Answer:
There is a 62% probability that the student will be awarded at least one of the two scholarships.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that the student gets a scolarship from Agency A.
B is the probability that the student gets a scolarship from Agency B.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that the student will get an scolarship from agency A but not from agency B and [tex]A \cap B[/tex] is the probability that the student will get an scolarship from both agencies.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
What is the probability that the student will be awarded at least one of the two scholarships?
This is
[tex]P = a + b + (A \cap B)[/tex]
We have that:
[tex]A = 0.55, B = 0.40[/tex]
If the student is awarded a scholarship from Agency A, the probability that the student will be awarded a scholarship from Agency B is 0.60.
This means that:
[tex]\frac{A \cap B}{A} = 0.6[/tex]
[tex]A \cap B = 0.6A = 0.6*0.55 = 0.33[/tex]
----------
[tex]A = a + (A \cap B)[/tex]
[tex]0.55 = a + 0.33[/tex]
[tex]a = 0.22[/tex]
--------
[tex]B = b + (A \cap B)[/tex]
[tex]0.40 = b + 0.33[/tex]
[tex]b = 0.07[/tex]
Answer:
[tex]P = a + b + (A \cap B) = 0.22 + 0.07 + 0.33 = 0.62[/tex]
There is a 62% probability that the student will be awarded at least one of the two scholarships.
The probability that the student will be awarded at least one of the two scholarships is 0.73.
To calculate the probability that the student will be awarded at least one of the two scholarships, we can use the following formula:
P(at least one scholarship) = 1 - P(no scholarships)
The probability that the student will not be awarded a scholarship from either agency is the product of the probability that the student will not be awarded a scholarship from Agency A and the probability that the student will not be awarded a scholarship from Agency B.
The probability that the student will not be awarded a scholarship from Agency A is
1 - 0.55 = 0.45.
The probability that the student will not be awarded a scholarship from Agency B is
1 - 0.40 = 0.60.
Therefore, the probability that the student will not be awarded a scholarship from either agency is
0.45 * 0.60 = 0.27.
Therefore, the probability that the student will be awarded at least one of the two scholarships is
1 - 0.27 = 0.73.
Another way to calculate this probability is to use the following formula:
P(at least one scholarship) = P(scholarship from Agency A) + P(scholarship from Agency B) - P(scholarship from both agencies)
We already know the probability that the student will be awarded a scholarship from each agency. The probability that the student will be awarded a scholarship from both agencies is the product of the probability that the student will be awarded a scholarship from Agency A and the probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A.
The probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A is 0.60.
Therefore, the probability that the student will be awarded a scholarship from both agencies is
0.55 * 0.60 = 0.33.
Therefore, the probability that the student will be awarded at least one of the two scholarships is
0.55 + 0.40 - 0.33 = 0.73.
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10-kg batch of a food product has a moisture content of175% dry basis. Calculate the amount of water to beremoved from the product to reduce the moisture content to15% wet basi
Answer:
4.9kg of water
Step-by-step explanation:
Dry basis is the percentage ratio of moisture amount to amount of dry product which is 175%
Ratio of moisture amount to dry product = 175 : 100
Mass of food product = 10kg
Amount of moisture in food product = 175/175+100 × 10kg = 175/275 × 10kg = 6.4kg
Wet basis is the percentage ratio of the moisture amount to the total weight of material which is 15%
Ratio of moisture amount to total weight of material is 15 : 100
Amount of moisture in food product = 15/100 × 10kg = 1.5kg
Amount of water contained in the final product = 1.5kg
Amount of other components in the food product = 10kg - 6.4kg = 3.6kg
Total amount = 1.5kg + 3.6kg = 5.1kg
% of water = 1.5kg/5.1kg × 100 = 29%
A shelf contains n separate compartments. There are r indistinguishable marbles.
In how many ways can the r marbles be placed in the n compartments?
Answer:[tex]\frac{n!}{r!(n-r)!}[/tex]
ways
Step-by-step explanation:
Given that a shelf contains n separate compartments. There are r indistinguishable marbles
The marbles are identical so they can be placed in any order.
Let us consider the places available for placing these r marbles
No of compartments available =n
Marbles to be placed = r
Since marbles are identical and order does not matter
number of ways the r marbles can be placed in the n compartments
= nCr
=[tex]\frac{n!}{r!(n-r)!}[/tex]
the area of the region bounded by the curve y=e^2x the x axis the y axis and the line x=2 is equal toA) e^4/2 -e B) e^4/2 - 1 C) e^4/2 - 1/2 D) 2e^4 -e E) 2e^4 -2
The area of the region is bounded by the curve [tex]\rm y=e^2x[/tex] the x-axis the y axis, and the line x=2 is equal to [tex]\rm \dfrac{e^4}{2}-\dfrac{1}{2}[/tex].
Given that,
The area of the region bounded by the curve [tex]\rm y=e^2x[/tex],
We have to determine,
The x-axis the y axis and the line x=2 is equal to.
According to the question,
The area of the region bounded by the curve
[tex]\rm y=e^2x[/tex]
The area of the region bounded by the curve is determined by integrating the curve at x = 0 to x = 2.
Integrating the curve on both sides,
[tex]\rm Area=\int\limits^2_0 { e^{2x}} \, dx\\\\Area=[ \dfrac{e^{2x}}{2}]^2_0\\\\Area= [ \dfrac{e^{2(2)}}{2}- \dfrac{e^{2(0)}}{2}]\\\\Area = \dfrac{e^4}{2}-\dfrac{e^0}{2}\\\\Area = \dfrac{e^4}{2}-\dfrac{1}{2}[/tex]
Hence, The area of the region is bounded by the curve [tex]\rm y=e^2x[/tex] the x-axis the y axis, and the line x=2 is equal to [tex]\rm \dfrac{e^4}{2}-\dfrac{1}{2}[/tex].
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Let AA and BB be events such that P(A∩B)=1/73P(A∩B)=1/73, P(A~)=68/73P(A~)=68/73, P(B)=21/73P(B)=21/73. What is the probability of A ∪BA ∪B, P(A∪B)P(A∪B)?
Answer:
[tex] P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}[/tex]
Step-by-step explanation:
Let A and B events. We have defined the probabilities for some events:
[tex] P(A') =\frac{68}{73} , P(B) =\frac{21}{73} , P(A \cap B) =\frac{1}{73}[/tex]
Where A' represent the complement for the event A
The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]
So for this case we can solve for P(A) like this:
[tex] P(A) = 1-P(A') = 1-\frac{68}{73}=\frac{5}{73}[/tex]
And now we can find [tex] P(A \cup B)[/tex] using the total probability rul given by:
[tex] P(A \cup B) = P(A)+P(B) -P(A \cap B)[/tex]
And if we replace the values given we got:
[tex] P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}[/tex]
And that would be the final answer.
Multiple-choice questions have a special grading rule determined by your instructor. Assume that your instructor has decided to grade these questions in the following way:
If you submit an incorrect answer to a multiple-choice question with n options, you will lose 1/(n−1) of the credit for that question. Just like the similar multiple-choice penalty on most standardized tests, this rule is necessary to prevent random guessing.
If a multiple-choice question has five answer choices and you submit one wrong answer before getting the question correct, how much credit will you lose for that part of the question?
Answer:
The person will lose 0.25 credits.
Step-by-step explanation:
According to given condition, the deduction on wrong answer is:
Deduction = 1/(n-1)
where, n = no. of options in the question.
Therefore, if a person gives an incorrect answer to a multiple chice question which has 5 options. Then,
n = 5
and the deduction will be given as:
Deduction = 1/(5-1)
Deduction = 1/4
Deduction = 0.25 credit
The person will lose 0.25 credits.
What are the cylindrical coordinates of the point whose rectangular coordinates are (x=−3, y=5, z=2)(x=−3, y=5, z=2) ?
Answer:
(5.83, 2.11, 2)
Step-by-step explanation:
To convert from rectangular coordinates to cylindrical coordinates we use
[tex]x=rcos(u)[/tex]
[tex]y=rsin(u)[/tex]
[tex]r=\sqrt{x^2+y^2}[/tex]
Therefore (-3,5,2):
[tex]r=\sqrt{(-3)^2+5^2}=5.83[/tex]
[tex]cosu=x/r=-3/5.83=-0.51[/tex]
[tex]u=2.11 radians[/tex]
So the coordinates are (5.83, 2.11, 2)
One positive number exceeds three times another positive number by 5. The product of the two numbers is 68. Find the number
Answer:12
Step-by-step explanation:
Let the numbers be a and b
a-3b=5 ........equation (1)
There product is :
ab=68 ..........equation (2)
From equation (1)
a-3b=5
Make a the subject of the formula
a=5+3b. ........equation (3)
Substitute equation (3) into equation (2)
(5+3b)b=68
3(b^2)+3b-68=0
The product is -204b^2(17 and -12)
(b+17)(b-12)=0
b=-17 or b=12
So the number is 12
First and another number is 4 and 17
Given that;Product of two number = 68
Find:Numbers
Computation:Assume;
First number = a
So, Another number = 3a + 5
[a][3a + 5] = 68
3a² + 5a = 68
3a² + 5a - 68 = 0
3a² + 17a - 12a - 68 = 0
3a(a - 4) + 17(a - 4)
So,
a = 4
First number = 4
So, Another number = 3a + 5 = 3(4) + 5 = 17
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In Exercises 1–6, write the first five terms of the sequence whose th term is given.
1. A_n = 3^n
2. A_n = (-2/5)^n
3. A_n = sin npi/2
4. A_n = 3n/n + 4
5. A_n = (-1)^n+1(2/n)
6. A_n = 2 + 2/n - 1/n^2
Answer:
Step-by-step explanation:
1. Un = 3^ n
U1 = 3, U2 = 9, U3 = 27, U4 = 81, U5 = 243
2. Un = (-2/5)^n
U1 = -2/5, U2 = 4/25, U3 = 8/125, U4 = 16/625, U5 = 32/3125
3. Un = sin npi/2
U1 = 1, U2 = 0, U3 = -1, U4 = 0, U5 = -1
4. Un = 3n/n+4
U1 = 3/5, U2 = 1, U3 = 9/7, U4 = 3/2, U5 = 5/3
5. Un = (-1)^n+1(2/n)
U1 = 1, U2 = 2, U3 = -1/3, U4 = 3/2, U5 = -3/5
6. Un = 2 + 2/n - 1/n^2
U1 = 3, U2 = 11/4, U3 = 23/9, U4 = 39/16, U5 = 59/25
A simple linear regression analysis was conducted to predict the Exam 3 score of students in STA 2023 based on their Exam 1 score. The analysis yielded the following results:
y-^ = 50.57+0.4845x.
1. Which of the following is the best description of the slope of the line?
Group of answer choices:
O As the Exam1 score increases by 1 point, the student's Exam 3 grade will increase, on average, by 0.4845 points.
O As the Exam 1 score increases by 1 point, the student's Exam 3 grade will increase, on average, by 50.57 points.
O As the Exam 3 score increases by 1 point, the student's Exam 1 score will increase, on average by 0.4845 points.
O As the Exam 3 score increases by 1 point, the student's Exam 1 score will increase, on average by 50.57 points.
Answer:
Option A) As the Exam 1 score increases by 1 point, the student's Exam 3 grade will increase, on average, by 0.4845 points.
Step-by-step explanation:
We are given the following in the equation:
[tex]y(x) = 50.57+0.4845x[/tex]
where, above equation is a linear regression equation to predict the Exam 3 score of students in STA 2023 based on their Exam 1 score.
Here,
y is the dependent variable that is score of exam 3.
x is the independent variable that is the score of exam 1.
Comparing the given equation to a linear equation, we have,
[tex]y = mx + c[/tex]
Slope, m = 0.4845
Intercept, c = 50.57
We define the slope as rate of change.
If there is a increase in x by 1 unit, then,
[tex]y(x) = 50.57+0.4845x\\y(x+1) = 50.57+0.4845(x+1)\\y(x+1)-y(x) = 50.57+0.4845(x+1)-50.57-0.4845x\\y(x+1)-y(x) = 0.4845(x+1-x)\\y(x+1)-y(x) = 0.4845[/tex]
Thus, we can interpret the slope of the line as
Option A) As the Exam 1 score increases by 1 point, the student's Exam 3 grade will increase, on average, by 0.4845 points.
The probability of a radar station detecting an enemy plane is 0.85 and the probability of not detecting an enemy plane is 0.15. If 4 stations are in use, what is the standard deviation? Round your answer to nearest hundredth.
Answer:
The standard deviation is of 0.71 enemies planes detected.
Step-by-step explanation:
For example enemy plane, there are only two possible outcomes. Either it is detected, or it is not. So we use the binomial probability distribution to solve this problem.
The standard deviation of n trials with probability p, of the binomial probability distribution, is given by the following formula.
[tex]\sqrt{V(X)} = \sqrt{np(1-p))}[/tex]
In this problem, we have that:
[tex]p = 0.85, n = 4[/tex]
So
[tex]\sqrt{V(X)} = \sqrt{np(1-p))} = \sqrt{4*0.85*0.15} = 0.71[/tex]
The standard deviation is of 0.71 enemies planes detected.
Use the identities A = A ∩ S and S = B ∪ B and a distributive law to prove that:
1) A = (A ∩ B) ∪ (A ∩ B).
2) If B ⊂ A then A = B ∪ (A ∩ B).
Answer:
Given that
A = A ∩ S -- (1)
S = B ∪ B -- (2)
To prove:
A = (A ∩ B) ∪ (A ∩ B)
(A ∩ B) ∪ (A ∩ B)
= [(A∪A) ∩ (A∪B)] ∩ [(B∪A) ∩ (B∪B)]
=[A ∩ (A∪B)] ∩ [(A∪B) ∩ S]
=A ∩ (A∪B)
=A
Hence proved.
2) If B ⊂ A then A = B ∪ (A ∩ B)
R.H.S = B ∪ (A ∩ B)
= (B ∪ A) ∩ (B∪B) --(3)
As B is subset of A so
(B ∪ A) = A
From (2)
(B ∪ B) = S
(3) becomes
=A ∩ S
from (1)
A ∩ S = A
Hence proved
The same hospital has $40 million in current assets and $30 million in current liabilities. During a 30 - day month, it incurred total operating expenses of $10 million. How many days of working capital did it maintain this period
Answer:
[tex]Days\ of\ working\ Capital= 30\ days[/tex]
During a 30 - day month,days of working capital are 30 days.
Step-by-step explanation:
Formula for days of working capital we are going to us is:
[tex]Days\ of\ working\ Capital=(Current\ Assets\ -Current\ Liabilities) / \frac{Operation\ Expense}{Time Period}[/tex]In our case:
Current Assets= $40 million.
Current liabilities=$30 million.
Operation Expense= $10 million.
Time Period=30 days Month.
[tex]Days\ of\ working\ Capital=\frac{\$40\ million-\$30\ million}{\frac{\$10 million}{30}} \\Days\ of\ working\ Capital= 30\ days[/tex]
During a 30 - day month,days of working capital are 30 days.
The horizontal axis in the sampling distribution of ModifyingAbove p with caret represents all possible sample proportions from a simple random sample of size n. Bold (a )What percent of sample proportions results in a 99% confidence interval that includes the population proportion? Bold (b )What percent of sample proportions results in a 99% confidence interval that does not include the population proportion?
Answer: (a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.
(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.
Step-by-step explanation:
(a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.
Explanation: If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals.
(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.
Explanation: The 99% of the confidence intervals includes the population proportion value, it means, the remaining (100% – 99%) 1% of the intervals does not includes the population proportion.
If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals and 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.
Answer:
(a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.
(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.
Step-by-step explanation:
(a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.
Explanation: If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals.
(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.
Explanation: The 99% of the confidence intervals includes the population proportion value, it means, the remaining (100% - 99%) 1% of the intervals does not includes the population proportion.
If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals and 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.
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The percent yield is calculated by dividing the actual yield by the theoretical yield times 100. True False
The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. This calculation gives the percentage of the theoretical yield that is actually obtained in the reaction.
Explanation:The percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100. This calculation gives the percentage of the theoretical yield that is actually obtained in the reaction. The formula for percent yield is:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Actual and theoretical yields can be expressed as masses or molar amounts as long as they are in the same units. The percent yield allows us to quantify the efficiency of a reaction and determine how much product was obtained compared to the maximum potential.
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The density of a fluid is given by the empirical equation rho=70.5 exp(8.27 X 10^-7 P) where rho is density (lbm/ft^3) and P is pressure (lbf/in.^2) a) What are the units of 70.5 and 8.27 x 106-7? b) Calculate the density in g/cm3 for a pressure of 9.00 x10^6 N/m^2. e) Derive a formula for rho (g/cm^3) as a function of P(N/m^2).
Answer:
(a) The unit of 70.5 is lbm/ft^3 and the unit of 8.27×10^-7 is in^2/lbf
(b) density = 0.1206g/cm^3
(c) rho = 0.1206exp(8.27×10^-7P)
Step-by-step explanation:
(a) The unit of 70.5 is the same as the unit of rho which is lbm/ft^3. The unit of 8.27×10^-7 is the inverse of the unit of P (lbf/in^2) because exp is found of a constant. Therefore, the unit of 8.27×10^-7 is in^2/lbf
(b) P = 9×10^6N/m^2
rho = 70.5exp(8.27×10^-7× 9×10^6) = 70.5exp7.443 = 70.5×1.71 = 120.6kg/m^3
rho = 120.6kg/m^3 × 1000g/1kg × 1m^3/10^6cm^3 = 0.1206g/cm^3
(c) Formula for rho (g/cm^3) as a function of P (N/m^2) is
rho = 0.1206exp(8.27×10^-7P) (the unit of 0.1206 is g/cm^3)
Simplify 2x-(-3x+9).
Answer:
5x-9
Step-by-step explanation:
phil paid $32 for 10 pounds of jellybeans. how mich did phil pay per pound
Answer:3,2
Step-by-step explanation:32/10
Phil paid $3.20 per pound for the jellybeans.
To find out how much Phil paid per pound of jellybeans, we need to divide the total cost by the total weight of the jellybeans. Phil paid $32 for 10 pounds of jellybeans, so the calculation is as follows:
Cost per pound = Total cost / Total weight
Cost per pound = $32 / 10 pounds
Cost per pound = $3.20
Therefore, Phil paid $3.20 per pound for the jellybeans.
In a statistical test, we have a choice of a left-tailed test, a right-tailed test, or a two-tailed test. Is it the null hypothesis or the alternate hypothesis that determines which type of test is used
Answer:
Alternate Hypothesis determines which type of test is used.
Step-by-step explanation:
We have find the hypothesis that helps us o determine the nature of test.
There are three types of test two tailed test, lest tailed test or right tailed testIt is the alternative hypothesis that states the nature of test.The null hypothesis always state that the sample belongs to the population and always shows equality to the population.The alternate hypotheses states that whether the sample belongs to the population or not and tell that there is a significant difference between the two variables.For example, for a one sample test of population mean we have the following alternate and null hypothesis.[tex]\text{Two tailed test}\\H_{0}: \mu = \mu_0\\H_A: \mu \neq\mu_0\\\\\text{Left tailed test}\\H_{0}: \mu = \mu_0\\H_A: \mu < \mu_0\\\\\text{Right tailed test}\\H_{0}: \mu = \mu_0\\H_A: \mu > \mu_0[/tex]
The mean hours of sleep that students get per night is 7 hours, the standard deviation of hours of sleep is 1.7 hours, and the distribution is approximately normal. Complete the following sentence. For about 95% of students, nightly amount of sleep is between ______.a. 5.3 and 8.7 hrsb. 5 and 9 hrsc. 3.6 and 10.4 hrsd. 1.9 and 12.1 hrs
Answer:
c. 3.6 and 10.4 hrs
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviations of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean 7, standard deviation 1.7.
For about 95% of students, nightly amount of sleep is between
7 - 2*1.7 = 7 - 3.4 = 3.6 hours
7 + 2*1.7 = 7 + 3.4 = 10.4 hours
So the correct answer is:
c. 3.6 and 10.4 hrs
Final answer:
To determine the range of sleep for 95% of students, the empirical rule is used. By adding and subtracting twice the standard deviation from the mean, we find that approximately 95% of students get between 3.6 and 10.4 hours of sleep. Thus, the correct answer is option c.
Explanation:
To calculate the range of hours that includes approximately 95% of the student's sleep times, we will use the empirical rule (68-95-99.7 rule) because the distribution of sleep hours is approximately normal. This rule states that for a normal distribution, approximately 95% of the data falls within two standard deviations from the mean. Since we know the mean is 7 hours and the standard deviation is 1.7 hours, we need to add and subtract twice the standard deviation from the mean to find the range.
Calculate the lower end of the range: 7 - 2(1.7) = 7 - 3.4 = 3.6 hours.Calculate the upper end of the range: 7 + 2(1.7) = 7 + 3.4 = 10.4 hours.Therefore, for about 95% of students, the nightly amount of sleep is between 3.6 and 10.4 hours, which corresponds to option c.
There are six photocopying machines in a college office. During August 2004, these machines produced 2567, 5456, 3769, 2245, 6678, and 3398 copies. Find the mean, median, and mode of the number of copies produced by these machines.
Answer:mean= 2905.83, median=3398, mode=5456
Step-by-step explanation:
The mean:
Mean=summation of six machine produced/ n
Mean=(2567+5456+3769+2245+3398)/6
Mean=17435/6
Mean=2905.8333333
b. The median
Firstly we have to rearranged the machine product in order:
2245, 2567, 3398, 3769, 5456
So 3398 is at the middle, so median is 3398
c. The mode
The machine produces the highest number (frequency) is mode. So the mode is 5456
Eighty-five mall customers were randomly surveyed across the state to determine if the live entertainment provided had increased the amount of money they spent. Can the probability be found by using the binomial probability formula?a. No. The trials are fixed, but the events are independent and the 5% guideline does not apply to this situation.b. Yes. The events are dependent; however, the 5% guideline can be applied to this situation.c. Yes. The outcomes can be classified into two categories: the trials are fixed, and the events are independent.d. No. The events are dependent.
Answer:
c. Yes. The outcomes can be classified into two categories: the trials are fixed, and the events are independent.
Since we can calculate the following probabilities:
p= probability that live entertainment had increased the amount of money they spent
q =probability that live entertainment had not increased the amount of money they spent
n = 85
And independence is satisfied.
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
In order to apply the binomial distirbution we need to satisfy some conditions given here:
1) Independence between the trials of the experiment
2) The number of observations n is fixed for this case n=85
3) Each observation represents one of two outcomes "success" or "failure" and the probability of success is defined.
So then based on this and on the information given we can conclude that the best option for this case is:
c. Yes. The outcomes can be classified into two categories: the trials are fixed, and the events are independent.
Since we can calculate the following probabilities:
p= probability that live entertainment had increased the amount of money they spent
q =probability that live entertainment had not increased the amount of money they spent
n = 85
And independence is satisfied.
What’s the surface area 3m 5m 4m
Answer:
94m²
Step-by-step explanation:
the surface area of 3m, 5m, 4m
these means its a rectangular prism with a length, width, height
the rectangular prism has 2 sides each ( meaning for the area of one side, you multiply by 2
3 x 5 = 15 x 2 = 30m²
3 x 4 = 12 x 2 = 24m²
4 x 5 = 20 x 2 = 40m²
surface area = 30 + 24 + 40 = 94m²
94m² is the answer
the surface area of 3m, 5m, 4m
This means it's a rectangular prism with a length, width, height
the rectangular prism has 2 sides each ( meaning for the area of one side, you multiply by 2
3 x 5 = 15 x 2 = 30m²
3 x 4 = 12 x 2 = 24m²
4 x 5 = 20 x 2 = 40m²
surface area = 30 + 24 + 40 = 94m²
What is a surface sample?
surface area is the total area covered by all faces of a 3D object. For example, if you need to find the amount of paint you need to paint a cube, the surface to which the paint is applied is that surface. It is always measured in square units.
The surface area of a right-angled prism is given by the following equation: SA = 2 (lh + wh + lw) A unit of the square.
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Suppose a researcher compiled a data set consisting of the following variables for a sample of 100 retired men. For each variable, select whether it is discrete or continuous. Variable Discrete Continuous The number of countries ever visited The number of sons or daughters Shoe size Body temperature
Answer:
a) Discrete Variable
b) Discrete Variable
c) Discrete Variable
d) Continuous Variable
Step-by-step explanation:
We have to identify the given variable as discrete r continuous.
Discrete Variables:
They are expressed in whole numbers.They are counted not measured.They cannot take any value within an interval.Continuous Variables:
They are expressed in decimal numbers.They are measured not counted.They cannot take any value within an interval.a) The number of countries ever visited
Since number of countries will always be expressed in whole numbers and not decimals. Also, they will always be counted and not measured. Thus, it is a discrete variable.
b) The number of sons
Since number of sons will always be expressed in whole numbers and not decimals. Also, they will always be counted and not measured. Thus, it is a discrete variable.
c) Shoe size
Shoe size are expressed in whole number. The underlying measure is length of feet which is a continuous variable but shoe size are always given in whole number. Thus, they cannot take any value within an interval. Thus, it is a discrete variable.
d) Body temperature
Body temperature can be expressed in decimal. A Body temperature of 42.5 makes sense. Thus, they can take any value within an interval. Also, it is measured not counted. Thus, it is a discrete variable.
One hundred random samples, each of size 25, are obtained from the Normal distribution with mean 0 and standard deviation 1 using Minitab. Subsequently, the 1-Sample Z procedure in Minitab is used (with the same confidence level) to obtain a confidence interval from each sample. Out of the 100 intervals thus obtained, 89 include the number 0.
Estimate the confidence level (in percentage terms) used to generate the 100 intervals using a 95% confidence interval.
Answer:
- 0.196 < μ < +0.196
Step-by-step explanation:
Given Data:n = 100
σ = 1
X bar (x⁻) = 0
1 - ∝ =95%
To Find:μ = ?
Formula:= X bar (x⁻) ± Z ∝/₂ * σ/√n
Solution:Finding ∝:
To find ∝, we will convert the confidence interval i.e. 95% into decimal by dividing it with 100 and subtracting it from 1.
1 - ∝ = 95%
1 - ∝ = 95/100
1 - ∝ = 0.95
∝ = 1 - 0.95
∝ = 0.05 -----(1)
We have to calculate ∝/2 so dividing both sides of (1) with 2
∝/2 = 0.05/2
∝/2 = 0.025
We will find the value of 0.025 in z-table. The value is:
Z ∝/2 = 1.96
For further calculations, we will use the value of Z∝/2.
Putting all values in the formula.
= 0 ± 1.96 * 1 / √100
We can also write this in the following form:
0 - 1.96/ √100 < μ < 0 + 1.96 / √100
- 1.96 / 10 < μ < 1.96 / 10 ∴ √100 = 10
- 0.196 < μ < + 0.196
The confidence level used to generate the 100 intervals can be estimated as 89%.
Explanation:The confidence level used to generate the 100 intervals can be estimated using a 95% confidence interval. Since 89 out of the 100 intervals include the number 0, we can consider that 89% of the intervals would include the true population mean, which is 0. Therefore, the estimated confidence level is 89%.
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A sorting algorithm takes 1 second to sort n =1000 items.
1) How many operations will be performed if the sorting algorithm is O(n2) (approximately)?
2) How long will it take to sort 10,000 items if the sorting algorithm is O(n2)?
3) How much time will one operation take if the sorting algorithm is O(n2)?
Answer:
1.1,000,000,
2, 1 minute 40 secs
3.10^-6 secs
Step-by-step explanation:
sorting algorithm takes 1 second to sort n =1000 items.
1) How many operations will be performed if the sorting algorithm is O(n2) (approximately)?
2) How long will it take to sort 10,000 items if the sorting algorithm is O(n2)?
3) How much time will one operation take if the sorting algorithm is O(n2)?
algorithm takes time proportional to n^2,
1. then 1,000^2=1,000,000,
2. if it takes 1 secs to generate 1000 items
yhen n^2=1000^2=1000000 and 10,000^2=100,000,000.
Dividing by 100. Therefore, the sorting algorithm would take
1 minute and 40 seconds to sort 10,000 items.
3. How much time will one operation take if the sorting algorithm is O(n2)?
1/1000^2
10^-6 secs to sort 1 operations
A O(n^2) sorting algorithm will perform about 1,000,000 operations in 1 second for 1000 items. It will take around 100 seconds for 10,000 items. The time taken per operation is roughly 1 microsecond.
Explanation:This question is about Big O Notation, a concept used in Computer Science for analyzing an algorithm's running time by characterizing the number of operations it will perform as a function of the input size (n).
If the sorting algorithm is O(n^2), this means the number of operations grows quadratically with n. Therefore, for every increase in n, the operations performed will square. So for n=1000, and it takes 1 second, that means the algorithm performs about 1,000,000 (1000^2) operations per second. If you are sorting 10,000 items and if the time complexity of the sorting algorithm is O(n^2), then, by the definition of big O notation, it will take roughly 100 times longer than sorting 1,000 items. Therefore, it will take approximately 100 seconds. To find out how much time per operation, divide the number of operations by the time taken. So for 1,000,000 operations in 1 second, it is approximately 1 microsecond per operation. Learn more about Big O Notation here:
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Find the area of the region in the first quadrant bounded by the line yequals4x, the line xequals4, the curve yequalsStartFraction 4 Over x EndFraction , and the x-axis.
Answer:
Step-by-step explanation:
First of all you need to sketch the region you are looking here. The sketch is in the attachment.
This area can be separated into two. The one below the line defined by [tex]y=4x[/tex] and one below the curve [tex]y=\frac{4}{x}[/tex]
After that you should find the point of intersection of the two curves so that integral limits can be defined. This is done by equating the two expressions.
[tex]4x=\frac{4}{x}\\4x^2=4\\x^2=1\\x=\pm1[/tex]
So the first integral has the limits [tex]x=0[/tex] and [tex]x=1\\[/tex], while the second one is defined with [tex]x=1[/tex] and the limit [tex]x=4[/tex].
You now just add the two integrals and that's your area:
[tex]\int_0^14xdx + \int_1^4\frac{4}{x}dx=7.54518[/tex]
Use the following frequency distribution (of SAMPLE DATA) in answering the questions: class 0-9 f=10; class 10-19 f=20; class 20-29 f=40; class 30-39 f=20; class 40-49 f=10.
1) How many classes are there in this frequency distribution?
2) What is the midpoint of the 3rd class?
3) What is the total number of observations in the frequency distribution?
4) What is the relative frequency of the 3rd class?
5) n the formula for the standard deviation in this problem do you use "N' or "n-1"?
6) What is the value of the mean?
7) What is the value of the standard deviation?
Answer:
1. Five classes in the frequency distribution
2.The mid point of the third class is 24.5
3.The total number of frequency is 100
4.The relative frequency of the third class is (20/100=0.2)
5. We use n-1 because we are dealing with a sample data
6. The mean is 24.5
7. The value of the standard deviation is 11
Step-by-step explanation:
See attached picture for the solutions
Consider the following data which describe the amount of time in minutes students spend studying for a quiz: 10, 11, 11, 12, 12, 14, 15, 18, 19, 20, 22, 24, 39, 40, 41, 44, 46, 50, 52, 52, 53, 55, 70. What numbers make up the leaf of the first stem if you were splitting the stems?
A) 0,1,1,2,2,4,5,8,9
B) 0,1,2,4,5,8,9
C) 0,1,1,2,2,4
D) 0, 1, 2, 4
Answer: A) 0,1,1,2,2,4,5,8,9
Step-by-step explanation:
We know that in a stem leaf plot ,
The stem represents the tens value of the term and leaves represent the ones values of the data.
Given data of the amount of time in minutes students spend studying for a quiz:
10, 11, 11, 12, 12, 14, 15, 18, 19, 20, 22, 24, 39, 40, 41, 44, 46, 50, 52, 52, 53, 55, 70.
Here , the least tens value is 1. (10, 11, 11, 12, 12, 14, 15, 18, 19)
So the first stem would have value.
Then the leaf of the first stem if you were splitting the stems contains all the ones-values corresponding to tens value as 1 (10, 11, 11, 12, 12, 14, 15, 18, 19).
= 0 , 1, 1,2, 2, 4 , 5 , 8 , 9
Hence, the correct answer is A) 0,1,1,2,2,4,5,8,9
In a stem-and-leaf plot for the given data, the leaves for the first stem (1, or numbers in the 10s) would be 0,1,1,2,2,4,5,8,9, hence the answer is Option A.
Explanation:In a stem-and-leaf plot, the data is organized by place value. The stem represents the tens digit, and the leaf represents the ones digit. Considering the given data set which ranges from 10 to 70, the first stem represents '1', indicating a range of 10s.
For the values in the 10s: 10, 11, 11, 12, 12, 14, 15, 18, 19, the corresponding leaf units would be 0, 1, 1, 2, 2, 4, 5, 8, 9. So, the correct answer is Option A: 0,1,1,2,2,4,5,8,9 which are the units of the numbers in the 10s place.
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Suppose p" must approximate p with relative error at most 10-3 . Find the largest interval in which p* must lie for each value of p.
Answer:
[tex][p-|p|*10^{-3} \, , \, p+|p|* 10^-3][/tex]
Step-by-step explanation
The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is
r = |p*-p|/|p|
we want r to be at most 10⁻³, thus
|p*-p|/|p| ≤ 10⁻³
|p*-p| ≤ |p|* 10⁻³
therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]
Find the equation of the sphere if one of its diameters has endpoints (-1, -6, -8) and (0, -4, -5)
Answer:
Step-by-step explanation:
Be the points Pa=(-1,-6,-8) ; Pb=(0,-4,-5) we have calculate the middle point or center
[tex]c=(\frac{x1+x2}{2}, \frac{y1+y2}{2}, \frac{z1+z2}{2})=(\frac{-1}{2}, -5,\frac{-13}{2})[/tex]
Now we must to find d=r (view graph)
[tex]r=\sqrt{(Cx-x2)^{2}+(Cy-y2)^{2}+(Cz-z2)^{2}}\\ r=\sqrt{(\frac{-1}{2} )^{2}+(-5+4)^{2}+(\frac{-13}{2}+5 )^{2}}\\r=\sqrt{\frac{1}{4} +1+\frac{9}{4}}=\sqrt{\frac{14}{4}}=r^{2}=\frac{14}{4}[/tex]
We find the canonical sphere equation
[tex](x-h)^{2}+(y-k)^{2}+(z-l)^{2}=r^{2}\\(x+\frac{1}{2})^{2}+(y+5)^{2}+(z+\frac{13}{2} )^{2}=\frac{14}{4}\\x^{2}+x+y^{2}+10y+z^{2}+13z+64=0[/tex]
Note: The Pa=(-1,-6,-8) can also be used in c