length of 3 units and one side with a length
of 4 units, then the third side has a length of 5 units."
Moira's claim is not true. Give an example of a right triangle with sides of length 3, 4, and a
number other than 5. Explain your reasoning.
If the two sides with length 3 and 4 are the two legs, then the missing side, i.e. the hypotenuse, is indeed 5.
But it could also be the case that 4 is the hypotenuse and 3 is one of the legs. In this case, the missing side is the other leg, so we calculate it using
[tex]\sqrt{4^2-3^2}=\sqrt{16-9}=\sqrt{7}[/tex]
So, a right triangle with legs [tex]\sqrt{7}[/tex] and 3 has an hypotenuse of 4.
Moira's statement correctly applies the Pythagorean theorem, implying it is impossible to have a right triangle with sides of 3, 4, and a value other than 5 and still conform to the requirements of a right-angled triangle.
The Pythagorean theorem, where a2 + b2 = c2 applies, and c represents the length of the hypotenuse, while a and b represent the lengths of the other two sides. However, the question asked to provide an example that opposes Moira's claim by asking for a right triangle with sides of length 3, 4, but with a hypotenuse different from 5. This is a trick question because, according to the Pythagorean theorem, if a triangle has sides of 3 and 4 units, the hypotenuse must indeed be 5 units to satisfy the equation 32 + 42 = 52 (9 + 16 = 25). Therefore, there cannot be a right-angle triangle with the sides 3, 4, and a number other than 5, as it would violate the Pythagorean theorem.
A Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. A salesman suggests that he try a statistics software package that gets students more involved with computers, predicting that it will increase students' scores. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly. The professor will have to pay for the software only if he chooses to continue using it. In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.a. What is the test statistic?b. What is the P-value?c. What is the appropriate conclusion?i. Fail to reject H_0. The change is not statistically significant. The software does not appear to improve exam scores.ii. Fail to reject H_0. The change is statistically significant. The software does appear to improve exam scores.iii. Reject H_0. The change is not statistically significant. The software does not appear to improve exam scores.iv. Reject H_0. The change is statistically significant. The software does appear to improve exam scores.
Answer:
Reject [tex]H_0[/tex] . The change is statistically significant. The software does appear to improve exam scores.
Step-by-step explanation:
We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.
In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.
Let [tex]\mu[/tex] = mean scores on the final exam.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 114 points {means that the mean scores on the final exam does not increases after using software}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 114 points {means that the mean scores on the final exam increase significantly after using software}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average points = 117 points
s = sample standard deviation = 8.7 points
n = sample of students = 218
So, test statistics = [tex]\frac{117-114}{\frac{8.7}{\sqrt{218} } }[/tex] ~ [tex]t_2_1_7[/tex]
= 5.091
Now, P-value of the test statistics is given by the following formula;
P-value = P( [tex]t_2_1_7[/tex] > 5.091) = Less than 0.05%
Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, 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 the change is statistically significant. The software does appear to improve exam scores.
The test statistic is 3.448 and the P-value is 0.0014. The appropriate conclusion is that the software does appear to improve exam scores.
Explanation:The test statistic can be calculated using the formula:
test_statistic = (sample_mean - population_mean) / (sample_standard_deviation / sqrt(sample_size))
Substituting in the given values:
test_statistic = (117 - 114) / (8.7 / sqrt(218)) = 3.448
The P-value can be calculated using a statistical software or calculator. For this question, the P-value is 0.0014
Since the P-value is less than 0.05 (significance level), we can reject the null hypothesis. The appropriate conclusion is: iv. Reject H_0. The change is statistically significant. The software does appear to improve exam scores.
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Find the number that makes the ratio equivalent to 1:9. 3:
The ratio 3:x equivalent to 1:9 results in x being equal to 27.
To find the number that makes the given ratio equivalent to 1:9 we can set up a proportion using the information given which is a ratio of 3 to an unknown number (let's call it x).
First we write down the two ratios as fractions and set them equal to each other to find x:
1/9 = 3/x
We then cross-multiply to solve for x:
1 × x = 9 × 3
This gives us:
x = 27
So, the number that makes the ratio equivalent to 1:9 when compared to 3 is 27.
1. A sample of n = 9 scores has ex = 108. What is the sample mean?
The value of the sample mean is 12.
Given
A sample with a mean of n = 9 has [tex]\rm \sum x=108[/tex].
What is formula is used to calculate the sample mean?The formula is used to calculate sample mean is;
[tex]\rm Sample \ mean = \dfrac{\sum x}{n}\\\\[/tex]
Substitute all the values in the formula
Then,
The sample mean is;
[tex]\rm Sample \ mean = \dfrac{\sum x}{n}\\\\\rm Sample \ mean = \dfrac{108}{9}\\\\ \rm Sample \ mean = 12[/tex]
Hence, the value of the sample mean is 12.
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A pine tree is 64 feet 3 inches tall. How many inches tall is the pine tree?
Answer:
3 inches you are right
Step-by-step explanation:
An object is taken from a freezer at negative 8 degrees Upper C. Let t be the time in hours after the object was taken from the freezer. At time t the average temperature of the object is increasing at the rate of Upper T prime (t )equals 20 e Superscript negative 0.4 t degrees Celsius per hour. Find the temperature of the object at time t.
Answer:
[tex]T(t)=-50e^{-0.4t}+42[/tex]
Step-by-step explanation:
We are given that
[tex]T(0)=-8^{\circ} C[/tex]
[tex]T'(t)=20e^{-0.4t}[/tex]
Integrating on both sides
[tex]T(t)=20\int e^{-0.4 t}[/tex]
[tex]T(t)=\frac{20}{-0.4}e^{-0.4t}+C[/tex]
Using the formula
[tex]\int e^{ax}dx=\frac{e^{ax}}{a}+C[/tex]
Substitute t=0 and T(0)=-8
[tex]-8=-50+C[/tex]
[tex]C=-8+50=42[/tex]
Substitute the value of C
[tex]T(t)=-50e^{-0.4t}+42[/tex]
An incoming college student took her college’s placement exams in French and mathematics. In French, she scored 85, and in math 80. The overall results for both exams are approximately normal. The mean French test score was 72 with a SD of 12, while the mean math test score was 68 with a SD of 8. On which exam did she do better as compared with the other incoming college students? Compute the z-scores and the percentiles for each exam to support your answer
Answer:
In Math, she scored in the 93rd percentile, which is higher than the French percentile. So she did better on the Math exam as compared with the other incoming college students
Step-by-step explanation:
Z-score:
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.
On which exam did she do better as compared with the other incoming college students?
On the exam for which she had the higher z-score.
Franch:
Scored 85.
Mean 72, SD = 12. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{85 - 72}{12}[/tex]
[tex]Z = 1.08[/tex]
[tex]Z = 1.08[/tex] has a pvalue of 0.8810, so her French score is in the 88th percentile.
Math:
Scored 80
Mean 68, SD = 8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 68}{8}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332.
In Math, she scored in the 93rd percentile, which is higher than the French percentile. So she did better on the Math exam as compared with the other incoming college students
Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.
Let W1 be the set of all polynomials of the form p(t)=at2, where a is in ℝ.
Let W2 be the set of all polynomials of the form p(t)=t2+a, where a is in ℝ.
Let W3 be the set of all polynomials of the form p(t)=at2+at, where a is in ℝ.
Answer:
1. Yes.
2. No.
3. Yes.
Step-by-step explanation:
Consider the following subsets of Pn given by
1.Let W1 be the set of all polynomials of the form [tex]p(t)=at^2[/tex], where a is in ℝ.
2.Let W2 be the set of all polynomials of the form [tex]p(t)=t^2+a[/tex], where a is in ℝ.
3. Let W3 be the set of all polynomials of the form [tex]p(t)=at^2+at[/tex], where a is in ℝ.
Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:
- The 0 vector of V is in W.
- Given v,w in W then v+w is in W.
- Given v in W and a a real number, then av is in W.
So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.
- First property:
Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.
For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.
- Second property:
W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials
[tex]p_1 (t) = at^2, p_2(t)=bt^2[/tex].
We must check that p_1+p_2(t) is in W1.
Note that
[tex]p_1(t)+p_2(t) = at^2+bt^2 = (a+b)t^2[/tex]
Since a+b is another real number, we have that p1(t)+p2(t) is in W1.
W3. Consider two elements in W3. Say [tex]p_1(t) = a(t^2+t), p_2(t)= b(t^2+t)[/tex]. Then
[tex]p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)[/tex]
So, again, p1(t)+p2(t) is in W3.
- Third property.
W1. Consider an element in W1 [tex] p(t) = at^2[/tex]and a real scalar b. Then
[tex] bp(t) = b(at^2) = (ba)t^2)[/tex].
Since (ba) is another real scalar, we have that bp(t) is in W1.
W3. Consider an element in W3 [tex] p(t) = a(t^2+t)[/tex]and a real scalar b. Then
[tex] bp(t) = b(a(t^2+t)) = (ba)(t^2+t)[/tex].
Since (ba) is another real scalar, we have that bp(t) is in W3.
After all,
W1 and W3 are subspaces of Pn for n= 2
and W2 is not a subspace of Pn.
W1 is a subspace of ℙn, W2 is not a subspace of ℙn, and W3 is a subspace of ℙn.
Explanation:For a set to be a subspace of ℙn, it must satisfy three conditions: the zero vector must be in the set, the set must be closed under addition, and the set must be closed under scalar multiplication.
W1:To determine if W1 is a subspace of ℙn, we check if it satisfies the three conditions:
The zero vector is a polynomial of the form p(t) = 0t^2, which is in W1.If p(t) and q(t) are polynomials of the form p(t) = at^2 and q(t) = bt^2, then their sum is (a+b)t^2, which is also in W1.If p(t) is a polynomial of the form p(t) = at^2 and c is a scalar, then c * p(t) = (ca)t^2, which is in W1.Therefore, W1 is a subspace of ℙn.
W2:The zero vector is a polynomial of the form p(t) = t^2 + 0, which is in W2.If p(t) and q(t) are polynomials of the form p(t) = t^2 + a and q(t) = t^2 + b, then their sum is (t^2 + a) + (t^2 + b) = 2t^2 + (a + b), which is not in the form required by W2. Therefore, W2 is not closed under addition and is not a subspace of ℙn.W3:The zero vector is a polynomial of the form p(t) = 0t^2 + 0t, which is in W3.If p(t) and q(t) are polynomials of the form p(t) = at^2 + at and q(t) = bt^2 + bt, then their sum is (a + b)t^2 + (a + b)t, which is also in W3.If p(t) is a polynomial of the form p(t) = at^2 + at and c is a scalar, then c * p(t) = (ca)t^2 + (ca)t, which is in W3.Therefore, W3 is a subspace of ℙn.
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41:02
Machine 1 can complete a task in x hours while an upgraded machine (machine 2) needs 9 fewer hours. The plant manager
knows the two machines will take at least 6 hours, as represented by the inequality:
Which answer choice includes all solutions to the inequality and identifies which interval(s) contain viable completion times of
machine 1?
(0.3] U (9,18]; both contain viable times for machine 1
0 (0.3] U (9,18); only (9,18] contains viable times for machine 1
0 (-0,0) U (3,9) U (18,co); no interval contains viable times for machine 1
0 (-0,0) U[3,9U (18,co); but only (18,co) contains viable times for machine 1
6:36 PM
041 122/2020
O
Type here to search
Answer:the answer is B
Step-by-step explanation:
Answer:
B 0 (0.3] U (9,18); only (9,18] contains viable times for machine 1
Step-by-step explanation:
The graph is 0<x<3 or 9<x<18
What is the solution to the equation x^3 = 25
Final answer:
The solution to the equation x^3 = 25 involves taking the cube root of both sides, yielding an approximate solution of x ≈ 2.924, rounded to three decimal places.
Explanation:
The question asks to find the solution to the equation x^3 = 25. This equation requires us to find a value of x such that, when cubed, equals 25. To solve this, we take the cube root of both sides of the equation, which gives us x = √[3]{25}. This cube root can be simplified using a calculator or estimated for a rough solution.
Thus, the solution to the equation is x ≈ 2.924 (rounded to three decimal places). It's important to note that in this case, we are only looking for real numbers as solutions. Complex solutions aren't considered for this specific problem.
The solution to the equation x^3 = 25 is found by taking the cube root of both sides. The approximate numerical solution is x ≈ 2.924.
The solution to the equation x^3 = 25 involves finding a number x that, when cubed, equals 25. We can solve this by taking the cube root of both sides of the equation. This operation will give us the principal root of x. The cube root of 25 is not a whole number, so we will end up with a decimal answer. To solve for x, we proceed as follows:
Write the equation: x^3 = 25.Take the cube root of both sides: x = √325.Calculate the cube root: x ≈ 2.924... (this is an approximation).The actual value of x is a decimal number that, because of the nature of cube roots, cannot be expressed as an exact fraction or simple square root. The approximate numerical solution to the given cubic equation is x ≈ 2.92401773821287.
To help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom 15% of the students in the district, based on the scores on a reading achievement exam given in the child's dominant language. The reading-score for the pooled students in the district is approximately normally distributed with a mean of 122 points, and standard deviation of 18 points. a. Find the probability of students score above 140 points? b. What is the 15th percentile of the students eligible for the program?
Answer:
a) [tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel and we got:
[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]
b) [tex]z=-1.036<\frac{a-122}{18}[/tex]
And if we solve for a we got
[tex]a=122 -1.036*18=103.35[/tex]
So the value of height that separates the bottom 15% of data from the top 85% is 103.35.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(122,18)[/tex]
Where [tex]\mu=122[/tex] and [tex]\sigma=18[/tex]
We are interested on this probability
[tex]P(X>140)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel and we got:
[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]
Part b
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X<a)=0.15[/tex] (a)
[tex]P(X>a)=0.85[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.15 of the area on the left and 0.85 of the area on the right it's z=-1.036. On this case P(Z<-1.036)=0.15 and P(z>-1.036)=0.85
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.15[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.15[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.036<\frac{a-122}{18}[/tex]
And if we solve for a we got
[tex]a=122 -1.036*18=103.35[/tex]
So the value of height that separates the bottom 15% of data from the top 85% is 103.35.
What is the circumference of a circle with a radius of 80 inches?(Use 3.14 for pie)
Final answer:
The circumference of a circle with an 80-inch radius is 502.4 inches, calculated using the formula C = 2πr with π approximated as 3.14.
Explanation:
The circumference of a circle with a radius of 80 inches can be calculated using the formula C = 2πr, where π (Pi) is approximately 3.14 and r is the radius. For given radius of 80 inches, the calculation would be:
C = 2 × 3.14 × 80 inches
C = 502.4 inches.
Therefore, the circumference of the circle is 502.4 inches.
30 cm
30 cm
26 cm
find the area
Answer:
23400 cm
Step-by-step explanation:
30 cm x 30 cm x 26 cm = 23,400 cm
In a certain Algebra 2 class of 22 students, 5 of them play basketball and 11 of them play baseball. There are 3 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball
Final answer:
The probability that a randomly chosen student from the Algebra 2 class plays basketball or baseball is 13/22, which is approximately 0.59 or 59% when calculated using the principle of inclusion and exclusion in probability.
Explanation:
To calculate the probability that a student chosen randomly from the Algebra 2 class plays basketball or baseball, you need to use the principle of inclusion and exclusion in probability theory. The total number of students in the class is 22. There are 5 students who play basketball and 11 who play baseball, but since 3 students play both sports, these students have been counted twice in the total of 5+11=16. Therefore, the correct number of students who play at least one of the sports is 5+11-3=13.
The probability that a randomly chosen student plays either basketball or baseball is hence the number of students who play either one or both sports divided by the total number of students in the class. This gives us 13/22. To convert this fraction to a decimal, we can divide 13 by 22, which gives us approximately 0.59. Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is approximately 0.59, or 59%.
what is the expression 5 square root of -4 + 4 square root of -25 in standard form
Answer:
30i
Step-by-step explanation:
[tex]5 \sqrt{ - 4} + 4 \sqrt{ - 25} \\ = 5 \sqrt{4 \times - 1} + 4 \sqrt{25 \times - 1} \\ = 5 \sqrt{4} \times \sqrt{ - 1} + 4 \sqrt{25} \times \sqrt{ - 1} \\ = 5 \times 2 \times i + 4 \times 5 \times i...( \because \sqrt{ - 1} = i) \\ = 10i + 20i \\ = 30i[/tex]
A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by
cin()-2+sin(t/4) lb/gal. 40-18000, sin(t)- cos(t)-46324e( )
The complete question is;
A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine the amount of salt in the tank at any time
t, if the concentration in the inflow is variable and given by
c(t) = 2 + sin(t/4) lb/gal.
Answer:
dA/dt = (8 + 4sin(t/4)) - (A_t/100)
Step-by-step explanation:
Rate is given as;
dA/dt = R_in - R_out
R_in = (concentration of salt inflow) x (input rate of brine)
So, R_in = (2 + sin(t/4)) x 4 = (8 + 4sin(t/4))
The solution is being pumped out at the same rate, thus it is accumulating at the same rate.
After t minutes, there will be 400 + (0 x t) gallons left = 400 gallons left
Thus,
R_out =(concentration of salt outflow) x (output rate of brine)
R_out = (A_t/400) x 4 = A_t/100
Thus,
A_t = (100/t)(8t - 16cos(t/4) - 60)
Since we want to find the amount of salt, A(t), let's integrate;
Thus, A = 8t - 16cos(t/4) - (A_t/100)t
A = 60 from the question. Thus,
60 = 8t - 16cos(t/4) - (A_t/100)t
Let's try to make A_t the subject;
(A_t/100)t = 8t - 16cos(t/4) - 60
A_t = (100/t)(8t - 16cos(t/4) - 60)
For 100 births, P(exactly 5555 girls)=0.0485 and P(5555 or more girls)=0.184. Is 5555 girls in 100 births a significantly high number of girls? Which probability is relevant to answering that question? Consider a number of girls to be significantly high if the appropriate probability is 0.05 or less.
Answer:
[tex]P(X \geq 55) = 0.184 \geq 0.05[/tex], so 55 girls in 100 births is not a significantly high number of girls
The appropriate probability to determine if a value x is significantly high is the probability of being equal or higher than x. It this probability is 0.05 or less, x is significantly high.
Step-by-step explanation:
A number x is said to be significantly high if:
[tex]P(X \geq x) \leq 0.05[/tex]
The appropriate probability to determine if a value x is significantly high is the probability of being equal or higher than x. It this probability is 0.05 or less, x is significantly high.
Is 55 girls in 100 births a significantly high number of girls?
We have that:
[tex]P(X \geq 55) = 0.184 \geq 0.05[/tex], so 55 girls in 100 births is not a significantly high number of girls
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is believed that the machine is underfilling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289. Assume the population is normally distributed. A level of significance of 0.01 will be used. State the null and alternative hypotheses.
Answer:
Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram {means that the machine is not under filling the bags}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram {means that the machine is under filling the bags}
Step-by-step explanation:
We are given that a manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting.
It is believed that the machine is under filling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289.
Let [tex]\mu[/tex] = mean weight bag filling capacity of machine.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram {means that the machine is not under filling the bags}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram {means that the machine is under filling the bags}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean = 429 grams
s = sample standard deviation = [tex]\sqrt{289}[/tex] = 17 grams
n = sample of bags = 21
So, test statistics = [tex]\frac{429-430}{\frac{17}{\sqrt{21} } }[/tex] ~ [tex]t_2_0[/tex]
= -0.269
Now at 0.01 significance level, the t table gives critical value of -2.528 at 20 degree of freedom for left-tailed test. Since our test statistics is higher than the critical value of t as -0.269 > -2.528, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that the machine is not under filling the bags.
Final answer:
The student's question involves setting up null and alternative hypotheses for a hypothesis test related to a bag filling machine's weight accuracy. The null hypothesis posits no difference from the 430-gram setting, while the alternative hypothesis suggests underfilling. The test uses a 0.01 significance level to determine whether to reject the null hypothesis.
Explanation:
The student is asking about hypothesis testing in the context of quality control for a bag filling machine. The null hypothesis (H0) assumes that there's no difference between the sample's mean weight and the population mean weight that the machine is set to, which is 430 grams; hence H0:
μ = 430 grams. The alternative hypothesis (Ha) suggests that the actual mean weight is less than the setting, given the concern about underfilling; thus Ha: μ < 430 grams.
For the given scenario, the bag sample's mean is 429 grams with a variance of 289 grams. Since a 0.01 level of significance will be used, if the test statistic calculated from the sample data falls into the critical region determined by this significance level, the null hypothesis would be rejected in favor of the alternative hypothesis.
How to find the perimeter of a quadrilateral
It's the same as any other shape. Find the lengths of the sides and add them up.
Answer:
To find the perimeter of a quadrilateral you would add the lengths of each side together.
Example:
The perimeter is the sum or all the lengths.
If there was a rectangle with a length of 5 cm and a width of 3 cm.
To work this out you would first multiply 5 by 2, which is 10. Then you would multiplying 3 by 2, which is 6. Then you would add 6 and 10, which is 16.
1) Multiply 5 By 2.
[tex]5*2=10[/tex]
2) Multiply 3 by 2.
[tex]3*2=6[/tex]
3) Add 10 and 6.
[tex]10+6=16[/tex]
A builder makes drainpipes that drop 1 cm over a horizontal distance of 30cm to prevent clogs a certain drainpipe needs to cover a horizontal distance of 700cm whats is the length of the drainpipe? round your answer to the nearest tenth of a centimeter
Step-by-step explanation:
Given a builder makes drainpipes that drop 1 cm over a horizontal distance of 30cm to prevent clogs. The ratio of the vertical drop to the horizontal distance covered is [tex]\frac{1}{30}[/tex] cm.
The angle of inclination,
tan(θ) = [tex]\frac{1}{30}[/tex]
θ = 1.9º
By trignometric ratio,
cos(θ) = [tex]\frac{horizontal distance}{length of the drainpipe}[/tex]
length of the drainpipe = [tex]\frac{700}{cos(1.91)}[/tex] = [tex]\frac{700}{0.999}[/tex]
length of the drainpipe = 700.7 cm
Answer:
The correct Answer for Khan Academy is 700.4
Step-by-step explanation:
Which of the following expressions are greater than 1? Choose all that apply.
A.
1675×5
B.
27×4
C.
33100×3
D.
1233×3
E.
1021×2
F.
319×5
Answer:
I think the answers are A, C, D, and E
Step-by-step explanation:
Consider the triangular pyramids shown.
Pyramid 1
Pyramid 2
Pyramid 1 is a triangular pyramid. The base has a base of 12 and height of 2.5. A side has a base of 12 and height of 13.1. 2 sides have a base of 6.5 and height of 14.
[Not drawn to scale]
Pyramid 2 is a triangular pyramid. The base has a base of 12 and height of 4.5. A side has a base of 12 and height of 6.5. 2 sides have a base of 7.5 and height of 8.
[Not drawn to scale]
The base of each pyramid is an isosceles triangle. Which has the greater surface area?
Pyramid 1 is greater by 58.6 square units.
Pyramid 1 is greater by 70.6 square units.
Pyramid 2 is greater by 12 square units.
Pyramid 2 is greater by 24 square units.
9514 1404 393
Answer:
(a) Pyramid 1 by 58.6 square units
Step-by-step explanation:
The area of each triangle is given by the formula ...
A = 1/2bh
The total area is the area of all four triangles that make up the outer surface of the pyramid.
Pyramid 1
A = 1/2(12×2.5 +12×13.1 +2(6.5×14)) = 184.6 . . . square units
Pyramid 2
A = 1/2(12×4.5 +12×6.5 +2(7.5×8)) = 126 . . . square units
Pyramid 1 has the larger area by ...
184.6 -126 = 58.6 . . . square units